tag:blogger.com,1999:blog-37414104180967168272021-02-27T01:14:35.066-08:00Unizor - Creative Mind through Art of MathematicsUnizor is a site where students can learn the high school math (and in the future some other subjects) in a thorough and rigorous way. It allows parents to enroll their children in educational programs and to control the learning process.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.comBlogger455125tag:blogger.com,1999:blog-3741410418096716827.post-83531053116837609652021-02-22T09:32:00.001-08:002021-02-22T09:32:56.779-08:00<i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Relativity - Transformation of Space-Time Coordinates</u><br/>(notes to item #3 of Einstein's "Electrodynamics")<br /><br />The following is an example of how a system of linear equations can be used to derive formulas of special theory of relativity. Albert Einstein has derived these formulas in his "Electrodynamics" in a more physical, more intuitive way. The following is pure mathematics and, as such, causes much less problems in understanding.<br/><br/>Assumptions:<br />1. Assume we have two systems of coordinates, one stationary with coordinates {X,T} (assuming for simplicity all the movements will occur in one space dimension along X-axis and one time dimension) and another with coordinates {x,t} moving along the X-axis with constant speed V.<br />2. Assume that at time T=0 systems coincide (i.e. X=0, t=0 and x=0).<br />3. Assume that the speed of something, as measured in both stationary and moving systems is the same and equal to C regardless of the direction of the movement (that "something" is the light in vacuum, but it's physical characteristics are unimportant)<br />4. Assume further that we are looking for linear orthogonal (i.e. preserving the distance between points and angles between vectors) transformation of coordinates from (X,T) to (x,t) that satisfies the above criteria. What would this transformation be?<br /><br />Linear transformation from {X,T} system to {x,t} system should look like this:<br /><b><i>x = pX + qT<br />t = rX + sT<br /></i></b>where p, q, r and s are 4 unknown coefficients of transformation, which we are going to determine by constructing a system of 4 linear equations with them.<br /><br />We should not add any constants into above transformations since {X=0,T=0} should transform into {x=0,t=0}.<br /><br/>A. Notice that the property of orthogonality is needed to preserve geometry (i.e. no deformation) and, therefore, to preserve the form of all physical equations of motion. As it is well known, orthogonal transformations have determinant of the matrix of coefficients equal to 1, i.e. <br /><b><i>ps − qr = 1</i></b>. An unfamiliar with this property student can study this subject separately (we at Unizor plan to include this into corresponding topic on vectors).<br />The above is the first equation to determine unknown coefficients. It's not linear, it is rather quadratic, but the rest of the equations will be linear and that's why we included this particular problem in the topic dedicated to linear systems.<br />We need three more equations to determine all the unknown coefficients.<br /><br />B. Since moving system moves along X-axis with speed V, its beginning of coordinate (point x=0) must at the moment of time T be on a distance VT from the beginning of coordinates of a stationary system. Hence, if X=VT, x=0 for any T. From this and the first transformation equation x = pX + qT we derive:<br /><b><i>0 = pVT + qT</i></b> or<br/><b><i>0 = (pV + q)T</i></b>.<br /><br/>Since this equality is true for any T, <br /><b><i>pV + q = 0</i></b> <br />and, unconditionally, <b><i>q = −pV</i></b>.<br />This is the second equation for our unknown coefficients.<br /><br />C. Since the speed of light C is the same in both systems {X,T} and {x,t}, an equation of its motion in the stationary system must be X = CT and in the moving system x = Ct. Therefore, if X=CT, then x=Ct. Put X=CT into equations of transformation of coordinates. We get x = pCT + qT, t = rCT + sT. Substitute these expressions into x=Ct: <br /><b><i>pCT + qT = rC<sup>2</sup>T + sCT</i></b>.<br />Reduce by T, <br /><b><i>pC + q = rC<sup>2</sup> + sC</i></b>.<br />This is the third equation for unknown coefficients.<br /><br />D. Repeat the logic of a previous paragraph for the light moving in the opposite direction with a speed −C. We get, if X = −CT, then x = −Ct. Therefore, x = −pCT + qT, t = −rCT + sT and (since x = −Ct) <br /><b><i>−pCT + qT = rC<sup>2</sup>T − sCT</i></b>. <br />Reduce by T, <br /><b><i>−pC + q = rC<sup>2</sup> − sC</i></b>.<br />This the fourth equation for unknown coefficients.<br /><br />So, this is the system of equations for 4 unknown coefficients of transformation <b><i>p, q, r, s</i></b>:<br/>(a) <b><i>ps − qr = 1</i></b><br/>(b) <b><i>q = −pV</i></b><br />(c) <b><i>pC + q = rC<sup>2</sup> + sC</i></b><br />(d) <b><i>−pC + q = rC<sup>2</sup> − sC</i></b><br /><br/>It's not exactly linear, but it has sufficient number of linear equations (all but one) to solve it using the known methodology. Let's solve this system of equations by combining the methods of substitution and elimination. We will express unknown variables <b><i>q, r, s</i></b> in terms of <b><i>p</i></b> using equations (b), (c) and (d). Then we will substitute them into (a) to get an equation for <b><i>p</i></b>. Solving it will allow to evaluate all other unknowns.<br/><br/>E. From (c) and (d), adding and subtracting these equations, we get:<br /><b><i>2q = 2rC<sup>2</sup></i></b>, therefore <b><i>q = rC<sup>2</sup></i></b><br /><b><i>2pC = 2sC</i></b>, therefore <b><i>p = s</i></b><br /><br />Since from (b) q = −pV,<br /><b><i>−pV = rC<sup>2</sup></i></b> and <b><i>r = −pV/C<sup>2</sup></i></b>.<br /><br />Now all coefficients of a transformation are expressed in terms of one unknown coefficient <b><i>p</i></b>. To get the value of <b><i>p</i></b>, use the first equation (a).<br /><br />F. Substituting q, r and s, expressed in terms of p, into an equation (a) <b><i>ps − qr = 1</i></b>, we get:<br /><b><i>p<sup>2</sup> − (−pV)·(−pV)/C<sup>2</sup> = 1</i></b>, therefore <br /><b><i>p<sup>2</sup>·(1 − V<sup>2</sup>/C<sup>2</sup>) = 1</i></b> and <br /><b><i>p = 1/√</b><span style='text-decoration:overline;'><b>1−V<sup>2</sup>/C<sup>2</sup> </span></i></b>.<br /><br />From this all other coefficients of a transformation matrix are derived:<br /><b><i>q = −V/√</b><span style='text-decoration:overline;'><b>1−V<sup>2</sup>/C<sup>2</sup> </span><br />r = −(V/C<sup>2</sup>)/√</b><span style='text-decoration:overline;'><b>1−V<sup>2</sup>/C<sup>2</sup> </span><br />s = 1/√</b><span style='text-decoration:overline;'><b>1−V<sup>2</sup>/C<sup>2</sup> </span><br /></i></b><br />G. The final transformation matrix looks exactly like in Einstein's article on electrodynamics, but seems to be much simpler to arrive at and the derivation is strictly mathematical.<br /><b><i>x = (1/√</b><span style='text-decoration:overline;'><b>1−V<sup>2</sup>/C<sup>2</sup> </span>)·X − (V/√</b><span style='text-decoration:overline;'><b>1−V<sup>2</sup>/C<sup>2</sup> </span>)·T<br />t = ((−V/C<sup>2</sup>)/√</b><span style='text-decoration:overline;'><b>1−V<sup>2</sup>/C<sup>2</sup> </span>)·X + (1/√</b><span style='text-decoration:overline;'><b>1−V<sup>2</sup>/C<sup>2</sup> </span>)·T<br /></i></b><br />Traditionally, factor V/C is replaced with Greek letter β, which results in formulas:<br /><b><i>x = (1/√</b><span style='text-decoration:overline;'><b>1−β<sup>2</sup> </span>)·X + (−V/√</b><span style='text-decoration:overline;'><b>1−β<sup>2</sup> </span>)·T<br />t = ((−V/C<sup>2</sup>)/√</b><span style='text-decoration:overline;'><b>1−β<sup>2</sup> </span>)·X + (1/√</b><span style='text-decoration:overline;'><b>1−β<sup>2</sup> </span>)·T<br /></i></b><br />One more simplification is usually done by introducing Lorentz factor <b><i>γ</i></b> equals to <b><i>1/√</b><span style='text-decoration:overline;'><b>1−β<sup>2</sup> </span></i></b>:<br /><b><i>x = γX − γVT<br/> = γ(X − VT)</i></b><br /><b><i>t = −γVX/C<sup>2</sup> + γT<br/> = γ(T − VX/C<sup>2</sup>)</i></b><br /><br />The final form of transformation of coordinates in the Special Theory of Relativity is:<br><b><i>x = (X − VT)/√</b><span style='text-decoration:overline;'><b>1−(V/C)<sup>2</sup> </span></i></b><br /><b><i>t = (T − VX/C<sup>2</sup>)/√</b><span style='text-decoration:overline;'><b>1−(V/C)<sup>2</sup> </span></i></b><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-70108092140186369022021-02-08T08:39:00.002-08:002021-02-08T08:39:41.602-08:00Electronics - Diode: UNIZOR.COM - Physics4Teens - Electromagnetism - Usage<iframe width="480" height="270" src="https://youtube.com/embed/wEuL4rA7R_8" frameborder="0"></iframe><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Electronics: Diode</u><br/><br/>To the category of <b>electronic</b> devices we relate all devices that use electricity and whose primary purpose is not just to heat or to do mechanical work.<br/>Radio, television, computers, all kinds of non-mechanical switches and regulators, phones, hardware that runs Internet and many others devices belong to this category of electronic devices.<br/><br/>Obviously, we cannot talk about how all these devices work. Instead, we will spend some time to understand the basic components that are used in these devices.<br/><br/><i>Diodes</i> are one of the simplest electronic components that are present in practically all electronic devices.<br/>The main purpose of a <i>diode</i> is to let the electric current only in one direction. This process of allowing electric current to go only in one direction is called <i>rectifying</i> the current.<br/><br/>The origin of diodes lies in an observations made by scientists and engineers (Fleming, Edison and others) at the end of 19th century. Two wires that did not touch each other were placed inside a vacuum tube, one connected to a positive pole of a battery (<i>anode</i>) and another connected to a negative one (<i>cathode</i>). At normal temperature there was no electric current between them, because they did not touch each other. But, if the negative wire was heated, some electric current between these wires was observed, while heating the positive wire did not cause any current in the circuit.<br/><br/>The explanation of this phenomenon is simple. Heating increases the activity of elementary particles inside a negatively charged metal of a wire that has an excess of electrons. With this increased activity certain electrons escape from the surface of the metal and form some kind of electron cloud. This is called <i>thermionic emission</i>.<br/>In the presence of a positively charged wire some electrons from a cloud are attracted to a positive wire, thus forming a current. New electrons from a negative poll of a battery replace the escaped electrons, maintaining a fresh supply of new electrons, which enables a steady current.<br/><br/>Obviously, if a positive wire with its deficiency of electrons is heated, even if some electrons do escape because of high temperature, they will be repelled by a negative wire and attracted back to a positive one. No current would be formed.<br/><br/>Below is a schematic representation of a vacuum tube diode with ampere meter in a circuit showing the existence of electric current.<br/><img src='http://www.unizor.com/Pictures/DiodeTube.png' style='width:200px;height:200px;'><br/><br/>The symbol for a diode in electronic schemas is<br/><img src='http://www.unizor.com/Pictures/DiodeSymbol.png' style='width:200px;height:100px;'><br/><br/>The primary usage of diodes is to <i>rectify</i> the alternating current, where they allow the current to go only in one direction, thus converting AC to DC.<br/>They are also used in signal isolation, filtering and mixing.<br/>Vacuum tube diodes are used now only in high capacity <i>rectifiers</i> with semiconductors based diodes used in all the electronic devices we usually deal with.<br/><br/>Let's analyze the process of rectifying AC using diodes.<br/>As we know, the current in a regular AC circuit is sinusoidal, changing the direction and the value.<br/>If a diode is included into a circuit in sequence, the current in one direction will go through, while it will be prohibited to go in the opposite direction.<br/>This causes the alternating current to change from a regular sinusoidal wave-like behavior into irregular direct current.<br/><img src='http://www.unizor.com/Pictures/DiodeRectifying.jpg' style='width:200px;height:200px;'><br/><br/>The irregularity of the current in a circuit can be improved by using a <i>bridge rectifier</i> built from 4 diodes as follows<br/><img src='http://www.unizor.com/Pictures/BridgeDiodes.png' style='width:200px;height:170px;'><br/><br/>If the AC generator produces positive charge at point A and negative at point B, the flow of electrons is in the direction BEFNMDCA. The electric current is defined as going against the flow of electrons in the opposite direction ACDMNFEB.<br/><br/>If the AC generator produces positive charge at point B and negative at point A, the flow of electrons is in the direction ACFNMDEB. The electric current is defined as going against the flow of electrons in the opposite direction BEDMNFCA.<br/><br/>As you see, the direction of a current in both cases is from point M to point N, regardless of polarity of a generator contacts.<br/><br/>The AC current rectified by diodes that form a bridge is better than rectified by a single diode, but still is quite irregular, comparing with DC current from a battery.<br/>Additional improvements can be achieved by splitting a current into two separate lines and putting a capacitor on one of them to change the phase of the oscillations and then combining signals together.<br/>Here is how the combination works.<br/><img src='http://www.unizor.com/Pictures/TwoRectifiers.png' style='width:200px;height:250px;'><br/><br/>The combined <nobr><i><b><font color=blue>Signal 1+2</font></b></i></nobr> still has some irregularity, but is much more stable than each of its components <nobr><i><b><font color=green>Signal 1</font></b></i></nobr> or <nobr><i><b><font color=brown>Signal 2</font></b></i></nobr>.<br/><br/>In general, by combining currents, shifted by phase relative to each other, helps improving the stability of the flow of electrons. Real life rectifiers are build on this principle.<br/><br/>Vacuum tube diodes have been largely replaced by semiconductors, but the principle of their work is very similar.<br/><br/> Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-30428960586444014992021-02-04T08:01:00.002-08:002021-02-04T08:01:48.270-08:00Electric Devices: UNIZOR.COM - Physics4Teens - Electromagnetism<iframe width="480" height="270" src="https://youtube.com/embed/LSr04WyOwWM" frameborder="0"></iframe><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Electric Devices</u><br/><br/>In this chapter we will discuss different usages of electricity by grouping them according to certain criteria.<br/><br/>Two major groups to differentiate are <b>electric</b> and <b>electronic</b> devices.<br/>To the category of <b>electric</b> devices we relate devices that use electricity for two primary purposes - to produce mechanical work, like rotation, and to produce heat, like in electric stove, including producing light by means of heating, like in incandescent lamps.<br/><br/>Regardless of such a simple definition of this category, the number of devices in this group is enormous, and these devices are the first ones invented to make our lives easier. Examples of these electric devices are a subject of this lecture.<br/><br/>To the category of <b>electronic</b> devices we relate all other devices that use electricity and whose primary purpose is not just to heat or to do mechanical work. They will be discussed in the next lecture.<br/><br/><br/><i>Mechanical Work</i><br/><br/>The easiest and most common example of the usage of electricity to produce mechanical work is an electric motor. This is a device that converts electricity into rotation.<br/><br/>For alternating current numerous one-phase and three-phase motors are used all around us.<br/>The primary motion they produce is rotation, which in some cases is converted into other forms of motion.<br/><br/>They pump water, rotate fans, work in refrigerators, rotate wheels of electric trains, lift elevator cabins, drill for oil and gas, operate machinery at manufacturing plants, move construction cranes.<br/><br/>Direct current in most cases comes from batteries and is used in direct current electric motors, like the one that starts the car engine, rotates a hard disk in a computer, rotates the battery powered electric drill etc.<br/><br/>Electric wall clock is another example of using electricity to move the wheels of a clock. Some electric clocks work off alternating current, some off direct one.<br/><br/>Washing machine has an electric pump to deal with water pumped in and out and another motor to rotate the drum.<br/><br/>Just as an illustration, let's calculate the technical characteristics of a motor that should supply water to a building, where I live in.<br/><br/>The water is pumped to the roof tank, then it flows down to all apartments.<br/><br/>We have <nobr><i>12 floors</i></nobr>, each about <nobr><i>3 meters</i></nobr> high, <nobr><i>200 apartments</i></nobr>, each apartment needs about <nobr><i>100 liters</i></nobr> of water during <nobr><i>3 hours</i></nobr> in the morning.<br/>So, the pump should pump <nobr><i>200·100=20,000 liters</i></nobr> of water to the height <nobr><i>12·3=36 meters</i></nobr> during <nobr><i>3 hours</i></nobr> time.<br/><br/>This allows us to calculate work <i>W</i> performed during this time and power <i>P</i> of the pump needed to perform this work.<br/><br/>Each liter of water has a mass of <nobr><i>1 kg</i></nobr> and, therefore, the weight of <nobr><i>9.8 N</i></nobr>.<br/><i><b>W = 9.8·20,000·36 =<br/>= 7,056,000 J</b> (joules)</i><br/><br/>Since the time to do this work is <nobr><i>T = 3 hours</i></nobr> and each hour has <nobr><i>3,600 seconds</i></nobr>, the power of the motor is<br/><i><b>P=W/T=7,056,000/(3·3,600)≅<br/>≅ 653 J/sec</b> (watts)</i><br/><br/>Usually, we need some excess of power to prevent shortage during some extra work requirements and to account for losses of power in the motor itself due to friction and heating, so a motor of about <nobr><i>1000 watt (1 kilowatt)</i></nobr> should suffice, if we allow it to work without interruption.<br/><br/>In practice, the motor should start and stop periodically, depending on the level of water in the tank, so it has to pump faster than water is consumed and we need a more powerful motor, say <nobr><i>1.5 KW</i></nobr>.<br/>With voltage to such a motor at the level of <nobr><i>220V</i></nobr> the current flowing through this motor is<br/><i><b>I = 1500W/220V ≅ 6.8A</b></i><br/><br/>In addition, considering that things break and we need an uninterrupted water supply, we need the same pump with the motor of the same power to be ready to automatically pick up the load in case the main pump breaks.<br/>That makes a design a bit more complicated with two pumps working in parallel, alternating their work and, in case one breaks, another working alone. This requires some electronic switching mechanism.<br/><br/><br/><i>Heat</i><br/><br/>Electric heater and incandescent lamp represent this group of electric devices. They warm and light up our homes.<br/><br/>We use electric stove to prepare our food.<br/><br/>Electric hair drying fan is an example of a combined mechanical (to push the air) and heating (to heat up the air) electric device.<br/><br/>Drying machine uses electricity to rotate a drum, produce heat and blow it into the drum using a fan.<br/><br/>For illustration, let's do some calculations related to incandescent lamps.<br/>Consider a lamp with marked power consumption of <nobr><i>P=100W</i></nobr> and voltage <nobr><i>U=120V</i></nobr>.<br/>Here we are talking about alternating current and, therefore, all characteristics are <i>effective</i>.<br/><br/>The effective electric current running through it is<br/><i><b>I = 100W/120V ≅ 0.8333A</b></i><br/>The resistance of the spiral in this lamp is<br/><i><b>R = 120V/0.8333A = 144Ω</b></i><br/><br/>Obviously, we can check that<br/><i><b>P = U²/R = I²·R</b></i><br/>With given voltage in the circuit, the power consumed by an incandescent lamp will be more when the resistance is less. That's why a lamp consuming <i>100W</i> has a thicker spiral (with less resistance) than a lamp consuming <i>40W</i> for the same voltage.<br/><br/>Another interesting example of using electricity to produce heat is welding. This is a process when an electric arc between two electrodes is formed and used to melt metal.<br/><br/>There are many different types of welding machines. An important characteristic is the electric current going through the arc formed between electrodes. Usually it's in hundreds of amperes, like <i>500A-1000A</i> with voltage in the range <i>30V-60V</i>.<br/>That makes the power consumption of a welding machine to be somewhere from <i>15KW</i> to <i>60KW</i>, which is a lot, comparing to a power of about <i>1.5KW</i> needed for a water pump described above.<br/><br/>These characteristics fluctuate as the welding process goes, they depend on the length of an electric arc and materials used as electrodes.<br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-37577092315479179512021-01-29T06:48:00.000-08:002021-01-29T06:48:41.195-08:00Electricity to Consumers: UNIZOR.COM - Physics4Teens - Electromagnetism ...<iframe style="background-image:url(https://i.ytimg.com/vi/yFqUetqW4Vs/hqdefault.jpg)" width="480" height="270" src="https://youtube.com/embed/yFqUetqW4Vs" frameborder="0"></iframe><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Distribution to Consumers</u><br/><br/>When we described the <b>grid</b> in the previous lecture, we concentrated on main principles of its work - synchronization of the generators of electric energy with the grid. It might make a wrong impression that the grid is one gigantic wire that have certain voltage, frequency and phase, and all the generators must adhere to this standard, when connected to it.<br/>The real situation is more complex.<br/><br/>Considering the grid covers large distances, we must maintain the high voltage in it to avoid waste of energy to heat (hundreds of thousands of volts). Most consumers, however, need relatively lower voltage (no more than a few hundred volts).<br/><br/>To connect each consumer to a grid through a powerful and very expensive transformer is impractical. Instead, we combine a group of geographically close consumers into it's own grid, connected at one point to a main ultra high voltage grid through a transformer that lowers the voltage, and then we connect consumers to this second level grid.<br/><br/>For example, the whole city can form this secondary grid with lower voltage than in the main ultra high voltage grid. Thus formed, this secondary grid can go to each street with lower voltage better suitable for consumption. And, because this grid is relatively localized, there will be no big loss of energy to distribute electricity at a lower voltage.<br/><br/>Making the picture even more complex, we can arrange third level grid lowering the voltage for each building in the city from street level of the secondary grid to a lower voltage building level that goes to each apartment.<br/><br/>Yet another complication can be introduced by connecting other generators to a grid. Since our grid now consists of many grids connected via transformers from main ultra high voltage to second level with lower voltage in the city and third level for each building, we can introduce generators into each grid, and the only requirement for this is to make sure that each generator conforms to a corresponding grid's voltage, frequency and phase.<br/><br/>For example, a building's management decided to put solar panels on the roof. They generate electricity for a building and, if there is excess of energy, it goes to a grid for general consumption. That, probably, is the third level grid that serves this building, it has it's own characteristics, and the output of solar panels must conform to these characteristics.<br/><br/>Similarly, a city decided to build a power plant working on burning the garbage. This power plant produces electricity that should go to a grid that serves the whole city, which we call the second level grid. The voltage produced by this plant is higher than the one in any building level grid but not as high as in main ultra high voltage grid that supplies the whole city.<br/><br/>The overall picture of distribution of electricity consists of different grids of different voltage, connected through transformers, having producers and consumers in each grid. Each consumer of electricity should have its parameters to be the same as the grid it's connected to (voltage and frequency). Each producer of electricity should be synchronized with the grid it's connected to in voltage, frequency and phase.<br/><br/>On the picture below we have schematically displayed the generation of electricity at <i>13,000V</i>, transformers that increase the voltage to <i>600,000V</i> to transmit along the long distances, transformers that decrease the voltage to <i>7,200V</i> at the entrance to a city that supplies this electricity to buildings and, finally, transformers that decrease the voltage to <i>240V</i> before entering buildings.<br/>Inside the buildings this voltage is distributed to individual apartments.<br/><img src='http://www.unizor.com/Pictures/GridA2Z.png' style='width:200px;height:170px;'><br/>In practice there are more devices participating in the grid, stabilizing the voltage, sensing the abnormal conditions, controlling different functions of the grid, protecting the grid against disastrous conditions, attacks or human errors etc.<br/><br/>The grid is constantly changing as new sources of energy come on line, new consumers are attached to a grid, new more efficient maintenance devices are introduced Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-88132076282570838052021-01-24T08:52:00.003-08:002021-01-24T08:52:45.433-08:00Electric Grid: UNIZOR.COM - Physics4Teens - Electromagnetism - Distribution<iframe width="480" height="270" src="https://youtube.com/embed/R_jVITpgmZM" frameborder="0"></iframe><br/><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>The Grid</u><br/><br/>The <i><b>grid</b></i> is an extremely important solution to most problems with electric power interruption due to different technical issues at power plants that generate electricity.<br/><br/>Consider a simple analogy of distributing water to apartments in a large building from a tank on a roof.<br/>If you have one tank, and it needs repair or cleaning, the water to all apartments must be shut off while the work performed.<br/>If, however, you have two water tanks on the roof both connected to a common distribution pipe, from which the water is flowing to all apartments, we can just close the connection of the tank to be cleaned to a distribution pipe, leaving another tank operational and water supply to apartments uninterrupted.<br/><br/>This same principle is used in combining many generating electricity power plants to a common distribution wiring, thus assuring uninterrupted power supply. This system of interconnected power plants forms a <i><b>grid</b></i> that feeds those consumers of electricity connected to it, assuring their uninterrupted work.<br/><br/>Obviously, with electricity it's much more complex than with water supply.<br/><br/>Let's enumerate problem we have to resolve when connecting different electric generators to a common power distribution system.<br/><br/><i>Direct Current</i><br/><br/>Let's connect two batteries generating direct current parallel to each other and power up a lamp.<br/><img src='http://www.unizor.com/Pictures/ParallelBatteries.png' style='width:200px;height:265px;'><br/>The proper connection requires the same <i>polarity</i> (positive pole of one battery is connected to positive pole of another, negative - to negative) and the same <i>voltage</i> generated by these batteries. Only then there will be no electric current between the batteries, only from a battery to a device consuming the electricity (a lamp in this case).<br/>These two conditions, similar <i>polarity</i> and equal <i>voltage</i>, are necessary and sufficient conditions to successfully connect two batteries in parallel. The overall energy capacity of these two batteries will be twice as big. They will last twice as long on the same load (one lamp) or they can double the load (have two lamps parallel to each other) and serve the same time as one battery on a single load.<br/><br/><br/><i>Alternating Current</i><br/><br/>Analogously, two generators of alternating current (AC) must have the same output voltage, if connected in parallel. Otherwise, there will be an unnecessary electric current between them, which diminishes their usefulness.<br/><br/>But for AC generators there are more characteristic parameters than just a voltage. Voltage varies as a sinusoid with time and is characterized by <i>amplitude</i>, <i>frequency</i> and <i>phase</i>.<br/><br/>All three parameters must be the same for a proper parallel connection of two generators. Their output voltages, as functions of time, must coincide exactly to each other. And this is a big challenge to build a grid with many different generators, each contributing their part in overall power supply.<br/><img src='http://www.unizor.com/Pictures/ParallelGenerators.png' style='width:200px;height:170px;'><br/><br/>The above considerations dictate strict restrictions on how any new source of electricity, like a new power plant or a new solar panel are hooked to a <i><b>grid</b></i>.<br/><br/>First of all, the output of a generator at the point of connection to a grid must be alternating. Solar panels, for example, produce direct current, so, before connecting to a grid, the DC electricity must be converted to AC. Special devices called <i>inverters</i> provide this type of conversion, assuring the frequency of voltage oscillation to be that of the grid.<br/><br/>Then, depending on a point of connection to a grid, the output voltage must be equalized with that of the grid at the connection point. This can be done with proper <i><b>transformers</b></i>.<br/><br/>Finally, we have to adjust the phase to synchronize the output of the generator with the phase of a grid. This can be achieve by adjusting the speed of a generator's rotor while monitoring the phase difference on a special sensor until proper <i><b>synchronization</b></i> is achieved.<br><br/>Overall, the connection to a grid is a sophisticated process that requires special devices, tools, instrumentation and care.<br/>There are many controls that monitor, adjust and maintain the regime of work of a grid. In many ways it's automated, computer controlled and reliable. However, human errors do happen and some of them result in significant distortions of power supply to large areas and affecting a lot of people. As an example, the 2003 blackout in Ohio resulted in power loss across Eastern United States and even some areas in Canada.<br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-54438069816225472312021-01-22T08:16:00.000-08:002021-01-22T08:16:37.062-08:00Electricity In Transit: UNIZOR.COM - Physics4Teens - Electromagnetism - ...<iframe width="480" height="270" src="https://youtube.com/embed/6PiSthNGN1Q" frameborder="0"></iframe><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Electricity in Transit</u><br/><br/>Let's discuss how electricity is delivered from the power plants to consumers.<br/><br/>The only way to deliver the electricity from the place it's generated to a place it's used is via electric wires.<br/>Since the distance between the power plant and a consumer can be substantial, may be even hundreds of kilometers, the problem of losses of electric energy in transit because of wire resistance is extremely important.<br/><br/>Examine a simple electric circuit consisting of a source of electricity (generator) and a consumer (like an electric motor).<br/>In theory, we have four places where electric energy is spent:<br/>(a) inside a generator due to internal resistance,<br/>(b) inside a wire from a generator to a motor due to wire resistance,<br/>(c) inside a motor due to useful work the electricity does and internal resistance,<br/>(d) inside a wire from a motor back to a generator due to wire resistance.<br/><img src='http://www.unizor.com/Pictures/ElectricityTransit.png' style='width:200px;height:60px;'><br/><br/>Obviously, only in a motor the energy is spent with some useful purpose, in other places the energy is spent just because it's unavoidable, that is wasted.<br/><br/>Amount of energy wasted to heat per unit of time due to wire resistance <i><b>W</b><sub>wire</sub></i> depends on the current running through wire <i><b>I</b><sub>wire</sub></i> and the wire resistance <i><b>R</b><sub>wire</sub></i> according to a formula<br/><i><b>W</b><sub>wire</sub><b> = I<sup>2</sup></b><sub>wire</sub><b>·R</b><sub>wire</sub></i><br/><br/>For practical example, let's calculate the amount of energy wasted in some long piece of copper wire.<br/>The resistance of a wire depends on <i>resistivity</i> of material it's made of <i>ρ</i>, is proportional to the length of a wire <i>L</i> and is inversely proportional to its cross-section area <i>A</i><br/><i><b>R</b><sub>wire</sub><b> = ρ·L<font size=4>/</font>A</b></i><br/><br/>For copper the resistivity is approximately<br/><i>ρ≅1.70·10<sup>−8</sup> Ω·m</i>.<br/>Assume, the combined wire length to and from a consumer of electricity is<br/><i>L = 1 km = 1000 m</i><br/>and its diameter is<br/><i>D = 2 mm = 2·10<sup>−3</sup> m</i>,<br/>which gives the area of its cross-section<br/><i>A = π·D²/4 ≅ 3.14·10<sup>−6</sup> m²</i><br/>Then the resistance of this peace of wire is<br/><i>R<sub>wire</sub> ≅ 5.4 Ω</i><br/><br/>For example, we are supposed to run an electric motor working at voltage<br/><i>U<sub>mot</sub>=220 volt</i><br/>and delivering power of<br/><i>W<sub>mot</sub>=2.2 kilowatt</i>.<br/><br/>Then the current it requires is<br/><i>I<sub>mot</sub> = W<sub>mot</sub>/U<sub>mot</sub> = 10 A</i><br/><br/>The current <i>I<sub>mot</sub>=10 A</i> must go through the wire<br/><i>I<sub>wire</sub> = I<sub>mot</sub> = 10 A</i><br/>Then the amount of energy wasted to heat in the copper wire of resistance <i>R<sub>wire</sub>=5.4 Ω</i> per unit of time (a second) is<br/><i>W<sub>wire</sub> = I²<sub>wire</sub>·R<sub>wire</sub> = 540 W</i><br/><br/>For a price of about $0.1 per kilowatt this amounts to about $0.054 per second. For 24 hours uninterrupted work the financial waste amounts to $4,665, and that is every day of operation of one motor, which is absolutely unacceptable.<br/><br/>Reducing the resistance of a wire by making it thicker or using multiple parallel wires has its practical limitations because of cost of wires. Therefore, our solution to reduce the energy wasted to heat due to wire resistance, while staying within reasonable limits with the cost of a wire, must be related to reducing the current <i>I</b><sub>wire</sub></i> running through a wire without reduction of power that is supposed to be delivered to consumers of electricity.<br/><br/>This can be accomplished by using transformers.<br/>Immediately after generation, the alternating current is directed to a <i>transformer</i> that increases the voltage and proportionally decreases the amperage.<br/><br/>At the output of this transformer the voltage reaches thousands of volts - from low voltage of 1000V to ultra high voltage above 800,000V, depending on the length of wires from generators to consumers.<br/>This high voltage electricity is delivered to consumers, where another transformers reduce the voltage to standard needed to run all their different devices. <br/><br/>Consumers of electricity get the voltage required to run their equipment, but the current running in the long wires between generators and consumers is low, thus reducing waste of electric energy.<br/><br/>Consider an example above with a motor that needs <i>W<sub>mot</sub>=2.2 kilowatt</i> of electricity at voltage <i>U<sub>mot</sub>=220 volt</i> and, therefore, requires <i>I<sub>mot</sub>=10 A</i> electric current.<br/>If, instead of transmitting electricity with these parameters, we increase the voltage by a <i>transformer</i> before sending it to long wires to, say, <i>2200V</i>, thus proportionally reducing the amperage by the same factor, our amperage will be<br/><i>A = W/U = 2200/2200 = 1A</i><br/>Reducing the amperage from <i>10A</i> to </i>1A</i> reduces the energy waste by a factor of <b>100</b> because the heat formula depends on a square of amperage.<br/><br/>The distribution of electricity, therefore, should include transformers that increase the voltage before sending electricity along long wires and decrease it wherever it's needed for usage by consumers.<br/>With this modification the picture that corresponds to practical aspects of distribution of electricity looks like this<br/><img src='http://www.unizor.com/Pictures/ElectricityTransformed.png' style='width:200px;height:80px;'><br/><br/>To increase the electrical systems' reliability, improve the energy balancing and make sure of uninterrupted power supply, the sources of electrical energy (electric power plants and other installations producing electric energy) are combined into a network called the <i><b>grid</b></i>.<br/>The principles of this networking are a subject of the next lecture.<br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-65267088255934775682021-01-10T09:25:00.000-08:002021-01-10T09:25:15.934-08:00Electricity at Power Plants: UNIZOR.COM - Physics4Teens - Electromagneti...<iframe width="480" height="270" src="https://youtube.com/embed/6R9a4AX-qrc" frameborder="0"></iframe><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Electricity at Power Plants</u><br /><br/>Addressing distribution of electricity, we will primarily discuss the way electricity, produced at the power plants, is delivered to consumers.<br/>We will concentrate on the process of distribution of electricity generated from the kinetic energy of rotating turbines, as the most quantitatively significant source of electricity.<br/><br/>Three stages of this distribution are<br/>(a) at the power plant<br/>(b) in transit<br/>(c) at consumers.<br/><br/>This lecture is about what's going on at the power plant that produces the electricity from the kinetic energy of rotating turbines.<br/><br/>Turbines at electric power plants are rotated because of a flow of steam or water, or wind. Turbines are acting as <i>rotors</i> in the electric power generator, while the electricity is produced in <i>stators</i> based on the principles of electromagnetic induction.<br/><br/><br/><i>Hydroelectric Stations</i><br/><br/>Let's estimate theoretically an amount of energy that a hydroelectric station can produce.<br/><img src='http://www.unizor.com/Pictures/HydroelectricPlant.jpg' style='width:200px;height:160px;'><br/><br/>First important component in generating electricity at a hydroelectric station is falling water. We can have falling water by building a dam on a river, like Hoover Dam on Colorado river or use natural difference in the level of water of the waterfalls, like at Niagara falls.<br/><br/>Having the water at two levels, we should direct the flow from top to bottom onto turbines through pipes. Amount of electricity that can be generated obviously depends on the amount of potential energy water at the top has relatively to the bottom level. As the water falls through the pipes onto turbine, its potential energy is converted into kinetic energy of moving water. This is how much energy we can use to generate electricity. It depends on the amount of water flowing through pipes and the height difference between the top and the bottom levels.<br/><br/>Let's assume that the amount of water falling down through pipes from top to bottom level is <nobr><i><b>M</b> (kg/sec)</i></nobr> and the difference in height from top to bottom is <nobr><i><b>H</b> (m)</i>.</nobr><br/>That means that the amount of energy falling water is losing per unit of time is<br/><i><b>P</b><sub>water</sub><b> = M·g·H</b> (J/sec </i>or<i> W)</i><br/><br/>Then we have to solve a purely technical problem to convert this energy into rotational energy of turbines.<br/><img src='http://www.unizor.com/Pictures/HydroTurbine.jpg' style='width:200px;height:135px;'><br/><br/>Different designs of turbines have been used and tested during a long time of using hydroelectric power. Contemporary turbines are pretty efficient in this process of conversion, but still far less than 100% effective. Losses of energy always exist, and we need some coefficient of efficiency of a turbine to get exact amount of rotational energy produced by falling water.<br/>Let's assume that <i><b>k</b></i> is such a coefficient. It has a value from 0 (absolutely ineffective conversion) to 1 (full energy amount of falling water is converted into rotational energy of a turbine). Then the amount of rotational energy produced by turbines per unit of time is<br/><i><b>P</b><sub>turbine</sub><b> = k·M·g·H</b></i><br/><br/>Next step is to convert rotational energy of turbines into electric energy.<br/>This is done by generators, which we discussed in previous lectures.<br/><img src='http://www.unizor.com/Pictures/RotorStator.jpg' style='width:200px;height:100px;'><br/><br/>The contemporary generators are pretty effective with norm being above 90%, so we can assume that the coefficient of effectiveness <i><b>k</b></i> introduced above encompasses both effectiveness of converting energy of falling water into rotation of turbines and conversion of rotation into electric energy.<br/>So, overall energy produced by a hydroelectric power station per unit of time (that is, the power produced) is<br/><i><b>P = k·M·g·H</b></i><br/>The hydroelectric power stations can be very large and can produce a lot of electric energy. The most powerful electric power stations are hydroelectric. The problem is, there are not too many rivers suitable for building hydroelectric power stations and an environmental effect of building a hydroelectric power station can be significant.<br/><br/>At the same time, the hydroelectric power stations are pretty efficient, the coefficient <i><b>k</b></i> in the formula above can be above 0.8, which means that about 80% of the power of water falling on turbines is effectively converted into electric power.<br/><br/><br/><i>Coal Burning Stations</i><br/><br/>Almost a third of electricity generated in the world is produced by fossil fuel burning power stations.<br/>Let's examine the coal burning power station.<br/><img src='http://www.unizor.com/Pictures/CoalBurningElectricPlant.jpg' style='width:200px;height:135px;'><br/><br/>The main steps of producing electricity by burning fossil fuel are<br/>(a) burning fossil fuel to boil water, converting chemical energy of burning fuel into kinetic energy of produced steam,<br/>(b) converting kinetic energy of steam into rotation of turbines,<br/>(c) converting rotational energy of turbines into electricity by generators.<br/><br/>Coal is a major source of fossil fuel with natural gas and oil following.<br/>Convenience of putting a coal burning electric power station anywhere should be weighed against environmental impact of such a plant.<br/><br/>Producing energy from burning coal is not a very efficient way to extract chemical energy. Significant portion of the energy produced by burning coal is wasted on each step and the overall efficiency of such a power station is about 40%. Most of the energy losses occur during the first stage of generating electricity - burning coal to boil water and produce steam.<br/>Some efficiency can be achieved by pulverizing coal to powder. However, the main product of burning fossil fuel - carbon dioxide <i><b>CO<sub>2</sub></b></i> - produces some unavoidable negative environmental effect.<br/><br/><br/><i>Nuclear Power Plants</i><br/><br/>The difference between a nuclear power plant and coal burning one is at the first stage to boil the water. While at coal burning plants the source of heat to boil water is burning coal, at the nuclear power plant the source of heat is energy released by breaking nuclei of heavy elements, like Uranium or Plutonium, into lighter components using neutrons.<br/><img src='http://www.unizor.com/Pictures/NuclearFission.jpg' style='width:200px;height:120px;'><br/><br/>The main steps of producing electricity in a nuclear power plant are<br/>(a) bombarding the enriched radioactive material (Uranium, Plutonium or other) with neutrons causing the nuclei of this material to break, releasing certain amount of heat to boil water getting steam,<br/>(b) converting kinetic energy of steam into rotation of turbines,<br/>(c) converting rotational energy of turbines into electricity by generators.<br/><br/>Efficiency of nuclear power plants is quite limited inasmuch as in coal burning power stations and is about 40%. That is, about 40% of the energy generated by heat is converted into electricity.<br/><br/>Of interest is a process of nuclear fission that produces the heat. Here is a simplified model of this process.<br/><br/>When a nucleus of Uranium-235 (92 protons + 143 neutrons) is bombarded with a neutron, it temporarily accepts this neutron inside, becoming Uranium-236 (92 protons + 144 neutrons).<br/>This isotope is not stable and a nucleus breaks into different parts. This is a complex process and parts might be different.<br/><br/>A typical scenario might be as follows.<br/>Broken parts are Barium (56 protons + 83 neutrons), Krypton (36 protons + 58 neutrons) and 3 neutrons are released to bombard other nuclei of Uranium 235, causing a <b>chain reaction</b>.<br/><br/>The combined mass of all parts is less than the mass of initial components. The remaining mass of an unstable nucleus of Uranium-236 is converted into radiation (heat, gamma-rays). The heat is used to boil the water, converting it into high energy steam to rotate the turbines.<br/><br/>The corresponding equations describing nuclear fission is:<br/><i><b><sup>1</sup>n<sub>0</sub> + <sup>235</sup>U<sub>92</sub> → <sup>236</sup>U<sub>92</sub> →<br/>→ </b>(fission)<b> →<br/>→ <sup>139</sup>Ba<sub>56</sub> + <sup>94</sup>Kr<sub>36</sub> + 3<sup>1</sup>n<sub>0</sub> + γ</b></i><br/><br/>In reality the process is much more complex because the broken parts of a nucleus might be different, themselves not stable and further emitting elementary particles.<br/>The process must be controlled by reducing the number of neutrons flying in all the directions after fission to prevent a nuclear explosion.<br/> Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-74465376178827381492020-12-27T07:19:00.000-08:002020-12-27T07:19:09.335-08:00Electricity from Solar Energy: UNIZOR.COM - Physics4Teens - Electromagne...<iframe width="480" height="270" src="https://youtube.com/embed/AIcuexrQQO8" frameborder="0"></iframe><br/><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Solar Energy → Electricity</u><br /><br/>Solar panels that produce electricity are not physically or chemically changed during this process. So, the only source of energy is the light, and the energy it carries is transformed by a solar panel into electricity, that is into movement of electrons.<br/>In this lecture we will analyze how the energy of light forces the electrons to move.<br/>The complete theory behind this process is based on principles of quantum mechanics and we will not be able to dive deep into this area, but a schematic description of the process will be presented.<br/><br/>Commercial solar panel are much more complex that might seem from the following explanation, but their engineering is not a subject of this lecture, which is only about the principle of generation of electricity from sun light.<br/><br/>Let's start with the most important component of a solar cell - silicon (<i><b>Si</b></i>).<br/>Silicon is a metalloid, it has crystalline structure and its atoms have four valence electrons on the outer most orbit, next down orbit contains eight electrons and the inner most - two electrons.<br/><br/>Silicon is one of the most frequently occurring elements on Earth and, together with oxygen, forms molecules of silicon dioxide <i><b>SiO<sub>2</sub></b></i> - main component of sand.<br/><br/>All silicon atoms are neutral since the number of electrons in each one and the number of protons in each nucleus are the same and equal to 14.<br/>Atoms are connected by their covalent bonds of valence electrons on the outer orbits into a lattice-like crystalline structure.<br/><br/>Picture below represents the crystalline structure of silicon<br/><img src='http://www.unizor.com/Pictures/SiliconCrystal.jpg' style='width:200px;height:200px;'><br/>For the purposes of this lecture we will represent inner structure of silicon atoms with only outer orbit of each atom with four valence electrons, as electrons on the inner orbits do not participate in the process of generating electricity from light.<br/><img src='http://www.unizor.com/Pictures/SiliconStructure.jpg' style='width:200px;height:200px;'><br/>The covalent bonds are quite strong, they do not easily release electrons. As a result, under normal conditions silicon is practically a dielectric.<br/>However, if we excite the electrons sufficiently enough to break the covalent bonds, some electrons will move. For example, if we increase the temperature of a piece of silicon or put it under a bright sun light and measure its electrical resistance, we will observe the resistance diminishing.<br/>That's why silicon and similar elements are called <b>semiconductors</b>.<br/><br/>While excited electrons of any semiconductor decrease its electrical resistance, they don't produce electromotive force because the material as a whole remains electrically neutral.<br/>To build a solar cell that produces electricity under sun light we will introduce two kinds of "impurities" into crystalline lattice of silicon.<br/>One is an element with five valence electrons on the outer most orbit, like phosphorus (<i><b>P</b></i>). When it's embedded into a crystalline lattice of silicon, one electron on that outer orbit of phosphorus would remain not attached to any neighboring atom through a covalent bond.<br/><img src='http://www.unizor.com/Pictures/SiliconPhosphorus.jpg' style='width:200px;height:200px;'><br/>This creates the possibility for this electron to start traveling, replacing other electrons and pushing them out, which are, in turn, push out others etc. Basically, we create as many freelance electrons as many atoms of phosphorus we add to silicon base.<br/>The whole material is still neutral, but it has certain number of freelance electrons and the same number of stationary positive ions - those nuclei of phosphorus that lost electrons to freelancers.<br/>Silicon with such addition is called <b>n-type</b> (letter <b>n</b> for <i>negative</i>).<br/><br/>Another type of "impurity" that we will add to silicon is an element with three valence electrons on the outer orbit, like boron (<i><b>B</b></i>).<br/>When atoms of boron are embedded into a crystalline lattice of silicon, the overall structure of this combination looks like this<br/><img src='http://www.unizor.com/Pictures/SiliconBoron.jpg' style='width:200px;height:200px;'><br/>In this case the lattice has a deficiency of an electron that is traditionally called a "hole". Existence of a "hole" opens the opportunity for neighboring valence electrons to fill it, thus creating a "hole" in another spot. These "holes" behave like positively charged particles traveling inside silicon with added boron inasmuch as negatively charged electrons travel in silicon with added phosphorus.<br/>Silicon with such addition of boron is called <b>p-type</b> (letter <b>p</b> for <i>positive</i>).<br/><br/>Now imagine two types of "impure" silicon, <b>n-type</b> and <b>p-type</b>, contacting each other. In practice, it's two flat pieces (like very thin squares) on top of each other.<br/>Let's examine what happens in a thin layer of border between these different types of material.<br/><br/>Initially, both pieces of material are electrically neutral with n-type having free traveling electrons and equal number of stationary positive nuclei of phosphorus inside a crystalline structure and with p-type having traveling "holes" and equal number of stationary negative nuclei of silicon inside a crystalline structure.<br/><br/>As soon as contact between these two types of material is established, certain exchange between electrons of the n-type material and "holes" of the p-type takes place in the border region called <b>p-n junction</b>. Electrons and "holes" in this border region combine, thus reconstituting the lattice.<br/>This process is called <b>recombination</b>.<br/><br/>The consequence of this process of recombination is that n-type material near the border loses electrons, thus becoming positively charged, while the p-type gains the electrons, that is loses "holes", thus becoming negatively charged.<br/><br/>Eventually, the diffusion between n-type and p-type materials stops because negative charge of the p-type sufficiently repels electrons from the n-type. There will be some equilibrium between both parts.<br/><br/>If we introduce heat or bright sun light to electrons of n-type part, the diffusion will be longer and greater charge will be accumulated on both sides of the material - positive on the n-type and negative on the p-type.<br/><br/>Now the key point is to connect n-type side to p-type through some kind of electrical connection and extra electrons from the p_type will go to positively charged n-type, thus creating an electric current. This current is weak because only the diffusion between n-type and p-type in the border region is a contributing factor, but it's still the electric current.<br/><br/>Obviously, the more excited electrons in the n-type material are - the stronger diffusion inside the p-n junction is and the stronger current is produced. That's why, if sun light is used to excite the electrons, the more light falls on the n-type side - the stronger current is produced.<br/><br/>A pair of n-type and p-type materials form a <i>cell</i>. Since we are talking about using sun light to excite electrons, these cells are called <i>solar cells</i> or <i>photovoltaic cells</i>. Cells can be connected in a series increasing the produced electromotive force (voltage) or they can be connected in parallel to increase the electric current (amperage). <i>Solar panels</i> are sets of <i>solar cells</i> connected in series to increase the generated electromotive force (voltage).<br/><br/>Materials used to produce commercial solar cells can be different, not necessarily silicon with phosphorus or boron additions, but the main principle of using p-n junction between superconductors is the same for all.<br/><br/> Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-75461795306853161912020-12-20T09:47:00.000-08:002020-12-20T09:47:45.156-08:00Chemical Energy to Electricity: UNIZOR.COM - Physics4Teens - Electromagn...<iframe width="480" height="270" src="https://youtube.com/embed/Mai5bN-U4H8" frameborder="0"></iframe><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Chemical Energy → Electric Energy</u><br /><br/>To generate energy, the source components participating in the process should have more energy than the final components of the process. In case of generation of any form of energy from chemical energy we deal with chemical reaction, during which the inter-atomic energy of source components should exceed the inter-atomic energy of products of chemical reaction.<br/><br/>The simple form of such process is burning that produces heat. For example, burning methane <i><b>CH<sub>4</sub></b></i> in atmosphere that contains oxygen <i><b>O<sub>2</sub></b></i> is a chemical reaction described by<br/><i><b>CH<sub>4</sub> + 2O<sub>2</sub> = CO<sub>2</sub> + 2H<sub>2</sub>O</b></i><br/>The inter-atomic energy of one molecule of methane and two molecules of oxygen must be greater than inter-atomic energy of a molecule of carbon dioxide and two molecules of water, otherwise there will not be any energy released as heat.<br/><br/>To generate electric energy from chemical we need the same type of inequality: the inter-atomic energy of primary components of the chemical reaction must be greater than inter-atomic energy of the resulting components.<br/>The way how this excess of energy represented depends on the chemical reaction. In the process of burning the excess of energy is in a form of heat, in case of the generating of electricity the excess of energy is in a form of electric current that has certain voltage and amperage and, therefore, is a carrier of energy.<br/><br/>Transformation of chemical energy into electricity is typically occurring in <b>batteries</b>.<br/>There are different types of batteries, and in this lecture we will mention a few with some details of how they work.<br/><br/>Consider a <b>lead-acid battery</b> used in many cars.<br/>In a simplified way it has three major components: solid <i>anode</i> (negative electrode) made of <i>lead <b>Pb</b></i>, solid <i>cathode</i> (positive electrode) made of <i>lead dioxide <b>PbO<sub>2</sub></b></i> and liquid <i>electrolyte</i> in-between them containing <i>sulfuric acid <b>H<sub>2</sub>SO<sub>4</sub></b></i> diluted in <i>water <b>H<sub>2</sub>O</b></i>.<br/><img src='http://www.unizor.com/Pictures/LeadAcidBattery.png' style='width:200px;height:190px;'><br/>To understand why electricity is generated by this device, let's look inside the atomic structure of its components and analyze what happens when there is a load (like a lamp) connected to its terminals.<br/><br/>Recall the classical planetary model of an atom.<br/>There are protons and neutrons forming its nucleus and electrons circulating around this nucleus on different orbits.<br/>Those electrons on outer most orbits are less attached to a host nucleus and many of them can fly around, potentially getting attached by other host nuclei.<br/>This creates certain amount of "free" positively (with deficit of electrons) charged <i>ions</i> called <b>cations</b> and negatively (with access of electrons) charged <i>ions</i> called <b>anions</b>.<br/><br/>Analogously, molecules are also not completely stable and can lose atoms, if inter-atomic forces are not strong enough.<br/>Applied to molecular structure of any substance, typical contents not only contains electrically neutral molecules, but also molecules with certain missing or extra atoms or electrons - <i>ions</i>.<br/><br/>Consider sulfuric acid inside a battery pictured above.<br/>Structural composition of atoms of each molecule of this acid can be represented as<br/><img src='http://www.unizor.com/Pictures/SulfuricAcid.jpg' style='width:200px;height:100px;'><br/>Many of electrically neutral molecules of sulfuric acid <b>H<sub>2</sub>SO<sub>4</sub></b> are partially breaking losing a positive ions (nuclei with a single proton) of hydrogen but retaining their electrons, which can be represented as<br/><i><b>H<sub>2</sub>SO<sub>4</sub> → 2H<sup> +</sup> + SO<sub>4</sub><sup>2−</sup></b></i><br/>This can be explained by the fact that hydrogen nucleus contains only one proton, this is the lightest and electrically weakest nucleus and, being so light and volatile, it easily breaks its inter-atomic links with the main molecule of sulfuric acid. The resulting <i>cations <b>H<sup> +</sup></b></i> and <i>anions <b>SO<sub>4</sub><sup>2−</sup></b></i> are actually responsible for burning and corrosion caused by sulfuric acid. In batteries, however, these ions are responsible for producing electricity.<br/>Here is how.<br/><br/>The negative <i>anode</i> of a lead-acid battery is made of lead (chemical symbol <i><b>Pb</b></i>), which, when going into chemical reactions, exhibits either 2 or 4 links to other atoms in the molecules.<br/>The negative ion (<i>anion</i>) <i><b>SO<sub>4</sub><sup>2−</sup></b></i>, produced during the breaking of the molecules of sulfuric acid, chemically reacts with led producing <i>lead sulfate</i>, and releasing two electrons left after ions of hydrogen broke free from the molecule of sulfuric acid as follows:<br/><i><b>Pb+SO<sub>4</sub><sup>2−</sup> → PbSO<sub>4</sub> + 2e<sup>−</sup></b></i><br/>Structural composition of atoms of each molecule of <i>lead sulfate</i> can be represented as<br/><img src='http://www.unizor.com/Pictures/LeadSulfate.png' style='width:200px;height:130px;'><br/><br/>Keep an eye on two electrons <i><b>2e<sup>−</b></i> produced at <i>anode</i>. They are produced on the surface of a lead <i>anode</i> and accumulated inside it up to a certain concentration that gives certain negative charge to an <i>anode</i>. These electrons will be the ones that produce the electric current, when some load (like a lamp) is connected to terminals of a battery.<br/><br/>Meanwhile, at the <i>cathode</i> terminal of a led-acid battery another reaction goes on between its main component <i>lead dioxide <b>PbO<sub>2</sub></b></i>, having the following structure<br/><img src='http://www.unizor.com/Pictures/LeadDioxide.png' style='width:200px;height:90px;'><br/>and electrolyte that still contains unused positive ions of hydrogen <i><b>2H<sup>+</sup></b></i> from the reaction on an <i>anode</i> and another pair of ions, positive <i><b>2H<sup>+</sup></b></i> and negative <i><b>SO<sub>4</sub><sup>2−</sup></b></i> of the sulfuric acid.<br/><br/>Now two reactions simultaneously go on at the surface of a <i>cathode</i>.<br/>First of all, lead dioxide connects with negative ion of sulfuric acid, producing lead sulfate and releasing two negative ions of oxygen:<br/><i><b>PbO<sub>2</sub> + SO<sub>4</sub><sup>2−</sup> →<br/>→ PbSO<sub>4</sub> + 2O<sup>−</sup></b></i><br/>Secondly, remaining two positive ions of hydrogen from a molecule of sulfuric acid participating in the above reaction and two positive ions of hydrogen from the reaction on an <i>anode</i> meet two negative ions of oxygen from the above reaction, forming two molecules of water <i><b>H<sub>2</sub>O</b></i> with two electrons still missing:<br/><i><b>4H<sup>+</sup> + 2O<sup>−</sup> → 2H<sub>2</sub>O<sup>2+</sup></b></i><br/><br/>Concentration of these molecules of water missing two electrons creates a positive charge on the <i>cathode</i> terminal of a battery. When this concentration reaches certain level, positive ions of hydrogen cannot reach a <i>cathode</i>, ions of hydrogen are not readily breaking off the molecules of sulfuric acid and reaction stops at certain level of positive charge, unless we connect <i>anode</i> and <i>cathode</i> through some external electric load to allow accumulated on an <i>anode</i> electrons compensate missing electrons on a <i>cathode</i>.<br/><br/>So, we have a shortage of two electrons on a <i>cathode</i> terminal of a battery to form electrically neutral molecules of lead dioxide and water, but these two electrons can travel through an outside load between the terminals of a battery from its <i>anode</i>.<br/>The overall reaction on a <i>cathode</i> looks like<br/><i><b>PbO<sub>2</sub> + 4H<sup>+</sup> + SO<sub>4</sub><sup>2−</sup> + 2e<sup>−</sup> →<br/>→ PbSO<sub>4</sub> + 2H<sub>2</sub>O</b></i><br/><br/>As a result of these chemical reactions electrons released by a reaction on an <i>anode</i> travel to a <i>cathode</i> through external load, thus creating an <b>electric current</b>.</br>In an absence of an external load reactions on terminals of a battery continue until some limit of concentration of negative charge on an <i>anode</i> and positive charge on a <i>cathode</i> is reached, after which the reactions stops as further ionization is prevented by accumulated charges.<br/><br/>This is how chemical energy (inter-molecular links) is converted into electrical energy in a lead-acid battery.<br/><br/>The above described lead-acid battery is capable to work in reverse, to accumulate chemical energy, if external electromotive force is applied to its terminals. In this case all reactions go in reverse, that's how car battery is charged by alternator, when a car is in motion.<br/>The overall picture of the lead-acid battery to discharge and charge is as follows<br/><img src='http://www.unizor.com/Pictures/LeadAcidProcess.png' style='width:200px;height:300px;'><br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-5405065304407990502020-12-07T08:00:00.000-08:002020-12-07T08:00:27.890-08:00Electricity from Kinetic Energy: UNIZOR.COM - Physics4Teens - Electromag...<iframe width="480" height="270" src="https://www.youtube.com/embed/9h04wQ904J8" frameborder="0"></iframe><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Kinetic Energy → Electric Energy</u><br /><br/>The main principle used in converting kinetic energy into electric is the principle of electromagnetic induction.<br/>Recall the Faraday's Law that defines the induced EMF as being proportional to a rate of changing the magnetic flux<br/><i><b>EMF = −</b>d<b>Φ/</b>d<b>t</b></i><br/><br/>We can achieve a variable magnetic flux by either rotating the wire frame in the permanent magnetic field or rotating the magnetic field inside the wire frame. The corresponding designs were discussed earlier in this course.<br/><br/>In this lecture we will talk about how we can make a rotation of a rotor in the electric generator.<br/><br/>The simplest form of generating a rotational movement is if we already have some mechanical movement, so all we need is to transform one form of motion (usually, along some trajectory) into a rotational one.<br/><br/>As the first example of such purely mechanical device, consider a propeller.<br/><img src='http://www.unizor.com/Pictures/Propeller.jpg' style='width:200px;height:150px;'><br/>It can be used to convert the flow of water in <b>hydroelectric plants</b> or the flow of wind through a <b>wind turbine</b> into a rotation. Once we have a rotational motion of the rotor in an electric generator, the rest goes along the previously described way of generation of electricity according to the laws of induction.<br/>It can be a generation of alternating current, including three-phase one, or direct current. These were discussed in details in previous lectures.<br/><br>In other cases we do not have already available motion that we can transform into a rotation, but we can artificially create one, using some other form of energy.<br/>The common process <b>thermal power station</b> is to generate a flow of steam by heating the water or a flow of some kind of combustion gases. This can be done by using the burners that burn coal, oil or natural gas.<br/>Another way of heating is to use nuclear energy to heat the water by controlling the chain reaction inside the radioactive core of a nuclear reactor.<br/>In some cases the solar energy is used to heat the water to produce a flow of steam.<br/>Rarely used types of heat are geothermal and ocean thermal sources.<br/><br/>In all these cases some kind of <b>turbines</b> are used to convert the flow of moving substance (water, air, gases, steam) into a rotation.<br/>Steam turbines are just a more sophisticated type of propeller (or rather coaxial propellers), allowing to extract as much as possible energy from the steam flow.<br/><img src='http://www.unizor.com/Pictures/Turbine.jpg' style='width:200px;height:150px;'><br/><br/>Another form of generating electricity from heat and kinetic energy is internal combustion engines. The work of such engine results in a reciprocating motion of a piston, converted, using a connecting rod and a crankshaft, into rotation of a rotor of an electric generator that generates electricity.<br/><img src='http://www.unizor.com/Pictures/Crankshaft.jpg' style='width:200px;height:150px;'><br/><br/>In all the above cases the electricity is produced from kinetic energy of some substance, which is either readily available in nature (like water flowing along a river) or produced as a result of some process (like heating the water to produce a flow of steam).<br/><br/> Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-48975085707000515982020-11-22T09:12:00.002-08:002020-11-22T09:12:57.764-08:00AC Power: UNIZOR.COM - Physics4Teens - Electromagnetism - AC Ohm's Law<iframe width="480" height="270" src="https://www.youtube.com/embed/Una1qjDReO8" frameborder="0"></iframe><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>AC Power</u><br /><br/><b>Power</b> is a rate of work performed by some source of energy (like electric current) per unit of time. More precisely, it's a derivative of work performed by a source of energy, as a function of time, by time.<br/><br/>As we know from the properties of a <i>direct electrical current</i>, its power is<br/><i><b>P = U·I = I²R = U²/R</b></i><br/>where <i><b>U</b></i> is the voltage around a resistor of resistance <i><b>R</b></i> and <i><b>I</b></i> is the electric current going through this resistor.<br/>All values in the above expression are constant for <i>direct current</i>.<br/><br/><i>Alternating current</i> presents a problem of having the voltage and the current to be variable and dependent not only on the resistors, but also on the presence of inductors and capacitors in the circuit.<br/><br/>As described in the previous lecture, the alternating current in the circuit of a resistor, an inductor and a capacitor connected in a series to a generator of sinusoidal EMF equals to<br/><i><b>I(t) = (E<sub>0</sub> <font size=4>/</font>Z)·sin(ωt+φ) =<br/>= I<sub>0</sub>·sin(ωt+φ)</b></i><br/>where impedance <i><b>Z=√<span style='text-decoration:overline'>(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)²+R²</span></b></i>,<br/><i><b>tan(φ)=(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)<font size=4>/</font>R</b></i>.<br/><br/>Having expressions for generated EMF <i><b>E(t)=E<sub>0</sub>·sin(ωt)</b></i> and electric current in a circuit <i><b>I(t)=I<sub>0</sub>·sin(ωt+φ)</b></i>, we can find an instantaneous power<br/><i><b>P(t) = E(t)·I(t) =<br/>= E<sub>0</sub>·I<sub>0</sub>·sin(ωt)·sin(ωt+φ)</b></i><br/><br/>When people talk about voltage or amperage in the AC circuit, they understand that these characteristics are variable and, to be more practical, they use <i>effective voltage <nobr><b>E<sub>eff</sub> = E<sub>0</sub> <font size=4>/</font>√<span style='text-decoration:overline'>2</span></b></i></nobr> and <i>effective amperage <nobr><b>I<sub>eff</sub> = I<sub>0</sub> <font size=4>/</font>√<span style='text-decoration:overline'>2</span></b></i></nobr> of the electric current. The usage of these characteristics allows to calculate the power consumed by a resistor-only circuit during a period [<i><b>0,T</b></i>] of time (for <i><b>T</b></i> significantly greater than one oscillation of a current) without integrating a variable function <i><b>P(t)=E(t)·I(t)</b></i> on interval [<i><b>0,T</b></i>], but just performing a multiplication of constants:<br/><i><b>W<sub>[0,T]</sub> = E<sub>eff</sub> · I<sub>eff</sub> · T</b></i><br/><br/>Adding an inductor and a capacitor brings some complication because of a phase difference between EMF and a current. To express the power consumed by an AC circuit that includes a resistor, an inductor and a capacitor in terms of effective voltage and effective amperage, let's find the work performed by an electric current during a period of oscillation in terms of <i><b>E<sub>eff</sub></b></i> and <i><b>I<sub>eff</sub></b></i> and divide it by this period. The result would be an average power consumed by a circuit per time of one oscillation that we will call the <b>effective power</b> of a circuit.<br/><br/>The period of one oscillation with angular speed <i><b>ω</b></i> is <i><b>T=2π/ω</b></i>.<br/>The instantaneous power consumption by a circuit is<br/><i><b>P(t) = E(t)·I(t) =<br/>= E<sub>0</sub>·sin(ωt)·I<sub>0</sub>·sin(ωt+φ)</b></i><br/>The energy consumed by a circuit during one period of oscillation <i><b>T=2π/ω</b></i> equals to<br/><i><b>W<sub>[0,T]</sub> = <font size=5>∫</font><sub>[0,T]</sub>P(t)·</b>d<b>t</b></i><br/>where<br/><i><b>P(t)=E<sub>0</sub>·sin(ωt)·I<sub>0</sub>·sin(ωt+φ)</b></i><br/><br/>We can simplify the product of two trigonometric functions to make it easier to integrate:<br/><i><b>sin(x)·sin(y) =<br/>= (1/2)·</b></i>[<i><b>cos(x−y)−cos(x+y)</b></i>]<br/>Using this for <i><b>x=ωt</b></i> and <i><b>y=ωt+φ</b></i>, we obtain<br/><i><b>sin(ωt)·sin(ωt+φ) =<br/>= (1/2)·</b></i>[<i><b>cos(φ)−cos(2ωt+φ)</b></i>]<br/><br/>To find the power consumption during one period of oscillation <i><b>T</b></i>, we have to calculate the following integral<br/><i><b>W<sub>[0,T]</sub> = <font size=5>∫</font><sub>[0,T]</sub>P(t)·</b>d<b>t</b></i><br/>where period <i><b>T=2π/ω</b></i> and<br/><i><b>P(t) = E<sub>0</sub>·I<sub>0</sub>·<br/>·(1/2)·</b></i>[<i><b>cos(φ)−cos(2ωt+φ)</b></i>]<br/><br/>This integral can be expressed as a difference of two integrals<br/><i><b><font size=5>∫</font><sub>[0,T]</sub>E<sub>0</sub>·I<sub>0</sub>·(1/2)·cos(φ)·</b>d<b>t</b></i><br/>which, considering <i><b>cos(φ)</b></i> is a constant for a given circuit, is equal to<br/><i><b>E<sub>0</sub>·I<sub>0</sub>·(1/2)·cos(φ)·T =<br/>= E<sub>0</sub>·I<sub>0</sub>·(1/2)·cos(φ)·2π/ω</b></i><br/>and<br/><i><b><font size=5>∫</font><sub>[0,T]</sub>E<sub>0</sub>·I<sub>0</sub>·(1/2)·cos(2ωt+φ)·</b>d<b>t</b></i><br/>which is equal to zero because integral of a periodical function <i><b>cos(x)</b></i> over any argument interval that equals to one or more periods equals to zero.<br/>The same can be proven analytically<br/><i><b><font size=5>∫</font><sub>[0,T]</sub>cos(2ωt+φ)·</b>d<b>t =<br/>= sin(2ωt+φ)/(2ω)<font size=5>|</font><sub>[0,T]</sub> =<br/>= sin(2ω·2π/ω+φ)/(2ω) −<br/>− sin(φ)/(2ω) =<br/>= </b></i>[<b><i>sin(4π+φ)−sin(φ)</b></i>]<b><i>/(2ω) = 0</b></i><br/><br/>Hence, the energy consumed by a circuit during one period of oscillation equals to<br/><i><b>W<sub>[0,T]</sub> = E<sub>0</sub>·I<sub>0</sub>·(1/2)·cos(φ)·2π/ω</b></i><br/>The average power consumption, that is the average rate of consuming energy that we will call <b>effective power</b>, equals to this amount of energy divided by time, during which it was consumed - one period of oscillation <i><b>T=2π/ω</b></i>:<br/><i><b>P<sub>eff</sub> = W<sub>[0,T]</sub><fonr size=4>/</font>T = <br/>= E<sub>0</sub>·I<sub>0</sub>·(1/2)·cos(φ)</b></i><br/>Since <nobr><i><b>E<sub>eff</sub> = E<sub>0</sub> <font size=5>/</font>√<span style='text-decoration:overline'>2</span></b></i></nobr> and <i><nobr><b>I<sub>eff</sub> = I<sub>0</sub> <font size=5>/</font>√<span style='text-decoration:overline'>2</span></b></i></nobr>, the last expression for power equals to<br/><i><b>P<sub>eff</sub> = E<sub>eff</sub>·I<sub>eff</sub>·cos(φ)</b></i><br/>where a phase shift <i><b>φ</b></i> depends on resistance and reactances of a circuit as follows<br/><i><b>tan(φ) = (X<sub>C</sub> − X<sub>L</sub>) <font size=5>/</font>R</b></i><br/><i><b>X<sub>C</sub> = 1/(ωC)</b></i> - capacitive reactance,<br/><i><b>X<sub>L</sub> = ωL</b></i> - inductive reactance,<br/><i><b>R</b></i> - resistance.<br/>The above formula is derived for RLC-circuit that contains a resistor or resistance <i><b>R</b></i>, a capacitor of capacitance <i><b>C</b></i> and an inductor of inductance <i><b>L</b></i><br/>Let's analyze different circuits and their effective power consumption rate.<br/><br/><i>R-Circuit</i><br/>R-circuit contains only a resistor. Therefore, both reactances <i><b>X<sub>C</sub></b></i> and <i><b>X<sub>L</sub></b></i> are zero and phase shift <i><b>φ</b></i> is zero as well. Since <i><b>cos(0)=1</b></i>, the effective power for this R-circuit is<br/><i><b>P<sub>eff</sub> = E<sub>eff</sub>·I<sub>eff</sub></b></i><br/>which fully corresponds to a power for a circuit with a direct current running through it.<br/><br/><i>RC-Circuit</i><br/>RC-circuit contains a resistor and a capacitor in a series. Reactance <i><b>X<sub>L</sub></b></i> is zero.<br/><i><b>P<sub>eff</sub> = E<sub>eff</sub>·I<sub>eff</sub>·cos(φ)</b></i><br/>where <i><b>tan(φ) = X<sub>C</sub> <font size=5>/</font>R</b></i><br/><br/><i>RL-Circuit</i><br/>RC-circuit contains a resistor and an inductor in a series. Reactance <i><b>X<sub>C</sub></b></i> is zero.<br/><i><b>P<sub>eff</sub> = E<sub>eff</sub>·I<sub>eff</sub>·cos(φ)</b></i><br/>where <i><b>tan(φ) = −X<sub>L</sub> <font size=5>/</font>R</b></i><br/>The negative values of <i><b>tan(φ)</b></i> is not important since function <i><b>cos(φ)</b></i> is even and <i><b>cos(φ)=cos(−φ)</b></i>.<br/><br/>Interestingly, if our circuit contains a resistor, but a capacitor and an inductor are in <i>resonance</i>, that is <i><b>X<sub>C</sub>=X<sub>L</sub></b></i>, the phase shift will be equal to zero, as if only a resistor is present in a circuit.<br/><br/><i>L-, C- and LC-Circuits</i><br/>If no resistor is present in the circuit (assuming the resistance of wiring is zero), the denominator in the expression<br/><i><b>tan(φ) = (X<sub>C</sub> − X<sub>L</sub>) <font size=5>/</font>R</b></i><br/>is equal to zero.<br/>Therefore, the phase shift is <i>φ=π/2=90°</i>, <i><b>cos(π/2)=0</b></i> and the power consumption is zero. So, only resistors contribute to a power consumption. Inductors and capacitors are not consuming any energy, they only shift the current phase relatively to a generated EMF. And, if an inductor and a capacitor are in <i>resonance</i>, there is no phase shift, they neutralize each other.<br/><br/> Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-15489290731687648772020-11-16T10:00:00.000-08:002020-11-16T10:00:22.080-08:00RLC Circuit Ohm Law: UNIZOR.COM - Physics4Teens - Electromagnetism - AC ...<iframe width="480" height="270" src="https://www.youtube.com/embed/2B1vv6ET4Fc" frameborder="0"></iframe><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>RLC Circuit Ohm's Law</u><br /><br/>This lecture combines the material of the previous two ones dedicated to the Ohm's Law in alternating current circuits.<br/>The first one was analyzing a circuit with a resistor and a capacitor.<br/>The second one analyzed a circuit with a resistor and an inductor.<br/>This lecture analyzes a circuit with all three elements - resistor, inductor and capacitor.<br/><br/>Consider a closed AC circuit with a generator with no internal resistance producing a sinusoidal electromotive force (<i>EMF</i> or <i>voltage</i>)<br/><i><b>E(t)=E<sub>0</sub>·sin(ωt)</b>(volt)</b></i><br/>where <i><b>E<sub>0</sub></b></i> is a peak EMF produced by an AC generator,<br/><i><b>ω</b></i> is angular speed of EMF oscillations and<br/><i><b>t</b></i> is time.<br/><img src='http://www.unizor.com/Pictures/RLC_circuit.png' style='width:200px;height:200px;'><br/>In this circuit we have a resistor with resistance <i><b>R</b></i>, an inductor with inductance <i><b>L</b></i> and a capacitor with capacitance <i><b>C</b></i> connected in series with EMF generator. This is RLC-circuit.<br/>Since the circuit is closed, the electric current <i><b>I(t)</b></i> going through a circuit is the same for all components of a circuit.<br/><br/>Assume that the voltage drop on a resistor (caused by its <i>resistance <b>R</b></i>) is <i><b>V<sub><font size=1>R</font></sub>(t)</b></i>, the voltage drop on an inductor (caused by its <i>inductive reactance <b>X<sub><font size=1>L</font></sub>=ω·L</b></i>) is <i><b>V<sub><font size=1>L</font></sub>(t)</b></i> and the voltage drop on a capacitor (caused by its <i>capacitive reactance <b>X<sub><font size=1>C</font></sub>=1/(ω·C)</b></i>) is <i><b>V<sub><font size=1>L</font></sub>(t)</b></i>.<br/><br/>Since a resistor, an inductor and a capacitor are connected in a series, the sum of these voltage drops should be equal to a generated EMF <i><b>E(t)</b></i>:<br/><i><b>E(t) = V<sub><font size=1>R</font></sub>(t) + V<sub><font size=1>L</font></sub>(t) + V<sub><font size=1>C</font></sub>(t)</b></i><br/><br/>As explained in the previous lectures the Ohm's Law localized around a resistor states that<br/><i><b>I(t) = V<sub><font size=1>R</font></sub>(t)<font size=4>/</font>R</b></i> or<br/><i><b>V<sub><font size=1>R</font></sub>(t) = I(t)·R</b></i><br/><br/>The inductance of an inductor <i><b>L</b></i>, voltage drop on this inductor caused by a self-induction effect <i><b>V<sub><font size=1>L</font></sub>(t)</b></i>, electromagnetic flux going through an inductor <i><b>Φ(t)</b></i> and electric current going through it <i><b>I(t)</b></i>, as explained in the lecture "<i>AC Inductors</i>" of the chapter "<i>Electromagnetism - Alternating Current Induction</i>", are in a relationship<br/><i><b>V<sub><font size=1>L</font></sub> = </b>d<b>Φ/</b>d<b>t = L·</b>d<b>I(t)/</b>d<b>t </b></i><br/><br/>The capacity of a capacitor <i><b>C</b></i>, voltage drop on this capacitor <i><b>V<sub><font size=1>C</font></sub>(t)</b></i> and electric charge accumulated on its plates <i><b>Q<sub><font size=1>C</font></sub>(t)</b></i>, according to a definition of a <i>capacitance <b>C</b></i>, are in a relationship<br/><i><b>C = Q<sub><font size=1>C</font></sub>(t)<font size=4>/</font>V<sub><font size=1>C</font></sub>(t)</b></i> or<br/><i><b>V<sub><font size=1>C</font></sub>(t) = Q<sub><font size=1>C</font></sub>(t)<font size=4>/</font>C</b></i><br/><br/>We have expressed both voltage drops <i><b>V<sub><font size=1>R</font></sub>(t)</b></i> and <i><b>V<sub><font size=1>L</font></sub>(t)</b></i> in terms of an electric current <i><b>I(t)</b></i>:<br/><i><b>V<sub><font size=1>R</font></sub>(t) = I(t)·R</b></i><br/><i><b>V<sub><font size=1>L</font></sub>(t) = L·I'(t)</b></i><br/>Since an electric charge <i><b>Q<sub><font size=1>C</font></sub>(t)</b></i> is also involved to express the voltage drop on a capacitor, we will use this electric charge as a main variable, using the definition of an electric current as a rate of change of electric charge<br/><i><b>I(t) = Q<sub><font size=1>C</font></sub>'(t)</b></i> and<br/><i><b>I'(t) = Q<sub><font size=1>C</font></sub>"(t)</b></i><br/><br/>Now all three voltage drops can be expressed as functions of <i><b>Q<sub><font size=1>C</font></sub>(t)</b></i> and, equating their sun to a generated EMF <i><b>E(t)</b></i>, we can the following differential equation<br/><i><b>E<sub>0</sub>·sin(ωt) = Q<sub><font size=1>C</font></sub>'(t)·R +<br/>+ L·Q<sub><font size=1>C</font></sub>"(t) + Q<sub><font size=1>C</font></sub>(t)<font size=4>/</font>C =<br/>= L·Q<sub><font size=1>C</font></sub>"(t) + R·Q<sub><font size=1>C</font></sub>'(t) +<br/>+ (1/C)·Q<sub><font size=1>C</font></sub>(t)</b></i><br/><br/>As we know, capacitive and inductive reactance are functionally equivalent to resistance, and they all have the same unit of measurement - <i>ohm</i>. Therefore, it's convenient, instead of capacitance <i><b>C</b></i> and inductance <i><b>L</b></i>, to use corresponding reactance <i><b>X<sub><font size=1>C</font></sub>=1/(ω·C)</b></i> and <i><b>X<sub><font size=1>L</font></sub>=ω·L</b></i>.<br/>So, we will substitute<br/><i><b>C = 1/(ω·X<sub><font size=1>C</font></sub>)</b></i><br/><i><b>L = X<sub><font size=1>L</font></sub>/ω</b></i><br/><br/>Substitute Q<sub><font size=1>C</font></sub>(t)=y(t) for brevity. Then our equation looks like<br/><i><b>(X<sub><font size=1>L</font></sub>/ω)·y"(t) + R·y'(t) +<br/>+ ω·X<sub><font size=1>C</font></sub>·y(t) = E<sub>0</sub>·sin(ωt)</b></i><br/><br/>Solving this equation for <i><b>y(t)=Q<sub><font size=1>C</font></sub>(t)</b></i> and differentiating it by time <i><b>t</b></i>, we will obtain the expression for an electric current <i><b>I(t)</b></i> in this circuit as a function of all given parameters and time.<br/><br/>First of all, let's find a particular solution of this differential equation.<br/>The form of the right side of this equation prompts to look for a solution in trigonometric form<br/><i><b>y(t) = F·sin(ωt) + G·cos(ωt)</b></i><br/>Then<br/><i><b>y'(t) =<br/>= ω·</b></i>[<i><b>F·cos(ωt)−G·sin(ωt)</b></i>]<i><b></b></i><br/><i><b>y"(t) =<br/>= −ω²·</b></i>[<i><b>F·sin(ωt)+G·cos(ωt)</b></i>]<i><b> = <br/>= −ω²·y(t)</b></i><br/><br/>Using this trigonometric representation of potential solution <i><b>y(t)</b></i>, the left side of our equation is<br/><i><b>(X<sub><font size=1>L</font></sub>/ω)·y"(t) + R·y'(t) +<br/>+ ω·X<sub><font size=1>C</font></sub>·y(t) =<br/>= −(X<sub><font size=1>L</font></sub>/ω)·ω²·y(t) + R·y'(t) +<br/>+ ω·X<sub><font size=1>C</font></sub>·y(t) =<br/>= −ω·X<sub><font size=1>L</font></sub>·y(t) + R·y'(t) +<br/>+ ω·X<sub><font size=1>C</font></sub>·y(t) =<br/>= R·y'(t) + ω·(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)·y(t) =<br/>= R·ω·</b></i>[<i><b>F·cos(ωt)−G·sin(ωt)</b></i>]<i><b> +<br/>+ ω·(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)·</b></i>[<i><b>F·sin(ωt)+<br/>+G·cos(ωt)</b></i>]<i><b> =<br/>= ω·</b></i>[<i><b>(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)·F−R·G</b></i>]<i><b>·sin(ωt)+<br/>+ω·</b></i>[<i><b>R·F+(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)·G</b></i>]<i><b>·cos(ωt)</b></i><br/><br/>Since this is supposed to be equal to <i><b>E<sub>0</sub>·sin(ωt)</b></i>, we have the following system of two linear equations with two variables <i><b>F</b></i> and <i><b>G</b></i><br/><i><b>ω·</b></i>[<i><b>(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)·F−R·G</b></i>]<i><b> = E<sub>0</sub></b></i><br/><i><b>ω·</b></i>[<i><b>R·F+(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)·G</b></i>]<i><b> = 0</b></i><br/>or<br/><i><b>(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)·F−R·G = E<sub>0</sub><font size=4>/</font>ω</b></i><br/><i><b>R·F+(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)·G = 0</b></i><br/><br/>Determinant of the matrix that defines this system is<br/><i><b>(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)²+R²</b></i>.<br/>It's always positive and usually is denoted as<br/><i><b>Z² = (X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)²+R²</b></i><br/>The value <i><b>Z</b></i> is called the <b>impedance</b> of a circuit and, as we will see, plays the role of a resistance for an entire RLC-circuit.<br/><br/>This system of two linear equations with two variables has a solution:<br/><i><b>F = (E<sub>0</sub>/ω)·(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)<font size=4>/</font>Z²</b></i><br/><i><b>G = −(E<sub>0</sub>/ω)·R<font size=4>/</font>Z²</b></i><br/><br/>Consider two expressions that participate in the above solutions <i><b>F</b></i> and <i><b>G</b></i>:<br/><i><b>(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)<font size=4>/</font>Z²</b></i> and <i><b>R<font size=4>/</font>Z²</b></i><br/>Since <i><b>Z² = (X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)²+R²</b></i>, we can always find an angle <i><b>φ</b></i> such that<br/><i><b>(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)<font size=4>/</font>Z = sin(φ)</b></i> and<br/><i><b>R<font size=4>/</font>Z = cos(φ)</b></i><br/><br/>Using this, we express the solutions to the above system as<br/><i><b>F = (E<sub>0</sub>/(ω·Z))·sin(φ)</b></i><br/><i><b>G = −(E<sub>0</sub>/(ω·Z))·cos(φ)</b></i><br/><br/>Now the solution of the differential equation for <i><b>y(t)=Q<sub><font size=1>C</font></sub>(t)</b></i> that we were looking for in a format<br/><i><b>y(t) = F·sin(ωt) + G·cos(ωt)</b></i><br/>looks like<br/><i><b>y(t)=(E<sub>0</sub>/(ω·Z))·sin(φ)sin(ωt) −<br/>− (E<sub>0</sub>/(ω·Z))·cos(φ)cos(ωt) =<br/>= −(E<sub>0</sub>/(ω·Z))·cos(ωt+φ)</b></i><br/><br/>As we noted before, differentiation of this function gives the electric current in the circuit <i><b>I(t)</b></i>:<br/><i><b>I(t) = y'(t) =<br/>= (E<sub>0</sub> <font size=4>/</font>(ω·Z))·ω·sin(ωt+φ) =<br/>= (E<sub>0</sub> <font size=4>/</font>Z)·sin(ωt+φ) =<br/>= I<sub>0</sub>·sin(ωt+φ)</b></i><br/>where <i><b>I<sub>0</sub>=E<sub>0</sub> <font size=4>/</font>Z</b></i> is an equivalent of the <b>Ohm's Law for an RLC-circuit</b>.<br/><br/>Similar relationship exists between effective voltage and effective current<br/><i><b>I<sub>eff</sub> = I<sub>0</sub> <font size=4>/</font>√<span style='text-decoration:overline'>2</span> =<br/>= E<sub>0</sub> <font size=4>/</font>(√<span style='text-decoration:overline'>2</span>·Z) = E<sub>eff</sub> <font size=4>/</font>Z</b></i><br/><br/>Here <b>impedance</b> <i><b>Z</b></i>, defined by resistance <i><b>R</b></i>, inductive reactance <i><b>X<sub><font size=1>L</font> </sub></b></i> and capacitive reactance <i><b>X<sub><font size=1>C</font> </sub></b></i> as<br/><i><b>Z = √<span style='text-decoration:overline'>(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)²+R²</span></b></i><br/>plays a role of a resistance in the RLC-circuit.<br/><br/>There is a phase shift <i><b>φ</b></i> of the electric current oscillations relative to EMF. It is also defined by the same characteristics of an RLC-circuit:<br/><i><b>(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)<font size=4>/</font>Z = sin(φ)</b></i><br/><i><b>R<font size=4>/</font>Z = cos(φ)</b></i><br/>Hence<br/><i><b>tan(φ) = (X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)<font size=4>/</font>R</b></i><br/>This makes phase shift positive or negative depending on the values of capacitive and inductive reactance.<br/>If <i><b>X<sub><font size=1>C</font></sub></b></i> is greater than <i><b>X<sub><font size=1>L</font></sub></b></i>, the phase shift is positive, if the reverse is true, the phase shift is negative.<br/>If the values of reactance are the same, that is <i><b>X<sub><font size=1>C</font></sub>=X<sub><font size=1>L</font></sub></b></i>, there is no phase shift. From definition of reactance, it happens when<br/><i><b>1<font size=4>/</font>(ω·C) = ω·L</b></i> or<br/><i><b>1<font size=4>/</font>(L·C) = ω²</b></i> or<br/><i><b>L·C = 1<font size=4>/</font>ω²</b></i><br/>This relationship between inductance, capacitance and angular speed of EMF oscillation is call <i>resonance</i>.<br/><br/>Expressions for electric current<br/><i><b>I(t) = (E<sub>0</sub> <font size=4>/</font>Z)·sin(ωt+φ) =<br/>= I<sub>0</sub>·sin(ωt+φ)</b></i><br/>in RLC-circuit, where impedance <i><b>Z=√<span style='text-decoration:overline'>(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)²+R²</span></b></i> and <i><b>tan(φ)=(X<sub><font size=1>C</font></sub>−X<sub><font size=1>L</font></sub>)<font size=4>/</font>R</b></i>, correspond to analogous formulas for R-, RC- and RL-circuits presented in the previous lectures. All it takes is to set the appropriate value of capacitance or inductance to zero.<br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-34796409266173961792020-11-10T09:06:00.003-08:002020-11-10T09:10:21.200-08:00Ohm's Law for RL_Circuit: UNIZOR.COM - Physics4Teens - Electromagnetism ...<iframe width="480" height="270" src="https://www.youtube.com/embed/pw31hSMBolw" frameborder="0"></iframe><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>RL Circuit Ohm's Law</u><br /><br/>Consider a closed AC circuit with a generator with no internal resistance producing a sinusoidal electromotive force (<i>EMF</i> or <i>voltage</i>)<br/><i><b>E(t)=E<sub>0</sub>·sin(ωt)</b>(volt)</b></i><br/>where <i><b>E<sub>0</sub></b></i> is a peak EMF produced by an AC generator,<br/><i><b>ω</b></i> is angular speed of EMF oscillations and<br/><i><b>t</b></i> is time.<br/>In this circuit we have a resistor with resistance <i><b>R</b></i> and an inductor with inductance <i><b>L</b></i> connected in series with EMF generator. This is RL-circuit.<br/>Since the circuit is closed, the electric current <i><b>I(t)</b></i> going through a circuit is the same for all components of a circuit.<br/><br/>Assume that the voltage drop on a resistor caused by its resistance is <i><b>V<sub>R</sub>(t)</b></i> and the voltage drop on an inductor caused by self-induction effect is <i><b>V<sub>L</sub>(t)</b></i>. Since a resistor and an inductor are connected in a series, the sum of these voltage drops should be equal to a generated EMF <i><b>E(t)</b></i>:<br/><i><b>E(t) = V<sub>R</sub>(t) + V<sub>L</sub>(t)</b></i><br/><br/>As explained in the previous lecture for an R-circuit, the Ohm's Law localized around a resistor states that<br/><i><b>I(t) = V<sub>R</sub>(t)<font size=4>/</font>R</b></i> or<br/><i><b>V<sub>R</sub>(t) = I(t)·R</b></i><br/><br/>The inductance of an inductor <i><b>L</b></i>, voltage drop on this inductor caused by a self-induction effect <i><b>V<sub>L</sub>(t)</b></i>, electromagnetic flux going through an inductor <i><b>Φ(t)</b></i> and electric current going through it <i><b>I(t)</b></i>, as explained in the lecture "<i>AC Inductors</i>" of the chapter "<i>Electromagnetism - Alternating Current Induction</i>", are in a relationship<br/><i><b>V<sub>L</sub> = </b>d<b>Φ/</b>d<b>t = L·</b>d<b>I(t)/</b>d<b>t </b></i><br/><br/>We have expressed both voltage drops <i><b>V<sub>R</sub>(t)</b></i> and <i><b>V<sub>L</sub>(t)</b></i> in terms of an electric current <i><b>I(t)</b></i>:<br/><i><b>V<sub>R</sub>(t) = I(t)·R</b></i><br/><i><b>V<sub>L</sub>(t) = L·I'(t)</b></i><br/><br/>Now we can substitute them into a formula for their sum being equal to a generated EMF<br/><i><b>E(t) = V<sub>R</sub>(t) + V<sub>L</sub>(t) =<br/>= I(t)·R + I'(t)·L</b></i><br/><br/>Since <i><b>E(t)=E<sub>0</sub>·sin(ωt)</b></i> we should solve the following differential equation to obtain function <i><b>I(t)</b></i><br/><i><b>E<sub>0</sub>·sin(ωt) = I(t)·R + I'(t)·L</b></i><br/><br/>Let's divide this equation by <i><b>L</b></i> and use simplified notation for brevity:<br/><i><b>y(t) = I(t)</b></i><br/><i><b>a = R<font size=4>/</font>L</b></i><br/><i><b>b = E<sub>0</sub>/</font>L</b></i><br/>Then our equation looks simpler<br/><i><b>y'(t)+a·y(t) = b·sin(ωt)</b></i><br/><br/>This exact differential equation was solved in the previous lecture dedicated to RC-circuit, notes for that lecture contain detail analysis of its solution<br/><i><b>y(t) = −b·cos(ωt+ψ)/√<span style='text-decoration:overline'>(a²+ω²)</span> + K</b></i><br/>where phase shift <i><b>ψ=arctan(a/ω)</b></i> and <i><b>K</b></i> is a constant that can be determined by initial condition.<br/><br/>Using original notation,<br/><i><b>I(t) = −(E<sub>0</sub>/</font>L)·cos(ωt+ψ)<font size=4>/</font><br/><font size=4>/</font>√<span style='text-decoration:overline'>(R/L)²+ω²</span> + K =<br/>= −E<sub>0</sub>·cos(ωt+ψ)<font size=4>/</font><br/><font size=4>/</font>√<span style='text-decoration:overline'>R²+(ω·L)²</span> + K =<br/>= −E<sub>0</sub>·cos(ωt+ψ)<font size=4>/</font>√<span style='text-decoration:overline'>R²+X<sub>L</sub>²</span> + K</b></i><br/>where <i><b>X<sub>L</sub>=ω·L</b></i> is <i>inductive reactance</i> of an inductor - a characteristic of an inductor functionally equivalent to a resistance for a resistor and measured in the same units - <i>ohm</i> and<br/><i><b>tan(ψ) = a/ω = R/(L·ω) = R/X<sub>L</sub></b></i><br/><br/>Let's apply some Trigonometry to simplify the above formula.<br/><i><b>−cos(ωt+ψ) =<br/>= −sin((π/2)−ωt−ψ) =<br/>= −sin((π/2)−ψ−ωt) =<br/>= sin(ωt−((π/2)−ψ)) =<br/>= sin(ωt−φ)</b></i><br/>where <i><b>φ=(π/2)−ψ</b></i> and, therefore, <i><b>tan(φ)=1/tan(ψ)=X<sub>L</sub>/R</b></i>.<br/><br/>Using phase shift <i><b>φ</b></i>, the equation for the current <i><b>I(t)</b></i> looks like<br/><i><b>I(t) = E<sub>0</sub>·sin(ωt−φ)<font size=4>/</font>√<span style='text-decoration:overline'>R²+X<sub>L</sub>²</span> + K</b></i><br/>where <i><b>tan(φ)=X<sub>L</sub>/R</b></i>.<br/><br/>The value of a constant <i><b>K</b></i> can be defined by initial conditions. Since we don't really know these conditions (like what is the value of a current at some moment in time), traditionally this constant is assigned the value of zero, motivating it by the fact that, if the generated EMF is oscillating between its minimum and maximum of the same magnitude with different signs, the current also will be "symmetrical" relative to zero level, which requires the value <i><b>K=0</b></i><br/><br/>The final version of the current in this RL-circuit is<br/><i><b>I(t) = E<sub>0</sub>·sin(ωt−φ)<font size=4>/</font>√<span style='text-decoration:overline'>R²+X<sub>L</sub>²</span> =<br/>= I<sub>0</sub>·sin(ωt−φ)</b></i><br/>where <i><b>I<sub>0</sub> = E<sub>0</sub><font size=4>/</font>√<span style='text-decoration:overline'>R²+X<sub>L</sub>²</span></b></i><br/><br/>The last issue is to analyze the <i>effective current</i> in this RL-circuit. Since for sinusoidal oscillations the effective current is by √<span style='text-decoration:overline'>2</span> less than peak amperage, the effective current is<br/><i><b>I<sub>eff</sub> = I<sub>0</sub> <font size=4>/</font>√<span style='text-decoration:overline'>2</span> =<br/>= E<sub>0</sub> <font size=4>/</font>(√<span style='text-decoration:overline'>X<sub>L</sub>²+R²</span>·√<span style='text-decoration:overline'>2</span>) =<br/>= E<sub>eff</sub> <font size=4>/</font>√<span style='text-decoration:overline'>X<sub>L</sub>²+R²</span></b></i><br/><b>This is the Ohm's Law for effective voltage and amperage in RC-circuit</b>.<br/><br/>It's important to notice that in the absence of a resistor (that is, <i><b>R=0</b></i>) the formula for <i><b>I(t)</b></i> transforms into<br/><i><b>I(t) = I<sub>0</sub>·sin(ωt−φ)</b></i><br/>where peak amperage <i><b>I<sub>0</sub>=E<sub>0</sub>/X<sub>L</sub></b></i> and phase shift <i><b>φ=arctan(∞)=π/2</b></i>, which corresponds to results obtained in a lecture "<i>AC Inductors</i>" of the "<i>Alternating Current Induction</i>" chapter of this course dedicated to only an inductor in the AC circuit, taking the current as given and deriving the EMF.<br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-45010285380571765202020-11-07T08:11:00.004-08:002020-11-10T09:11:25.120-08:00Ohm's Law for R- and RC-Circuits: UNIZOR.COM - Physics4Teens - Electromagnetism - AC O...<iframe width="480" height="270" src="https://www.youtube.com/embed/TJvdM-5k0Ek" frameborder="0"></iframe><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>R and RC Circuit Ohm's Law</u><br /><br/>In this chapter we will examine different aspects of the Ohm's Law as they occur in different alternating current (AC) circuits.<br/>Four different types of AC circuits will be considered in this and subsequent lectures:<br/>(a) R-circuit that contains only a resistor;<br/>(b) RC-circuit that contains a resistor and a capacitor;<br/>(c) RL-circuit that contains a resistor and an inductor;<br/>(d) RLC-circuit that contains a resistor, an inductor and a capacitor.<br/>The first two are subject to this lecture.<br/><br/><br/><i>R-circuit</i><br/><br/>Consider a closed AC circuit with a generator with no internal resistance producing a sinusoidal electromotive force (<i>EMF</i> or <i>voltage</i>)<br/><i><b>E(t)=E<sub>0</sub>·sin(ωt)</b>(volt)</b></i><br/>where <i><b>E<sub>0</sub></b></i> is a peak EMF produced by an AC generator,<br/><i><b>ω</b></i> is angular speed of EMF oscillations<br/>and <i><b>t</b></i> is time.<br/>In this circuit there is only a resistor with resistance <i><b>R</b>(ohm)</b></i>.<br/><br/>During any infinitesimal time interval the electric current going through a circuit depends only on the generated EMF and a resistance of a circuit components. Since the only resistive component is a resistance <i><b>R</b></i>, the electric current at any moment of time will obey the Ohm's Law for direct current.<br/><br/>Therefore the electric current <i><b>I(t)</b></i> in this R-circuit is<br/><i><b>I(t) = E(t)/R = E<sub>0</sub>·sin(ωt)/R =<br/>= (E<sub>0</sub>/R)·sin(ωt) = I<sub>0</sub>·sin(ωt)</b></i><br/>where <i><b>I<sub>0</sub> = E<sub>0</sub>/R</b></i> is a peak current.<br/>Now we can express the Ohm's Law for peak voltage and peak amperage in the R-circuit as<br/><i><b>I<sub>0</sub> = E<sub>0</sub> <font size=4>/</font>R</b></i><br/><br/>Since in many cases we are interested in <i>effective</i> voltage and <i>effective</i> electric current, and knowing that<br/><i><b>E<sub>eff</sub> = E<sub>0</sub> <font size=4>/</font>√<span style='text-decoration:overline'>2</span></b></i> and <i><b>I<sub>eff</sub> = I<sub>0</sub> <font size=4>/</font>√<span style='text-decoration:overline'>2</span></b></i>,<br/>the Ohm's Law for effective voltage and effective amperage in the R-circuit can be expressed in a form<br/><i><b>I<sub>eff</sub> = E<sub>eff</sub> <font size=4>/</font>R</b></i><br/><br/><br/><i>RC-circuit</i><br/><br/>Let's add a capacitor with capacity <i><b>C</b></i> (<i>farad</i>) to R-circuit, connecting it in series with a resistor. This is RC-circuit.<br/>As we know, alternating current goes through a capacitor, so we have a closed circuit of resistor <i><b>R</b></i> and capacitor <i><b>C</b></i> connected in a series with EMF generator that produces sinusoidal voltage <i><b>E(t)=E<sub>0</sub>·sin(ωt)</b></i>.<br/>The electric current <i><b>I(t)</b></i> going through a circuit is the same for all components of a circuit.<br/><br/>Assume that the voltage drop on a resistor is <i><b>V<sub>R</sub>(t)</b></i> and the voltage drop on a capacitor is <i><b>V<sub>C</sub>(t)</b></i>. Since a resistor and a capacitor are connected in a series, the sum of these voltage drops should be equal to a generated EMF <i><b>E(t)</b></i>:<br/><i><b>E(t) = V<sub>R</sub>(t) + V<sub>C</sub>(t)</b></i><br/><br/>As explained above for an R-circuit, the Ohm's Law localized around a resistor states that<br/><i><b>I(t) = V<sub>R</sub>(t)<font size=4>/</font>R</b></i> or<br/><i><b>V<sub>R</sub>(t) = I(t)·R</b></i><br/><br/>The capacity of a capacitor <i><b>C</b></i>, voltage drop on this capacitor <i><b>V<sub>C</sub>(t)</b></i> and electric charge accumulated on its plates <i><b>Q<sub>C</sub>(t)</b></i>, according to a definition of a <i>capacity</i>, are in a relationship<br/><i><b>C = Q<sub>C</sub>(t)<font size=4>/</font>V<sub>C</sub>(t)</b></i> or<br/><i><b>V<sub>C</sub>(t) = Q<sub>C</sub>(t)<font size=4>/</font>C</b></i><br/><br/>At the same time, by definition of the electric current, the electric current going through a capacitor is just a rate of change of electrical charge on its plates, that is, a derivative of an accumulated electric charge by time:<br/><i><b>I(t) = Q'<sub>C</sub>(t)</b></i><br/><br/>This is the same electric current that goes through a resistor, since the circuit is closed. Therefore, we can substitute this expression of an electric current into a formula for a voltage drop on a resistor<br/><i><b>V<sub>R</sub>(t) = Q'<sub>C</sub>(t)·R</b></i><br/><br/>We have expressed both voltage drops <i><b>V<sub>R</sub>(t)</b></i> and <i><b>V<sub>C</sub>(t)</b></i> in terms of an electric charge <i><b>Q<sub>C</sub>(t)</b></i> accumulated on the plates of a capacitor:<br/><i><b>V<sub>R</sub>(t) = Q'<sub>C</sub>(t)·R</b></i><br/><i><b>V<sub>C</sub>(t) = Q<sub>C</sub>(t)<font size=4>/</font>C</b></i><br/><br/>Now we can substitute them into a formula for their sum being equal to a generated EMF<br/><i><b>E(t) = V<sub>R</sub>(t) + V<sub>C</sub>(t) =<br/>= Q'<sub>C</sub>(t)·R + Q<sub>C</sub>(t)<font size=4>/</font>C</b></i><br/><br/>Since <i><b>E(t)=E<sub>0</sub>·sin(ωt)</b></i> we should solve the following differential equation to obtain function <i><b>Q<sub>C</sub>(t)</b></i><br/><i><b>E<sub>0</sub>·sin(ωt) =<br/>= Q'<sub>C</sub>(t)·R + Q<sub>C</sub>(t)<font size=4>/</font>C</b></i><br/><br/>Let's divide this equation by <i><b>R</b></i> and use simplified notation for brevity:<br/><i><b>y(t) = Q<sub>C</sub>(t)</b></i><br/><i><b>a = 1<font size=4>/</font>(R·C)</b></i><br/><i><b>b = E<sub>0</sub>/</font>R</b></i><br/>Then our equation looks simpler<br/><i><b>y'(t)+a·y(t) = b·sin(ωt)</b></i><br/><br/>Let's discuss how to solve (integrate) this differential equation.<br/>Recall the formula for a derivative of a product of two functions<br/><i><b>(x(t)·y(t))' =<br/>= x(t)·y'(t) + x'(t)·y(t)</b></i><br/><br/>Let's find such function <i><b>x(t)</b></i> that, if we use it as a multiplier of our differential equation, the left side will look like a complete derivative of <i><b>x(t)·y(t)</b></i>.<br/><br/>Multiplied by this function <i><b>x(t)</b></i>, our equation looks like this:<br/><i><b>x(t)·y'(t)+a·x(t)·y(t) =<br/>= b·x(t)·sin(ωt)</b></i><br/>Let's find function <i><b>x(t)</b></i> such that <i><b>a·x(t)</b></i> is <i><b>x'(t)</b></i>. Then the left side of the equation will be equal to <i><b>(x(t)·y(t))'</b></i>.<br/><br/>Obvious choice for such function is <i><b>e<sup>a·t</sub></b></i>. It's an intelligent guess, but it can be determined analytically.<br/>Indeed,<br/><i><b>a·x(t) = x'(t)</b></i><br/><i><b>a·x(t) = </b>d<b>x/</b>d<b>t</b></i><br/><i><b>a·</b>d<b>t = </b>d<b>x/x(t)</b></i><br/><i><b></b>d<b>(a·t) = </b>d<b>ln(x(t))</b></i><br/><i><b>a·t = ln(x(t)) + K<sub>1</sub></b></i><br/><i><b>x(t) = K<sub>2</sub>·e<sup>a·t</sub></b></i><br/>Here <i><b>K<sub>1</sub></b></i> and <i><b>K<sub>2</sub></b></i> are any constants, so let's use <i><b>K<sub>2</sub>=1</b></i>.<br/>Then <i><b>x(t)=e<sup>a·t</sup></b></i>.<br/><br/>Now the equation takes the following form<br/><i><b>e<sup>a·t</sup>·y'(t)+a·e<sup>a·t</sup>·y(t) =<br/>= b·e<sup>a·t</sup>·sin(ωt)</b></i><br/>which is equivalent to<br/><i><b></b></i>[<i><b>e<sup>a·t</sup>·y(t)</b></i>]<i><b>' = b·e<sup>a·t</sup>·sin(ωt)</b></i><br/><br/>The above equation should be integrated to get <i><b>e<sup>a·t</sup>·y(t)</b></i> and, finally <i><b>y(t)</b></i>.<br/><br/>The indefinite integral of the left side is <i><b>e<sup>a·t</sup>·y(t)</b></i>.<br/><br/>The indefinite integral of the right side can be found straight forward using Euler formula<br/><i><b>cos(φ)+i·sin(φ) = e<sup>i·φ</sup></b></i><br/>and following from it expressions for trigonometric functions<br/><i><b>cos(φ) = (e<sup>i·φ</sup>+e<sup>−i·φ</sup>)<font size=4>/</font>2</b></i><br/><i><b>sin(φ) = (e<sup>i·φ</sup>−e<sup>−i·φ</sup>)<font size=4>/</font>(2i)</b></i><br/><br/>The result of integration of the right side (without multiplier <i><b>b</b></i>) is<br/><i><b><font size=5>∫</font>e<sup>a·t</sup>·sin(ωt)·</b>d<b>t =<br/>= </b></i>[<i><b>e<sup>a·t</sup>/(a²+ω²)</b></i>]<i><b>·<br/>·</b></i>[<i><b>a·sin(ωt)−ω·cos(ωt)</b></i>]<i><b> + K<sub>3</sub></b></i><br/>(<i><b>K<sub>3</sub></b></i> is any constant)<br/><br/>Therefore,<br/><i><b>e<sup>a·t</sup>·y(t) = b·</b></i>[<i><b>e<sup>a·t</sup>/(a²+ω²)</b></i>]<i><b>·<br/>·</b></i>[<i><b>a·sin(ωt)−ω·cos(ωt)</b></i>]<i><b> + K<sub>3</sub></b></i><br/><br/>Reducing by <i><b>e<sup>a·t</sup></b></i> both sides, the expression for <i><b>y(t)=Q<sub>C</sub>(t)</b></i> looks like<br/><i><b>y(t) = </b></i>[<i><b>b/(a²+ω²)</b></i>]<i><b>·<br/>·</b></i>[<i><b>a·sin(ωt)−ω·cos(ωt)</b></i>]<i><b> + K<sub>4</sub></b></i><br/><br/>Consider two constants <i><b>a/√<span style='text-decoration:overline'>(a²+ω²)</span></b></i> and <i><b>ω/√<span style='text-decoration:overline'>(a²+ω²)</span></b></i>. It is convenient to represent them as <i><b>sin(φ)</b></i> and <i><b>cos(φ)</b></i> correspondingly for some angle <i><b>φ=arctan(a/ω)</b></i>.<br/>Then<br/><i><b>a·sin(ωt)−ω·cos(ωt) = <br/>= √<span style='text-decoration:overline'>(a²+ω²)</span>·<br/>·</b></i>[<i><b>sin(φ)·sin(ωt)−cos(φ)·cos(ωt)</b></i>]<br/><br/>The last expression equals to<br/><i><b>−√<span style='text-decoration:overline'>(a²+ω²)</span>·cos(ωt+φ)</b></i><br/>and our equation for <i><b>y(t)</b></i> looks like<br/><i><b>y(t) = −b·cos(ωt+φ)/√<span style='text-decoration:overline'>(a²+ω²)</span> +<br/>+ K<sub>4</sub></b></i><br/><br/>Since <i><b>y(t)</b></i> is the electric charge <i><b>Q<sub>C</sub>(t)</b></i> accumulated on the plates of a capacitor, its derivative is an electric current in the circuit:<br/><i><b>I(t) = b·ω·sin(ωt+φ)/√<span style='text-decoration:overline'>(a²+ω²)</span></b></i><br/><br/>Let's restore this equation to original constants<br/><i><b>a = 1<font size=4>/</font>(R·C)</b></i><br/><i><b>b = E<sub>0</sub>/</font>R</b></i><br/>The factor at <i><b>sin(ωt+φ)</b></i> in the equation above then is<br/><i><b>b·ω/√<span style='text-decoration:overline'>(a²+ω²)</span> =<br/>= E<sub>0</sub>·ω/</font>(R·√<span style='text-decoration:overline'>1/(R·C)²+ω²</span>) =<br/>= E<sub>0</sub>/</font>(√<span style='text-decoration:overline'>1/(ωC)²+R²</span>) =<br/>= E<sub>0</sub>/</font>(√<span style='text-decoration:overline'>X<sub>C</sub>²+R²</span>)</b></i><br/>where <i><b>X<sub>C</sub>=1/(ωC)</b></i> was defined in one of the previous lectures as <i>reactance</i> of a capacitor (or <i>capacitive reactance</i>).<br/><br/>The angle <i><b>φ=arctan(a/ω)</b></i> in terms of original constants is<br/><i><b>φ=arctan(1<font size=4>/</font>(R·C·ω)) =<br/>= arctan (X<sub>C</sub> <font size=4>/</font> R)</b></i><br/>This angle is called <i>phase shift</i> of the current from the voltage.<br/><br/>Now we have a formula for an electric current in the RC-cirucit that connects generated EMF, resistance of a resistor and capacity of a capacitor<br/><i><b>I(t) = E<sub>0</sub>·sin(ωt+φ)/√<span style='text-decoration:overline'>X<sub>C</sub>²+R²</span> =<br/>= I<sub>0</sub>·sin(ωt+φ)</b></i><br/>where <i><b>I<sub>0</sub>=E<sub>0</sub>/√<span style='text-decoration:overline'>X<sub>C</sub>²+R²</span></b></i> is the <i>peak amperage</i> of the electric current.<br/><b>This is the Ohm's Law for RC-circuit</b>.<br/><br/>The expression <i><b>√<span style='text-decoration:overline'>X<sub>C</sub>²+R²</span></b></i> plays in this case the same role as a resistance in case of direct current circuits.<br/><br/>It is important that there is a <i>phase shift <b>φ</b></i> of the oscillations of an electric current relatively to oscillation of generated EMF.<br/><br/>The last issue is to analyze the <i>effective current</i> in this RC-circuit. Since for sinusoidal oscillations the effective current is by √<span style='text-decoration:overline'>2</span> less than peak amperage, the effective current is<br/><i><b>I<sub>eff</sub> = I<sub>0</sub> <font size=4>/</font>√<span style='text-decoration:overline'>2</span> =<br/>= E<sub>0</sub> <font size=4>/</font>(√<span style='text-decoration:overline'>X<sub>C</sub>²+R²</span>·√<span style='text-decoration:overline'>2</span>) =<br/>= E<sub>eff</sub> <font size=4>/</font>√<span style='text-decoration:overline'>X<sub>C</sub>²+R²</span></b></i><br/><b>This is the Ohm's Law for effective voltage and amperage in RC-circuit</b>.<br/><br/>It's important to notice that in the absence of a resistor (that is, <i><b>R=0</b></i>) the formula for <i><b>I(t)</b></i> transforms into<br/><i><b>I(t) = I<sub>0</sub>·sin(ωt+φ)</b></i><br/>where peak amperage <i><b>I<sub>0</sub>=E<sub>0</sub>/X<sub>C</sub></b></i> and phase shift <i><b>φ=arctan(∞)=π/2</b></i>, which corresponds to results obtained in a lecture "<i>AC Capacitors</i>" of the "<i>Alternating Current Induction</i>" chapter of this course dedicated to only a capacitor in the AC circuit.<br/> Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-28467502599650030882020-10-26T05:39:00.002-07:002020-10-26T05:39:56.980-07:00Problems on LC Circuits: UNIZOR.COM - Physics4Teens - Electromagnetism -...<iframe width="480" height="270" src="https://www.youtube.com/embed/asNG-xoBhME" frameborder="0"></iframe><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Problems on LC Circuit</u><br /><br/><i>Problem A</i><br/><br/>Consider a circuit that contains an AC generator, an inductor of inductance <i><b>L</b></i> and a capacitor of capacity <i><b>C</b></i> in a series.<br/><img src='http://www.unizor.com/Pictures/LC_circuit.jpg' style='width:200px;height:135px;'><br/><br/>Generated EMF has frequency <i><b>f=50Hz</b></i> and <i>effective voltage</i> <i><b>E<sub>eff</sub>=220V</b> (volt)</i>.<br/>The <i>effective AC current</i> is <i><b>I<sub>eff</sub>=5A</b> (ampere)</i>.<br/>The <i>capacitance</i> of a capacitor is <i><b>C=10μF</b> (microfarad)</i>.<br/>Find inductance <i><b>L</b> (henry)</i> of an inductor?<br/><br/><i>Solution</i><br/><br/>Let's start with an expression for the AC current in the LC circuit in terms of generated EMF and <i>reactance</i> of the capacitor and inductor.<br/><br/>EDF generated by a source of electricity<br/><i><b>E(t) = E<sub>0</sub>·sin(ωt)</b></i><br/>where<br/><i><b>E<sub>0</sub></b></i> is a <i>peak voltage</i> on the terminals of a generator,<br><i><b>ω=2πf</b></i> is an <i>angular velocity</i> of a rotor in radians per second, where <i><b>f</b></i> is a frequency of generated EMF in number of cycles per second.<br/><br/>Alternating electric <i>current</i> in the circuit<br/><i><b>I(t) = I<sub>0</sub>·cos(ωt)</b></i><br/>where<br/><i><b>I<sub>0</sub> = E<sub>0</sub>/(X<sub>C</sub>−X<sub>L</sub>)</b></i> is <i>peak amperage</i>,<br/><i><b>X<sub>C</sub> = 1/(ω·C)</b></i> is <i>capacitive reactance</i>,<br/><i><b>X<sub>L</sub> = ω·L</b></i> is <i>inductive reactance</i><br/><br/>Effective voltage and effective amperage are less than, correspondingly, peak voltage and peak amperage by <i>√<span style='text-decoration:overline'>2</span></i>.<br/>Therefore, from the expression for AC current follows<br/><i><b>I(t) = I<sub>0</sub>·cos(ωt) =<br/>= E<sub>0</sub>·cos(ωt)/(X<sub>C</sub>−X<sub>L</sub>)</b></i><br/><i><b>I<sub>eff</sub> = I<sub>0</sub>/√<span style='text-decoration:overline'>2</span> =<br/>= E<sub>0</sub>/</b></i>[<i><b>√<span style='text-decoration:overline'>2</span>(X<sub>C</sub>−X<sub>L</sub>)</b></i>]<i><b> =<br/>= E<sub>eff</sub>/(X<sub>C</sub>−X<sub>L</sub>)</b></i><br/><br/>The unknown in this expression is <i><b>X<sub>L</sub></b></i> that can be found:<br/><i><b>X<sub>C</sub>−X<sub>L</sub> = E<sub>eff</sub>/I<sub>eff</sub></b></i><br/><i><b>X<sub>L</sub> = X<sub>C</sub>−E<sub>eff</sub>/I<sub>eff</sub></b></i><br/><br/>Since <i><b>X<sub>C</sub>=1/(ω·C)</b></i> and <i><b>X<sub>L</sub>=ω·L</b></i><i><b>L·ω = 1/(ω·C) − E<sub>eff</sub>/I<sub>eff</sub></b></i><br/><i><b>L = 1/(ω²·C) − E<sub>eff</sub>/(ω·I<sub>eff</sub>)</b></i><br/><br/>Substituting<br/><i><b>ω=2πf=2π·50Hz=314(1/sec)</b></i><br/><i><b>C=10μF=10<sup>−5</sup>F</b></i><br/><i><b>E<sub>eff</sub>=220V</b></i><br/><i><b>I<sub>eff</sub>=5A</b></i><br/>into above expression for <i><b>L</b></i>, we obtain (rounded to 0.001)<br/><i><b>L= 1/(314²·10<sup>−5</sup>)−220/(314·5)=<br/>= 0.874</b>H (henry)</i><br/><br/><br/><i>Problem B</i><br/><br/>Consider a circuit that contains a source of a noisy electrical signal, which is a combination of many sinusoidal waves with different frequencies, amplitudes and phase shifts, a resistor with resistance <i><b>R</b></i> and a capacitor with capacitance <i><b>C</b></i> connected parallel to a resistor.<br/><img src='http://www.unizor.com/Pictures/CapacitorParallel.jpg' style='width:200px;height:250px;'><br/><br/>The signal (electrical current) coming through a resistor will have higher frequencies weakened by a presence of a capacitor.<br/>Explain why.<br/><br/><i>Explanation</i><br/><br/>Reactance of a capacitor <i><b>X<sub>C</sub></b></i>, which functionally similar to a resistance of a resistor, is inversely proportional to a frequency of the voltage on its ends<br/><i><b>X<sub>C</sub> = 1/(ω·C) = 1/(2πf·C)</b></i>,<br/>where<br/><i><b>ω=2πf</b></i> is the angular velocity of oscillations,<br/><i><b>f</b></i> is a frequency of oscillations,<br/><i><b>C</b></i> is the <i>capacitance</i> of a capacitor.<br/ <br/>Therefore, the greater frequency - the less <i><b>reactance</b></i> of a capacitor.<br/>Since <i><b>reactance</b></i> of a capacitor is functionally similar to a resistance of a resistor, this circuit is similar to a circuit with two parallel resistors with one of them having lower resistance with higher frequencies of the oscillations of the electrical current.<br/><br/>Using this analogy, we see that higher frequency oscillations of the electric current going through a capacitor will meet less resistance than lower frequency oscillations. Therefore, higher frequency oscillations of the current will go easier through a capacitor, thus having less impact on a resistor.<br/><br/>More precisely, the current going through parallel resistors is inversely proportional to their resistance (see lecture "DC Ohm's Law" in this part of the course). This is true not only for constant direct current produced by constant EMF <i><b>E</b></i>, but also at any moment <i><b>t</b></i>, when generated EMF oscillates (even irregularly) as a function of time <i><b>E(t)</b></i>.<br/>Therefore, a capacitor will lower <i>reactance</i> will have higher current going through it, thus weakening the higher frequency oscillations of the current going through a resistor.<br/><br/><i>Problem C</i><br/><br/>Consider a circuit that contains a source of a noisy electrical signal, which is a combination of many sinusoidal waves with different frequencies, amplitudes and phase shifts, a resistor with resistance <i><b>R</b></i> and an inductor with inductance <i><b>L</b></i> connected parallel to a resistor.<br/><img src='http://www.unizor.com/Pictures/InductorParallel.jpg' style='width:200px;height:250px;'><br/><br/>The signal (electrical current) coming through a resistor will have lower frequencies weakened by a presence of a inductor.<br/>Explain why.<br/><br/><i>Explanation</i><br/><br/>Reactance of an inductor <i><b>X<sub>L</sub></b></i>, which functionally similar to a resistance of a resistor, is proportional to a frequency of the voltage on its ends<br/><i><b>X<sub>L</sub> = ω·L = 2πf·L</b></i>,<br/>where<br/><i><b>ω=2πf</b></i> is the angular velocity of oscillations,<br/><i><b>f</b></i> is a frequency of oscillations,<br/><i><b>L</b></i> is the <i>inductance</i> of an inductor.<br/ <br/>Therefore, the lower frequency - the less <i><b>reactance</b></i> of an inductor.<br/>Since <i><b>reactance</b></i> of an inductor is functionally similar to a resistance of a resistor, this circuit is similar to a circuit with two parallel resistors with one of them having lower resistance with lower frequencies of the oscillations of the electrical current.<br/><br/>Using this analogy, we see that lower frequency oscillations of the electric current going through an inductor will meet less resistance than higher frequency oscillations. Therefore, lower frequency oscillations of the current will go easier through an inductor, thus having less impact on a resistor.<br/><br/>More precisely, the current going through parallel resistors is inversely proportional to their resistance (see lecture "DC Ohm's Law" in this part of the course). This is true not only for constant direct current produced by constant EMF <i><b>E</b></i>, but also at any moment <i><b>t</b></i>, when generated EMF oscillates (even irregularly) as a function of time <i><b>E(t)</b></i>.<br/>Therefore, an inductor will lower <i>reactance</i> will have higher current going through it, thus weakening the lower frequency oscillations of the current going through a resistor.<br/><br/><i>SUMMARY</i><br/>Capacitors and inductors can be used to filter certain "useful" frequencies from the noisy signals.<br/><br/> Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com1tag:blogger.com,1999:blog-3741410418096716827.post-70429271092990169882020-10-07T08:59:00.001-07:002020-10-07T08:59:02.022-07:00Series LC Circuit: UNIZOR.COM - Physics4Teens - Electromagnetism - Alter...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Series LC Circuit</u><br /><br /><br /><br />Consider a circuit that contains an AC generator, an inductor of inductance <i><b>L</b></i> and a capacitor of capacity <i><b>C</b></i> in a series.<br /><br /><img src="http://www.unizor.com/Pictures/LC_circuit.jpg" style="height: 135px; width: 200px;" /><br /><br /><br />The current <i><b>I<sub>L</sub>(t)</b></i> going through an inductor is the same as the current <i><b>I<sub>C</sub>(t)</b></i> going through a capacitor. So, we will use an expression <i><b>I(t)</b></i> for both.<br /><br /><br /><br />The electromotive force (EMF) generated by an AC generator depends only <br />on its own properties and can be described as a function of time <i><b>t</b></i><br /><br /><i><b>E(t) = E<sub>0</sub>·sin(ωt)</b></i><br /><br />where<br /><br /><i><b>E<sub>0</sub></b></i> is a peak voltage on the terminals of a generator,<br /><br /><i><b>ω</b></i> is an angular velocity of a rotor in radians per second.<br /><br /><br /><br /><i>Inductance</i> <i><b>L</b></i> of an inductor and <i>capacity</i> <i><b>C</b></i> of a capacitor produce voltage drops <i><b>V<sub>L</sub>(t)</b></i> and <i><b>V<sub>C</sub>(t)</b></i> correspondingly.<br /><br /><br /><br />As we know, the voltage drop on an inductor is causes by self-induction <br />and depends on the rate of change (that is, the first derivative by <br />time) of a magnetic flux <i><b>Φ(t)</b></i> going through it<br /><br /><i><b>V<sub>L</sub>(t) = </b>d<b>Φ(t)/</b>d<b>t</b></i><br /><br />Magnetic flux, in turn, depends on a current going through a wire of an inductor <i><b>I(t)</b></i> and the inductor's inductance <i><b>L</b></i><br /><br /><i><b>Φ(t) = L·I(t)</b></i><br /><br />Therefore, the voltage drop on an inductor equals to<br /><br /><i><b>V<sub>L</sub>(t) = L·</b>d<b>I(t)/</b>d<b>t</b></i><br /><br /><br /><br />As we know, the amount of electricity <i><b>Q(t)</b></i> accumulated in a capacitor is proportional to voltage <i><b>V<sub>C</sub>(t)</b></i> applied to its plates (that is, voltage drop on a capacitor). The constant <i>capacity</i> of a capacitor <i><b>C</b></i><br /> is the coefficient of proportionality (see lecture "Electric Fields" - <br />"Capacitors" in this course) that depends on a type of a capacitor<br /><br /><i><b>C = Q(t)/V<sub>C</sub>(t)</b></i><br /><br />Therefore,<br /><br /><i><b>Q(t)=C·V<sub>C</sub>(t)</b></i><br /><br /><br /><br />Knowing the amount of electricity <i><b>Q(t)</b></i> accumulated in a capacitor as a function of time <i><b>t</b></i>, we can determine the electric current <i><b>I(t)</b></i> in a circuit, which is a rate of change (that is, the first derivative by time) of the amount of electricity<br /><br /><i><b>I(t) = </b>d<b>Q(t)/</b>d<b>t = C·</b>d<b>V<sub>C</sub>(t)/</b>d<b>t</b></i><br /><br /><br /><br />The sum of voltage drops on an inductor and a capacitor is supposed to be equal to EMF produced by an AC generator <i><b>E(t)=E<sub>0</sub>·sin(ωt)</b></i>, which is the final equation in our system:<br /><br /><i><b>V<sub>L</sub>(t) = L·</b>d<b>I(t)/</b>d<b>t</b></i><br /><br /><i><b>I(t) = C·</b>d<b>V<sub>C</sub>(t)/</b>d<b>t</b></i><br /><br /><i><b>E<sub>0</sub>·sin(ωt) = V<sub>L</sub>(t) + V<sub>C</sub>(t)</b></i><br /><br /><br /><br />To solve this system of three equations, including two differential ones, let's resolve the third equation for <i><b>V<sub>C</sub>(t)</b></i> and substitute it in the second one.<br /><br /><i><b>V<sub>C</sub>(t) = E<sub>0</sub>·sin(ωt) − V<sub>L</sub>(t)</b></i><br /><br /><i><b>I(t)=C·</b>d</i>[<b><i>E<sub>0</sub>·sin(ωt)−V<sub>L</sub>(t)</i></b>]<b><i>/</i></b><i>d<b>t</b></i><br /><br /><br /><br />In the last equation we can differentiate each component and, using symbol <i><b>'</b></i> for a derivative, obtain<br /><br /><i><b>I(t)=CωE<sub>0</sub>·cos(ωt)−C·V'<sub>L</sub>(t)</b></i><br /><br /><br /><br />Together with the first equation from the original system of three <br />equations above we have reduced the system to two equations (again, we <br />use symbol <i><b>'</b></i> for brevity to signify differentiation)<br /><br /><i><b>V<sub>L</sub>(t) = L·I'(t)</b></i><br /><br /><i><b>I(t)=CωE<sub>0</sub>·cos(ωt)−C·V'<sub>L</sub>(t)</b></i><br /><br /><br /><br />Substituting <i><b>V<sub>L</sub>(t)</b></i> from the first of these equations into the second, we obtain one equation for <i><b>I(t)</b></i>, which happens to be a differential equation of the second order (we will use symbol <i><b>"</b></i> to signify a second derivative of <i><b>I(t)</b></i> for brevity)<br /><br /><i><b>I(t)=CωE<sub>0</sub>·cos(ωt)−CL·I"(t)</b></i><br /><br />or in a more traditional for differential equation form<br /><br /><i><b>a·I"(t) + b·I(t) = E<sub>0</sub>·cos(ωt)</b></i><br /><br />where<br /><br /><i><b>a = L/ω</b></i><br /><br /><i><b>b = 1/(Cω)</b></i><br /><br /><br /><br />Without getting too deep into a theory of differential equations, notice that the one and only <b>known</b> function in this equation that depends on time <i><b>t</b></i> is <i><b>cos(ωt)</b></i>. It's second derivative also contains <i><b>cos(ωt)</b></i>. So, if <i><b>I(t)</b></i> is proportional to <i><b>cos(ωt)</b></i>, its second derivative <i><b>I"(t)</b></i> will also be proportional to <i><b>cos(ωt)</b></i> and we can find the coefficient of proportionality to satisfy the equation.<br /><br /><br /><br />Let's try to find such coefficient <i><b>K</b></i> that function <i><b>I(t)=K·cos(ωt)</b></i> satisfies our equation.<br /><br /><i><b>I'(t) = −ωK·sin(ωt)</b></i><br /><br /><i><b>I"(t) = −ω²K·cos(ωt)</b></i><br /><br /><br /><br />Now our differential equation for <i><b>I(t)</b></i> is<br /><br /><i><b>−a·ω²K·cos(ωt) + b·K·cos(ωt) = E<sub>0</sub>·cos(ωt)</b></i><br /><br /><br /><br />From this we can easily find a coefficient <i><b>K</b></i><br /><br /><i><b>K = E<sub>0</sub>/(b−aω²)</b></i><br /><br /><br /><br />Since <i><b>a=L/ω</b></i> and <i><b>b=1/(Cω)</b></i><br /><br /><i><b>K = E<sub>0</sub>/</b></i>{[<i><b>1/(Cω)</b></i>]<i><b> − Lω</b></i>}<br /><br /><br /><br />In the previous lectures we have introduced concepts of <i>capacitive reactance</i> <i><b>X<sub>C</sub>=1/(Cω)</b></i> and <i>inductive reactance</i> <i><b>X<sub>L</sub>=Lω</b></i>. Using these variables, the expression for coefficient <i><b>K</b></i> is<br /><i><b>K = E<sub>0</sub>/(X<sub>C</sub>−X<sub>L</sub>)</b></i><br /><br />Therefore,<br /><br /><i><b>I(t) = E<sub>0</sub>·cos(ωt)/(X<sub>C</sub>−X<sub>L</sub>)</b></i><br /><br />or<br /><br /><i><b>I(t) = I<sub>0</sub>·cos(ωt)</b></i><br /><br />where<br /><br /><i><b>I<sub>0</sub> = E<sub>0</sub>/(X<sub>C</sub>−X<sub>L</sub>)</b></i><br /><br /><br /><br />The last equation brings us to a concept of a <i>reactance</i> of the LC circuit<br /><br /><i><b>X<sub>C</sub>−X<sub>L</sub></b></i><br /><br />that is similar to <i>resistance</i> of regular resistors.<br /><br />Using a concept of <i>reactance</i>, the last equation resembles the Ohm's Law.<br /><br /><br /><br />Let's determine the voltage drops on a capacitor <i><b>V<sub>C</sub>(t)</b></i> and an inductor <i><b>V<sub>L</sub>(t)</b></i> using the expression for the current <i><b>I(t)</b></i>.<br /><br /><br /><br />Since <i><b>Q(t)=C·V<sub>C</sub>(t)</b></i> and <i><b>I(t)=Q'(t)</b></i>, we can find <i><b>V<sub>C</sub>(t)</b></i> by integrating <i><b>I(t)/C</b></i>.<br /><br /><i><b>V<sub>C</sub>(t) = <span style="font-size: large;">∫</span><sub>[0,t]</sub>I(t)·</b>d<b>t/C =<br /><br />= I<sub>0</sub>·sin(ωt)/(C·ω) =<br /><br />= X<sub>C</sub>·E<sub>0</sub>·sin(ωt)/(X<sub>C</sub>−X<sub>L</sub>)</b></i><br /><br /><br /><br /><i><b>V<sub>L</sub>(t) = L·I'(t) =<br /><br />= −L·E<sub>0</sub>·sin(ωt)·ω/(X<sub>C</sub>−X<sub>L</sub>) =<br /><br />= −X<sub>L</sub>·E<sub>0</sub>·sin(ωt)/(X<sub>C</sub>−X<sub>L</sub>)</b></i><br /><br /><br /><br />Let's check that the sum of voltage drops in the circuit <i><b>V<sub>L</sub>(t)</b></i> and <i><b>V<sub>C</sub>(t)</b></i> is equal to the original EMF generated by a source of electricity.<br /><br />Indeed,<br /><br /><i><b>V<sub>L</sub>(t) + V<sub>C</sub>(t) =<br /><br />=(X<sub>C</sub>−X<sub>L</sub>)·E<sub>0</sub>·sin(ωt)/(X<sub>C</sub>−X<sub>L</sub>)=<br /><br />= E<sub>0</sub>·sin(ωt)</b></i><br /><br /><br /><br /><i>Summary</i><br /><br /><br /><br />EDF generated by a source of electricity<br /><br /><i><b>E(t) = E<sub>0</sub>·sin(ωt)</b></i><br /><br />where<br /><br /><i><b>E<sub>0</sub></b></i> is a peak voltage on the terminals of a generator,<br /><br /><i><b>ω</b></i> is an angular velocity of a rotor in radians per second.<br /><br /><br /><br />Alternating electric current in the circuit<br /><br /><i><b>I(t) = I<sub>0</sub>·cos(ωt)</b></i><br /><br />where<br /><br /><i><b>I<sub>0</sub> = E<sub>0</sub>/(X<sub>C</sub>−X<sub>L</sub>)</b></i><br /><br /><i><b>X<sub>C</sub> = 1/(ω·C)</b></i><br /><br /><i><b>X<sub>L</sub> = ω·L</b></i><br /><br /><br /><br />Voltage drop on a capacitor<br /><br /><i><b>V<sub>C</sub>(t) = X<sub>C</sub>·E<sub>0</sub>·sin(ωt)/(X<sub>C</sub>−X<sub>L</sub>)</b></i><br /><br /><br /><br />Voltage drop on an inductor<br /><br /><i><b>V<sub>L</sub>(t) = −X<sub>L</sub>·E<sub>0</sub>·sin(ωt)/(X<sub>C</sub>−X<sub>L</sub>)</b></i><br /><br /><br /><br /><i>Phase Shift</i><br /><br /><br /><br />Notice that <i>cos(x)=sin(x+π/2)</i>. Graph of function <i>y=sin(x+π/2)</i> is shifted to the left by <i>π/2</i> relative to graph of <i>y=sin(x)</i>.<br /><br />Therefore, oscillations of the current <i><b>I(t)</b></i> in the LC circuit are ahead of oscillation of the EMF generated by a source of electricity <i><b>E(t)</b></i> by a <i>phase shift</i> of <i>π/2</i>.<br /><br /><br /><br />Oscillations of the voltage drop on a capacitor <i><b>V<sub>C</sub>(t)</b></i> in the LC circuit are synchronized (<i>in phase</i>) with generated EMF.<br /><br /><br /><br />Notice that <i>−sin(x)=sin(x+π)</i>. Graph of function <i>y=sin(x+π)</i> is shifted to the left by <i>π</i> relative to graph of <i>y=sin(x)</i>.<br /><br />Therefore, oscillations of the voltage drop on an inductor <i><b>V<sub>L</sub>(t)</b></i> in the LC circuit are ahead of oscillation of the EMF generated by a source of electricity <i><b>E(t)</b></i> by <i>π</i> (this is called <i>in antiphase</i>).<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-8522541312110155812020-10-02T07:04:00.001-07:002020-10-02T07:04:23.682-07:00AC Inductors: UNIZOR.COM - Physics4Teens - Electromagnetism<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Alternating Current and Inductors</u><br /><br /><br /><br />For the purpose of this lecture it's important to be familiar with the concept of a <i>self-induction</i> explained in "Electromagnetism - Self-Induction" chapter of this course.<br /><br /><br /><br />In this lecture we will discuss the AC circuit that contains an <i>inductor</i> - a wire wound in a reel or a solenoid, thus making multiple loops, schematically presented on the following picture.<br /><br /><img src="http://www.unizor.com/Pictures/AC_inductor.jpg" style="height: 150px; width: 200px;" /><br /><br /><br /><br />Both direct and alternating current go through an inductor, but, while <br />direct current goes with very little resistance through a wire, whether <br />it's in a shape of a loop or not, alternating current meets some <br />additional resistance when this wire is wound into a loop.<br /><br /><br /><br />Consider the following experiment.<br /><br /><img src="http://www.unizor.com/Pictures/SolenoidMagnetLamp.jpg" style="height: 285px; width: 200px;" /><br /><br /><br /><br />Here the AC circuit includes a lamp, an inductor in a shape of a solenoid and an iron rod fit to be inserted into a solenoid.<br /><br />While the rod is not inside a solenoid, the lamp lights with normal <br />intensity. But let's gradually insert an iron core into a solenoid. As <br />the core goes deeper into a solenoid, the lamp produces less and less <br />light, as if some kind of resistance is increasing in the circuit.<br /><br />This experiment demonstrates that inductors in the AC circuit produce <br />effect similar to resistors, and the more "inductive" the inductor - the<br /> more resistance can be observed in a circuit.<br /><br />The theory behind this is explained in this lecture.<br /><br /><br /><br />The cause of this resistance is <i>self-induction</i>. This concept was <br />explained earlier in this course and its essence is that variable <br />magnetic field flux, going through a wire loop, creates electromotive <br />force (EMF) directed against the original EMF that drives electric <br />current through a loop.<br /><br /><br /><br />Any current that goes along a wire creates a magnetic field around this <br />wire. Since the current in our wire loop is alternating, the magnetic <br />field that goes through this loop is variable. According to the <br />Faraday's Law, the variable magnetic field going through a wire loop <br />generates EMF equal to a rate of change of the magnetic field flux and <br />directed opposite to the EMF that drives the current through a wire, <br />thus resisting it.<br /><br /><br /><br />Magnetic flux <i><b>Φ(t)</b></i> going through inductor, as a function of time <i><b>t</b></i>, is proportional to an electric current <i><b>I(t)</b></i> going through its wire<br /><br /><i><b>Φ(t) = L·I(t)</b></i><br /><br />where <i><b>L</b></i> is a coefficient of proportionality that depends <br />on physical properties of the inductor (number of loop in a reel, type <br />of its core etc.) called <i>inductance</i> of the inductor.<br /><br /><br /><br />If the current is alternating as<br /><br /><i><b>I(t) = I<sub>max</sub>·sin(ωt)</b></i><br /><br />the flux will be<br /><br /><i><b>Φ(t) = L·I<sub>max</sub>·sin(ωt)</b></i><br /><br /><br /><br />According to Faraday's Law, self-induction EMF <i><b>E<sub>i</sub></b></i><br /> is equal in magnitude to a rate of change of magnetic flux and opposite<br /> in sign (see chapter "Electromagnetism - "Self-Induction" in this <br />course)<br /><br /><i><b>E<sub>i</sub>(t) = −</b>d<b>Φ/</b>d<b>t =<br /><br />= −L·</b>d<b>I(t)/</b>d<b>t =<br /><br />= −L·ω·I<sub>max</sub>·cos(ωt) =<br /><br />= −L·ω·I<sub>max</sub>·sin(ωt+π/2) =<br /><br />= −E<sub>imax</sub>·sin(ωt+π/2)</b></i><br /><br />where<br /><br /><i><b>E<sub>imax</sub> = L·ω·I<sub>max</sub></b></i><br /><br /><br /><br />The unit of measurement of <i>inductance</i> is <b>henry (H)</b> with <b>1H</b> being an inductance of an inductor that generates <b>1V</b> electromotive force, if the rate of change of current is <b>1A/sec</b>.<br /><br />That is,<br /><br /><i>henry = volt·sec/ampere = ohm·sec</i><br /><br /><br /><br />An expression <i><b>X<sub>L</sub>=L·ω</b></i> in the above formula for <i><b>E<sub>i</sub></b></i> is called <i>inductive reactance</i>. It plays the same role for an inductor as <i>resistance</i> for resistors.<br /><br />The units of the inductive reactance is <b>Ohm (Ω)</b> because<br /><br /><i>henry/sec = ohm·sec/sec = ohm</i>.<br /><br /><br /><br />Using this concept of <i>inductive reactance</i> <i><b>X<sub>L</sub></b></i> of an <i>inductor</i>, the time dependent induced EMF is<br /><br /><i><b>E<sub>i</sub>(t) = −X<sub>L</sub>·I<sub>max</sub>·sin(ωt+π/2) = −E<sub>imax</sub>·sin(ωt+π/2)</b></i><br /><br />and<br /><br /><i><b>E<sub>imax</sub> = X<sub>L</sub>·I<sub>max</sub></b></i>,<br /><br />which for <i>inductors</i> in AC circuit is an analogue of the Ohm's Law for <i>resistors</i>.<br /><br /><br /><br />What's most important in the formula<br /><br /><i><b>E<sub>i</sub>(t) = −E<sub>imax</sub>·sin(ωt+π/2)</b></i><br /><br /> and the most important property of an inductor in an AC circuit is <br />that, while the electric current in a circuit oscillates with angular <br />speed <i><b>ω</b></i>, the <b>voltage drop on an inductor oscillates with the same angular speed <i><b>ω</b></i> as the current, but its period is shifted in time by <i>π/2</i> relative to the current</b>.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-20606381892967132802020-09-29T13:51:00.001-07:002020-09-29T13:51:31.376-07:00UNIZOR.COM - Physics4Teens - Electromagnetism - Ohm's Law - Problems 4<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Direct Current - Ohm's Law - Problems 4</u><br /><br /><br /><br /><i>Problem A</i><br /><br /><br /><br />Given a circuit presented on a picture below.<br /><img src="http://www.unizor.com/Pictures/CapacitorDischarge.jpg" style="height: 240px; width: 200px;" /><br /><br />Initially, a red switch is in position <b><span style="color: red;">A</span></b> to fully charge a <i>capacitor</i> of capacity <b>C</b> from a battery producing a direct current with voltage <b>V</b>.<br /><br /><br /><br />When a capacitor is fully charged, a switch is moved to position <b><span style="color: red;">B</span></b>, disconnecting a capacitor from a battery and forming a new circuit that includes only a fully charged capacitor and a <i>resistor</i> of resistance <b>R</b>.<br /><br /><br /><br />When a switch is in position <b><span style="color: red;">B</span></b>, a <br />capacitor starts discharging its charge through a resistor. It's charge <br />will gradually diminish to zero, when all excess electrons on its one <br />plate will flow through a resistor to a plate with deficiency of <br />electrons.<br /><br /><br /><br />During this process of discharge the electric current in a circuit that <br />contains a capacitor and a resistor will change from some maximum value <br />in the beginning of this process to zero, when the discharge is <br />completed.<br /><br /><br /><br />Find the charge on a capacitor <i><b>Q(t)</b></i> and an electric current flowing trough a resistor <i><b>I(t)</b></i> as functions of time <i><b>t</b></i>.<br /><br /><br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />Assume, our switch is in position <b><span style="color: red;">A</span></b>, and we are at the charging stage, when the battery of voltage <i><b>V</b></i> is charging a capacitor of capacity <i><b>C</b></i>.<br /><br />The capacity of a capacitor is defined as the constant ratio of a charge<br /> accumulated by a capacitor to a voltage applied to its plate (see <br />"Capacitors" lecture in the "Electromagnetism - Electric Field" <br />chapter):<br /><br /><i><b>C = Q/V</b></i><br /><br />Therefore, the full charge of a capacitor at the end of the first stage of charging is <i><b>V·C</b></i>.<br /><br /><br /><br />Then we flip a switch into position <b><span style="color: red;">B</span></b>, starting the second stage - discharging of a capacitor through a resistor.<br /><br />At the beginning of this second stage a capacitor is fully charged. So, at time <i><b>t=0</b></i> its charge is<br /><br /><i><b>Q(0) = V·C</b></i><br /><br /><br /><br />The charge on a capacitor at any time produces a voltage between its plates<br /><br /><i><b>V(t) = Q(t)/C</b></i><br /><br />This voltage produces a current flowing through a resistor <i><b>I(t)</b></i> that, according to the Ohm's Law, should be equal to<br /><br /><i><b>I(t) = V(t)/R</b></i><br /><br />From the two equations above we conclude<br /><br /><i><b>Q(t)/C = I(t)·R</b></i><br /><br />This is our first equation that connects two time-dependent (that is, <br />functions of time) variables - an electric current in a circuit <i><b>I(t)</b></i> and a charge on a capacitor <i><b>Q(t)</b></i>.<br /><br /><br /><br />The second functional equation is, basically, a definition of an <br />electric current as the rate of electric charge flowing in a circuit <br />(that is, <i>amperage</i> is how much electricity in <i>coulombs</i> flows through a circuit per unit of time - a <i>second</i>).<br /><br />Mathematically speaking, an electric current is the first derivative of <br />an electric charge by time, taken with a sign that depends on the <br />direction of the change of the charge (<i>plus</i> if the charge is increasing and <i>minus</i> if decreasing):<br /><br /><i><b>I(t) = −</b>d<b>Q(t)/</b>d<b>t</b></i><br /><br />Considering the charge <i><b>Q(t)</b></i> is decreasing and, therefore, <br />its derivative is negative, while we would like the electric current to a<br /> be a positive number, we have to use a minus sign in this equation.<br /><br />This is our second functional equation (that happens to be differential)<br /> connecting two functions - an electric current in a circuit <i><b>I(t)</b></i> and a charge on a capacitor <i><b>Q(t)</b></i>.<br /><br /><br /><br />Now we have two functional equations, one of them is differential, and an initial condition:<br /><br /><i><b>Q(t)/C = I(t)·R</b></i><br /><br /><i><b>I(t) = −</b>d<b>Q(t)/</b>d<b>t</b></i><br /><br /><i><b>Q(0) = V·C</b></i><br /><br />It's up to our mathematical skills to solve this system of equations.<br /><br /><br /><br />First, we substitute <i><b>I(t)</b></i> from the second equation into the first, getting a differential equation for <i><b>Q(t)</b></i><br /><br /><i><b>Q(t)/C = −R·</b>d<b>Q(t)/</b>d<b>t</b></i><br /><br />This can be converted into<br /><br /><i>d<b>Q(t)/Q(t) = −</b>d<b>t/(R·C)</b></i><br /><br />or<br /><br /><i>d</i>[<i><b>ln(Q(t))</b></i>]<i><b> = </b>d</i>[<i><b>−t/(R·C)</b></i>]<br /><br /><br /><br />If differentials of two functions are equal, the functions themselves <br />are just separated by a constant that can be determined using the <br />initial condition. Let denote that constant as <i><b>K</b></i>.<br /><br /><i><b>ln(Q(t)) = −t/(R·C) + K</b></i><br /><br />or, applying an exponent to both sides of this equation,<br /><br /><i><b>Q(t) = e<sup>K</sup>·e<sup>−t/(R·C)</sup></b></i><br /><br /><br /><br />It's time to use the initial condition <i><b>Q(0)=V·C</b></i> to determine the multiplier <i><b>e<sup>K</sup></b></i>.<br /><br />For <i><b>t=0</b></i> the right side of an expression for <i><b>Q(t)</b></i> equals to <i><b>e<sup>K</sup></b></i>. Therefore, this multiplier equals to <i><b>V·C</b></i>.<br /><br />Therefore, the final expression for a charge on a capacitor as a function of time <i><b>Q(t)</b></i> is<br /><br /><i><b>Q(t) = V·C·e<sup>−t/(R·C)</sup></b></i><br /><br />So, a charge on a capacitor is exponentially diminishing.<br /><br /><br /><br />From the expression of <i><b>Q(t)</b></i> we can find the expression on an electric current going through a resistor, using the equation<br /><br /><i><b>I(t) = −</b>d<b>Q(t)/</b>d<b>t</b></i><br /><br />from which follows<br /><br /><i><b>I(t) = −</b>d</i>[<i><b>V·C·e<sup>−t/(R·C)</sup></b></i>]<i><b>/</b>d<b>t =<br /><br />= −V·C·</b>d</i>[<i><b>e<sup>−t/(R·C)</sup></b></i>]<i><b>/</b>d<b>t =<br /><br />= −(V·C)·(−1/(R·C))·e<sup>−t/(R·C)</sup> =<br /><br />= (V/R)·e<sup>−t/(R·C)</sup></b></i><br /><br /><br /><br /><br /><br /><i>Answer</i><br /><br /><br /><br /><i><b>Q(t) = (V·C)·e<sup>−t/(R·C)</sup></b></i><br /><br />The multiplier <i><b>V·C</b></i> is the initial full charge of a capacitor.<br /><br /><br /><br /><i><b>I(t) = (V/R)·e<sup>−t/(R·C)</sup></b></i><br /><br />The multiplier <i><b>V/R</b></i> is the current that would flow through a resistor, if there were no capacitor. This follows from the Ohm's Law.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-31010889866327480662020-09-08T09:07:00.001-07:002020-09-08T09:07:47.172-07:003-Phase AC Problem: UNIZOR.COM - Physics4Teens - Electromagnetism - Alte...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Problems on AC Induction</u><br /><br /><br /><br /><i>Problem A</i><br /><br /><br /><br />Three-phase generator has four wires coming out from it connected to a "star" with three <b>phase</b> wires carrying sinusoidal EMF shifted by 120°=2π/3 from each other and one <b>neutral</b> wire.<br /><br />The angular speed of a generator's rotor is <i><b>ω</b></i>.<br /><br />Assume that the peak difference in electric potential between each phase wire and a neutral one is <i><b>E</b></i>.<br /><br />Describe the difference in electric potential between each pair of phase wires as a function of time.<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />The difference in electric potential between phase wires and a neutral one can be described as<br /><br />Phase 1: <i><b>E<sub>1</sub>(t)=E·sin(ωt)</b></i><br /><br />Phase 2: <i><b>E<sub>2</sub>(t)=E·sin(ωt−2π/3)</b></i><br /><br />Phase 3: <i><b>E<sub>3</sub>(t)=E·sin(ωt+2π/3)</b></i><br /><br />The difference in electric potential between phase 1 wire and phase 2 wire can be represented as<br /><br /><i><b>E<sub>1,2</sub>(t) = E<sub>1</sub>(t) − E<sub>2</sub>(t)</b></i><br /><br />Similarly, the difference in electric potential between two other pairs of phase wires is<br /><br /><i><b>E<sub>2,3</sub>(t) = E<sub>2</sub>(t) − E<sub>3</sub>(t)</b></i><br /><br /><i><b>E<sub>3,1</sub>(t) = E<sub>3</sub>(t) − E<sub>1</sub>(t)</b></i><br /><br />Let's calculate all these voltages.<br /><br /><i><b>E<sub>1,2</sub>(t) = E·sin(ωt) − E·sin(ωt−2π/3) =<br /><br />= E·sin((ωt−π/3)+π/3) − E·sin((ωt−π/3)−π/3) =</b></i><br /><br />[substitute <i>φ=ωt−π/3</i>]<br /><br /><i><b>= E·sin(φ+π/3)−E·sin(φ−π/3) =<br /><br />= E·</b></i>[<i><b>sin(φ+π/3)−sin(φ−π/3)</b></i>]<br /><br />Let's simplify the trigonometric expression.<br /><br /><i><b>sin(φ+π/3) − sin(φ−π/3) =<br /><br />= sin(φ)·cos(π/3) +<br /><br />+ cos(φ)·sin(π/3) −<br /><br />− sin(φ)·cos(π/3) +<br /><br />+ cos(φ)·sin(π/3) =<br /><br />= 2cos(φ)·sin(π/3) =<br /><br />= √<span style="text-decoration: overline;">3</span>cos(φ) =<br /><br />= √<span style="text-decoration: overline;">3</span>cos(ωt−π/3)</b></i><br /><br />Therefore,<br /><br /><i><b>E<sub>1,2</sub>(t) = E√<span style="text-decoration: overline;">3</span>cos(ωt−π/3)</b></i><br /><br />As we see, the electric potential between phase 1 and phase 2 wires is also sinusoidal (since <i>cos(x)=sin(x+π/2), cos(ωt−π/3)</i> equals to <i>sin(ωt+π/6)</i>),<br /> but shifted in time, and its peak voltage is greater than the peak <br />voltage between a phase wire and a neutral one by a factor of √<span style="text-decoration: overline;">3</span>.<br /><br />Similar factor difference of √<span style="text-decoration: overline;">3</span> is between <b>effective voltages</b> of these pairs of wires.<br /><br />Analogous calculations for the other pairs of phase wires produce the following.<br /><br /><i><b>E<sub>2,3</sub>(t) = E·sin(ωt−2π/3) − E·sin(ωt+2π/3) =</b></i><br /><br />[substitute <i>φ=ωt</i>]<br /><br /><i><b>= sin(φ)·cos(2π/3) −<br /><br />− cos(φ)·sin(2π/3) −<br /><br />− sin(φ)·cos(2π/3) −<br /><br />− cos(φ)·sin(2π/3) =<br /><br />= −2cos(φ)·sin(2π/3) =<br /><br />= −√<span style="text-decoration: overline;">3</span>cos(φ) =<br /><br />= −√<span style="text-decoration: overline;">3</span>cos(ωt)</b></i><br /><br />Therefore,<br /><br /><i><b>E<sub>2,3</sub>(t) = −E√<span style="text-decoration: overline;">3</span>cos(ωt)</b></i><br /><br />Also the same factor difference of √<span style="text-decoration: overline;">3</span> relative to phase/neutral voltage.<br /><br />Finally, the third phase/phase voltage calculations produce the following.<br /><br /><i><b>E<sub>3,1</sub>(t) = E·sin(ωt+2π/3) − E·sin(ωt) =<br /><br />= E·sin((ωt+π/3)+π/3) − E·sin((ωt+π/3)−π/3) =</b></i><br /><br />[substitute <i>φ=ωt+π/3</i>]<br /><br /><i><b>= E·sin(φ+π/3)−E·sin(φ−π/3) =<br /><br />= E·</b></i>[<i><b>sin(φ+π/3)−sin(φ−π/3)</b></i>]<br /><br />Let's simplify the trigonometric expression.<br /><br /><i><b>sin(φ+π/3) − sin(φ−π/3) =<br /><br />= sin(φ)·cos(π/3) +<br /><br />+ cos(φ)·sin(π/3) −<br /><br />− sin(φ)·cos(π/3) +<br /><br />+ cos(φ)·sin(π/3) =<br /><br />= 2cos(φ)·sin(π/3) =<br /><br />= √<span style="text-decoration: overline;">3</span>cos(φ) =<br /><br />= √<span style="text-decoration: overline;">3</span>cos(ωt+π/3)</b></i><br /><br />Therefore,<br /><br /><i><b>E<sub>3,1</sub>(t) = E√<span style="text-decoration: overline;">3</span>cos(ωt+π/3)</b></i><br /><br />Again, the same factor difference of √<span style="text-decoration: overline;">3</span> relative to phase/neutral voltage.<br /><br />CONCLUSION<br /><br />The voltage between any two phase wires is in magnitude greater than phase to neutral voltage by √<span style="text-decoration: overline;">3</span>. Both are sinusoidal, but one is shifted in time relatively to another.<br /><br />The phase to phase <b>effective voltage</b> (which is by √<span style="text-decoration: overline;">2</span> less then peak voltage) also is greater than phase to neutral by the same √<span style="text-decoration: overline;">3</span>.<br /><br />EXAMPLES<br /><br />If phase to neutral effective voltage is 127V, the phase to phase effective voltage is 220V.<br /><br />If phase to neutral effective voltage is 220V, the phase to phase effective voltage is 380V.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-74959013741698563602020-08-27T08:59:00.001-07:002020-08-27T08:59:06.657-07:00Three-phase AC: UNIZOR.COM - Physics4Teens - Electromagnetism - Alternat...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Three Phases AC</u><br /><br /><br /><br /><i>The Basic Principle<br /><br /> of 3-Phase AC Generation</i><br /><br /><br /><br />Recall the process of generating alternating current (AC) using a pair <br />of permanent magnets and a wire frame (a coil) rotating around the axis <br />perpendicular to magnetic field lines.<br /><br />The pictures below represent the schematic design of such a system and a<br /> graph of an electromotive force (EMF) generated between the ends <i>a<sub>1</sub></i> and <i>a<sub>2</sub></i> of the wire frame<br /><br /><i><b>E<sub>a<sub>1</sub>a<sub>2</sub></sub> = E<sub>max</sub>·sin(ωt)</b></i><br /><br />where<br /><br /><i>E<sub>max</sub></i> is the maximum absolute value of EMF,<br /><br /><i>ω</i> is the angular speed of rotation of the wire frame,<br /><br /><i>t</i> is time.<br /><br /><img src="http://www.unizor.com/Pictures/3phase1.png" style="height: 120px; width: 200px;" /><br /><br /><br /><br /><img src="http://www.unizor.com/Pictures/3phase1sin.png" style="height: 120px; width: 200px;" /><br /><br /><br /><br />An obvious improvement to this design is to use the power of rotation <br />more efficiently by having two wire frames on the same axis positioned <br />perpendicularly to each other. In this case we can have two independent <br />sources of EMF with the only difference of one of them to be shifted in <br />time relatively to another by 1/2 of the time of rotation.<br /><br />This shift is related to a simple fact that at the moment one wire <br />frame, aligned along the magnetic line, crosses these magnetic lines <br />with the highest rate, while another wire frame, that is perpendicular <br />to magnetic field lines, moves along these lines without actual <br />crossing. Then the roles are changed, as the coils rotate.<br /><br />The EMF between the ends <i>b<sub>1</sub></i> and <i>b<sub>2</sub></i> of the second wire frame will then be<br /><br /><i><b>E<sub>b<sub>1</sub>b<sub>2</sub></sub> = E<sub>max</sub>·sin(ωt−π/2)</b></i><br /><br /><img src="http://www.unizor.com/Pictures/3phase2.png" style="height: 130px; width: 200px;" /><br /><br /><br /><br /><img src="http://www.unizor.com/Pictures/3phase2sin.png" style="height: 120px; width: 200px;" /><br /><br /><br /><br />Why stop at two wire frames? Let's have three coils positioned at 120° <br />relative to each other. Now we will have three independent sources of <br />EMF shifted in time from each other by 1/3 of the time of rotation (<b>phase</b><br /> shift) - the time needed by one wire frame to take the position between<br /> the magnet poles, previously taken by another wire frame.<br /><br />Three different EMF, therefore, will be equal to<br /><br /><i><b>E<sub>a<sub>1</sub>a<sub>2</sub></sub> = E<sub>max</sub>·sin(ωt)</b></i><br /><br /><i><b>E<sub>b<sub>1</sub>b<sub>2</sub></sub> = E<sub>max</sub>·sin(ωt−2π/3)</b></i><br /><br /><i><b>E<sub>c<sub>1</sub>c<sub>2</sub></sub> = E<sub>max</sub>·sin(ωt−4π/3) = E<sub>max</sub>·sin(ωt+2π/3)</b></i><br /><br /><img src="http://www.unizor.com/Pictures/3phase3.png" style="height: 140px; width: 200px;" /><br /><br /><br /><br /><img src="http://www.unizor.com/Pictures/3phase3sin.png" style="height: 120px; width: 200px;" /><br /><br /><br /><br /><br /><br /><i>Practical Implementation</i><br /><br /><br /><br />The design of a three phase generator, as depicted above, is just the <br />first try of an idea. If the magnet is fixed and three wire frames <br />(coils) are rotating between its poles, it presents a problem to connect<br /> these coils to transmit the generated electricity to consumers, we need<br /> sliding contacts, brushes and other impractical devices.<br /><br /><br /><br />In real life generators the three coils make up a stator - a fixed part <br />of a generator, while the magnet is rotating inside a circle of coils by<br /> external power (like steam, water, wind etc.), generating the <br />alternating current in the coils, which allows to make an electric <br />connection to coils fixed.<br /><br /><br /><br />At the first glance, three coils have three pairs of connections with <br />sinusoidal EMF generated in each pair and, to transfer AC electricity <br />from all coils to consumers, we seem to need three pairs of wires, two <br />from each coil - six wires altogether. This, however, can be improved by<br /> using the following technique.<br /><br />Let's connect ends <i>a<sub>2</sub></i>, <i>b<sub>2</sub></i> and <i>c<sub>2</sub></i><br /> of three coils together (see a picture below, it's a black wire at the <br />bottom connected to a black circle going around all coils) and see what <br />kind of resulting voltage will be observed on each end of the coils. <br />This is called a <b>star</b> connection of the generator's coils.<br /><br /><img src="http://www.unizor.com/Pictures/3phaseLoad.jpg" style="height: 160px; width: 200px;" /><br /><br />We know that the electric potential (EMF) on each of the above contacts <br />has a sinusoidal magnitude with a time shift by 1/3 of a period relative<br /> to each other. When we connect these three contacts, the potential at <br />the joint will be<br /><br /><i><b>E<sub>0</sub> = E<sub>max</sub>·sin(ωt) + E<sub>max</sub>·sin(ωt−2π/3) + E<sub>max</sub>·sin(ωt+2π/3) = E<sub>max</sub>·X</b></i><br /><br />where<br /><br /><i><b>X = sin(ωt) + sin(ωt−2π/3) + sin(ωt+2π/3) =<br /><br />= sin(ωt) + sin(ωt)·cos(2π/3) − cos(ωt)·sin(2π/3) + sin(ωt)·cos(2π/3) + cos(ωt)·sin(2π/3) =<br /><br />= sin(ωt) + sin(ωt)·(−1/2) − cos(ωt)·(√<span style="text-decoration: overline;">3</span>/2) + sin(ωt)·(−1/2) +<br />cos(ωt)·(√<span style="text-decoration: overline;">3</span>/2) = 0</b></i><br /><br />Therefore, <i><b>E<sub>0</sub> = 0</b></i><br /><br />There will be no difference in electrical potential between the joint and the ground.<br /><br /><br /><br />This fact enables to transmit all three phases of generated electricity along four wires - one from <i>a<sub>1</sub></i> (<b>phase 1</b>) contact, one from <i>b<sub>1</sub></i> (<b>phase 2</b>) contact, one from <i>c<sub>1</sub></i> (<b>phase 3</b>) contact and one <b>neutral</b> from a joint connection to <i>a<sub>2</sub></i>, <i>b<sub>2</sub></i> and <i>c<sub>2</sub></i>.<br /><br />The <b>neutral</b> wire is usually grounded since its electric potential is equal to zero.<br /><br /><br /><br />With this arrangement we still have an advantage of having three <br />independent phases of alternating current, but we need only four wires <br />to transmit it - three <b>phase</b> wires and one <b>neutral</b>.<br /><br />Connecting any device to any phase and a neutral wires, we will get a closed circuit with AC running in it.<br /><br /><br /><br /><br /><br /><i>Energy Consideration</i><br /><br /><br /><br />Obviously, putting two or three coils in a stator of a generator doubles<br /> or triples the energy output carried by outgoing wires. The Law of <br />Energy Conservation must work, so where is the energy is coming from?<br /><br /><br /><br />Recall the electromagnetic induction experiment described in the lecture<br /> "Faraday's Law" in the "Electromagnetic Induction" chapter of this <br />course with a wire moving in the uniform magnetic field.<br /><br /><img src="http://www.unizor.com/Pictures/Faraday.png" style="height: 111px; width: 200px;" /><br /><br />Since we physically move wire's electrons in one direction <br />perpendicularly to magnetic field lines, the Lorentz force pushes them <br />perpendicularly to both, the direction of the movement of a wire and the<br /> direction of the magnetic field lines, that is, along the wire, thereby<br /> creating an electric current between wire ends.<br /><br /><br /><br />Now electrons are moving with a wire in one direction and along the wire in another.<br /><br />The first movement maintains the electric current in the wire, but the <br />second, again, is a subject of the Lorentz force that pushes the <br />electrons perpendicularly to their direction, that is opposite to the <br />original direction of a wire movement.<br /><br />This force resists the movement of a wire in its original direction. We <br />have to perform work against this force to move the wire.<br /><br /><br /><br />Similar considerations are true in a case of a circular movement of a <br />wire frame in a magnetic field or, if wire coils are in a stator, the <br />force is needed to rotate the magnet in a rotor. That is, we have to <br />spend energy to generated the electricity, the rotor's rotation is <br />possible only if we apply the force against the Lorentz forces resisting<br /> this rotation. The magnetic field generated by the electric current in a<br /> wire coil of a stator resists the rotation of a magnet in a rotor.<br /><br /><br /><br />If we have more than one coil in a stator, each one resists the rotation<br /> of a rotor, so we have to spend proportionally more effort to rotate <br />the rotor.<br /><br />The Law of Energy Conservation works. The more coils we have in a stator<br /> - the more electricity is generated, but the more resistance to a <br />rotor's rotation needs to be overcome.<br /><br /><br /><br /><br /><br /><i>Three Phase AC Motor</i><br /><br /><br /><br />The lecture "AC Motors" of this chapter described the necessity of having a rotating magnetic field to make an AC motor.<br /><br />To achieve such a rotating field we had to resort to artificially create<br /> a second AC current with a phase shift by 90° using a capacitor or a <br />transformer.<br /><br />Most of household AC motors (like in a fan) work on this principle, they need only two wires, which are, as we can say now, a <b>phase</b> wire and a <b>neutral</b> one.<br /><br /><br /><br />Powerful industrial level AC motors (like in a water pump that works in a<br /> tall building to pump water to the roof tank) needs more power, and we <br />can use all three phases to create a rotating magnetic field.<br /><br /><br /><br />So, all four wires coming from the AC generator, three <b>phase</b> wires and one <b>neutral</b><br /> one go into an AC motor, whose principal construction very much <br />resembles the one described in the previous lecture. The only difference<br /> is, we already have three wires with AC phase shifted by 120° <br />relatively to each other. So, we have to position three wire coils in a <br />stator at 120° angles to each other, connect one end of each coil to a <br />corresponding <b>phase</b> wire and another end - to a common <b>neutral</b><br /> wire, and the rotating field is ready. Then it will work pretty much as<br /> it was described in the "AC Motors" lecture, but smoother because three<br /> phases make a smoother rotation of a magnetic field than two phases.<br /><br /><br /><br />At the end I would like to say again, that it was Nicola Tesla's genius <br />that created all the basic principles, based on which all the current AC<br /> motors are working now. His contributions to our industrial development<br /> are grossly underappreciated. Calling an electric car model "Tesla" is a<br /> late but well deserved tribute to his creativity.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-3118095052187790112020-08-23T13:14:00.001-07:002020-08-23T13:14:08.114-07:00AC MotorsUNIZOR.COM - Physics4Teens - Electromagnetism - Alternating Cur...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Alternating Current Motors</u><br /><br /><br /><br /><i>Alternating current (AC)</i> motors are used where the sources of electricity produce alternating current - in our homes, at industrial facilities etc.<br /><br />For example, AC motors work in refrigerators, water pumps, air conditioners, large fans and many other devices.<br /><br /><br /><br />The idea of AC motors belongs to Nicola Tesla, who invented it at the end of 19th century.<br /><br />And it was a brilliant idea!<br /><br /><br /><br />Before talking about AC motors, let's recall how DC motors are working. <br />Their basic design was presented in a lecture Physics 4 Teens - <br />Electromagnetism - Magnetism of Electric Current - DC Motors.<br /><br /><br /><br />The important feature of DC motor, which was the starting idea behind <br />its design, was the rotational momentum exerted by a magnetic field of a<br /> permanent magnet onto a wire frame with the direct electric current <br />going through it due to Lorentz force. Without a <i>commutator</i> or <br />any electronic switches the rotating frame would rotate until its plane <br />will reach a perpendicular position relative to magnetic lines, which <br />will be its equilibrium position.<br /><br /><br /><br />Let's start with this idea of an AC motor and try to make it work.<br /><br /><br /><br />Firstly, as in the case of DC motors, let's replace the permanent magnet<br /> in a stator with two coils around an iron cores - electromagnets that <br />create magnetic field. Notice, however, that the current in these <br />electromagnets is alternating, which means that the magnetic field <br />between these electromagnets is alternating as well in magnitude and <br />direction.<br /><br /><br /><br />Secondly, we will place a permanent magnet on an axis, so it freely rotates between the poles of the electromagnets<br /><br /><br /><br />What will happen if we switch on the alternating current in the electromagnets?<br /><br />Well, nothing noticeable. The problem is, the polarity of electromagnets<br /> will start switching with frequency of the AC - 50 or 60 oscillations <br />per second in usual commercial wiring, and a permanent magnet in-between<br /> will be forced in two opposite directions with the same frequency and <br />it will not start rotating.<br /><br /><br /><br />But let's manually start rotating the magnet in any direction strong <br />enough to force it to rotate with significant angular speed.<br /><br />At different positions the variable external magnetic field of <br />electromagnetic stator interferes with rotating permanent magnet rotor, <br />sometimes slowing it, sometimes speeding its rotation.<br /><br />In a short while the rotation of the permanent magnet rotor will <br />synchronize with variable external field of electromagnetic stator and <br />rotation will be maintained with the same angular speed as the frequency<br /> of AC in the coils of electromagnets.<br /><br /><br /><br />Let's analyze this rotation in steps after the synchronization is achieved.<br /><br />Let a pole facing the rotor of one electromagnet be X and an opposite pole of another electromagnet be Y.<br /><br />As AC changes its direction and magnitude, pole X is gradually changing <br />from North (X=N) to zero (X=0), to South (X=S), again to zero (X=0), <br />again to North (X=N) etc.<br /><br />At the same time pole Y of an opposite electromagnet is gradually <br />changing from South (Y=S) to zero (Y=0), to North (Y=N), again to zero <br />(Y=0), again to South (X=N) etc.<br /><br /><br /><br />Synchronously rotating permanent magnet of a rotor should with its North<br /> pole approach X=S and, simultaneously, with its South pole approach <br />Y=N.<br /><br /><img src="http://www.unizor.com/Pictures/ACelectromagnetSN.jpg" style="height: 150px; width: 200px;" /><br /><br />As a rotor approaches with its poles the poles of a stator, X and Y <br />should gradually weaken and, when the permanent magnet is fully aligned <br />along XY line, the magnitudes of the magnetic fields of electromagnets <br />should diminish to zero (X=0, Y=0).<br /><br /><br /><br />Permanent magnet rotor will pass this point by inertia and the polarity of the electromagnets switches, so now X=N and Y=S.<br /><br /><img src="http://www.unizor.com/Pictures/ACelectromagnetNS.jpg" style="height: 150px; width: 200px;" /><br /><br />That causes rotation to continue until the permanent magnet of a rotor again takes a position along XY line.<br /><br />By that time the electromagnets will be at X=0 and Y=0, rotor will <br />continue rotation by inertia, then AC switches the polarity of <br />electromagnets again and rotation continues in a similar manner <br />indefinitely.<br /><br /><br /><br />Our first version of an AC motor is functional, but it has two obvious disadvantages.<br /><br />One disadvantage is the usage of permanent magnet, they are very <br />expensive. We could not avoid it in a DC motor, but in an AC case we <br />might think that variable magnetic field might help to use the induction<br /> effect to avoid it.<br /><br />Another disadvantage is that switching the AC on does not really start <br />the rotation, we manually started it, and only then, if we gave a strong<br /> push, it started to rotate and maintained this rotation.<br /><br /><br /><br />Let's use a wire loop instead of a permanent magnet rotor. But, to act <br />as a permanent magnet, it needs an electric current running through the <br />wire, and we don't want any extra sources of electricity to connect to <br />it, it's complicated, needs a commutator, like in the original DC <br />motors.<br /><br />Instead, we will count on the induction effect created by an alternating<br /> current in a stator, which creates an alternating magnetic field, <br />which, in turn, induces the electric current in the wire loop of a <br />rotor.<br /><br /><br /><br />The induced electric current in a wire loop of our rotor appears when a <br />wire crosses the magnetic field lines. Magnetic field produced by a <br />stator is directed always along a center line XY and is changing in <br />magnitude from a maximum in one direction (from X to Y) to zero, to a <br />maximum in another direction (from Y to X), again to zero etc.<br /><br /><br /><br />The problem is, if the rotor is standing still, its wire does not cross <br />magnetic field lines. Therefore, no electric current will be produced in<br /> it, and there will be no rotation. Rotor needs an initial push <br />sufficient to synchronize its rotation with the frequency of alternating<br /> magnetic field to continue the rotation. Our second disadvantage is <br />still unresolved.<br /><br /><br /><br />Let's imagine that, besides two main electromagnets with poles X and Y, <br />we have another pair of auxiliary electromagnets in a stator, that are <br />positioned perpendicularly to line XY. Let their poles be A and B and <br />(very important!) the current in them, also sinusoidal, is shifted in <br />time relatively to a current in main electromagnets by 90°.<br /><br />So, when the magnetic field in one direction is maximum along line XY, <br />it's zero along line AB; then it gradually decreases along XY and <br />increases along AB until it's zero along XY and maximum along AB. This <br />cycle repeats itself indefinitely.<br /><br /><br /><br />Graphically, it can be represented as follows.<br /><br /><img src="http://www.unizor.com/Pictures/ACrotation.png" style="height: 300px; width: 200px;" /><br /><br />Graphs on the left and on the right of a picture represent the electric <br />current in each pair of electromagnets - XY and AB. Notice, they are <br />shifted by π/2=90°. The middle part of a picture represents the <br />direction of the magnetic field created by both pairs of electromagnets <br />of a stator.<br /><br />As the time goes, the magnetic field direction is rotating. The rotating<br /> magnetic field around a closed wire loop of a rotor causes the rotor's <br />wire to cross the magnetic field lines of a stator, which, in turn, <br />causes induction of electric current in a rotor's closed wire loop, <br />which will act now as a permanent magnet, which is supposed to follow <br />the rotation of the magnetic field around it.<br /><br />This causes the rotation of the rotor. The rotor's actual physical <br />rotation will follow the stator's magnetic field rotation, created <br />without any physical movement.<br /><br /><br /><br />The rotor cannot go after the stator's magnetic field rotation exactly <br />synchronously because then there will be no crossing of magnetic field <br />lines. So, the rotor, after it reached the same speed as the stator's <br />magnetic field rotation, will lose its rotational momentum and slow <br />down. Then the rotor's wire, rotating slower than the magnetic field <br />around it, will cross the magnetic field lines, there will be induction <br />current in it, it's magnetic properties will be restored and it will try<br /> to catch up with the stator's magnetic field rotation. This process <br />will continue as long as AC is supplied.<br /><br /><br /><br />An obvious improvement is to use multiple wire loops as a rotor with <br />common axis of rotation. That will cause more uniform rotation.<br /><br /><br /><br />What's remaining in our project of designing the AC motor is to create <br />another alternating current for the second pair of electromagnets in a <br />stator with a time delay of π/2=90° relatively to the main AC.<br /><br />This problem is resolved and described in the previous lecture about AC <br />capacitors. If we introduce another circuit fed from the same AC source,<br /> but with a capacitor in it, this circuit will have the alternating <br />current shifted in time exactly as we want. This current will go through<br /> the second pair of electromagnets and both pairs will create a <br />revolving magnetic field.<br /><br /><br /><br />The interaction between magnetic fields of a stator and a rotor is quite<br /> complex, when the rotor rotates. The magnetic flux going through a <br />rotor wire loop depends on a variable magnetic field of a stator and <br />variable area of a rotor wire loop in a direction of a magnetic field <br />lines of a stator. The exact calculations of this process are beyond the<br /> scope of this course.<br /><br />The main idea, however, is clear - <b>to create an AC motor we have to create a revolving magnetic field</b> - the great idea of Nicola Tesla. That can be accomplished by using known methods described above.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-75921051130765982892020-08-19T08:47:00.001-07:002020-08-19T08:47:43.493-07:00DC Motors: UNIZOR.COM - Physics4Teens - Electromagnetism - Magnetism of ...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Direct Current Motors</u><br /><br /><br /><br /><i>Direct current (DC)</i> motors are used where the sources of electricity produce direct current, like batteries.<br /><br />For example, DC motors work in car engine starter, computers, toys, drones and many other devices.<br /><br /><br /><br />We will concentrate on principles of their work without going into many <br />details. Basically, we will describe these motors as they were first <br />thought of by their inventors. Obviously, initial ideas were further <br />improved by many people and improvements are still introduced after more<br /> than 200 years after their invention, but the basic ideas are still <br />there.<br /><br /><br /><br />Let's start with the idea of the DC motor.<br /><br />The beginning of a development of a DC motor is associated with a simple<br /> experiment we described before - the one that demonstrated the Lorentz <br />force on a current in a magnetic field.<br /><br />If the direction of a current <i><b>I</b></i> is perpendicular to a direction of magnetic field lines <i><b>B</b></i>,<br /> the magnetic field "pushes" the conductor that carries this current in a<br /> direction perpendicular to both magnetic field lines and a current by <br />force <i><b>F</b></i>.<br /><br /><img src="http://www.unizor.com/Pictures/LorentzForce.jpg" style="height: 180px; width: 200px;" /><br /><br /><br /><br />The first modification of this experiment was to use a rectangular wire <br />frame instead of a straight line conductor and let it spin around an <br />axis. The force exerted by a magnetic field will act on both sides of a <br />wire frame that are perpendicular to magnetic lines in opposite <br />directions, creating a rotational momentum, so the frame with rotate.<br /><br /><img src="http://www.unizor.com/Pictures/WireLoopRotation.jpg" style="height: 120px; width: 200px;" /><br /><br />The force <i><b>F</b></i> of a magnetic field "pushes" segment <i><b>AB</b></i> up and segment <i><b>CD</b></i>, where the electric current goes in an opposite to segment <i><b>AB</b></i> direction, is "pushed" down by this force. This creates a rotational momentum.<br /><br /><br /><br />The problem is, the rotation will stop when the wire frame plane will be<br /> perpendicular to a direction of a magnetic field, because the forces <br />exerted by a magnetic field on both sides of a wire frame <i><b>AB</b></i> and <i><b>CD</b></i><br /> will no longer create a rotating momentum, they will act against each <br />other within the wire frame plane and after a short oscillation caused <br />by a rotational inertia the rotation stops.<br /><br /><br /><br />On the picture above, when the plane of a wire loop reaches the position perpendicular to the magnetic field force, the segment <i><b>AB</b></i> will be on the top and the magnetic field force <i><b>F</b></i> will "push" this segment up. Segment <i><b>CD</b></i> will be on the bottom and the magnetic field force <i><b>F</b></i><br /> will "push" this segment down. Both forces are acting within the same <br />plane with the axis of rotation and nullify each other. No rotational <br />momentum is created.<br /><br /><br /><br />Our next improvement is related to overcoming this problem.<br /><br />What happens if exactly at the moment when our wire frame plane is <br />perpendicular to magnetic field, when the forces exerted by a magnetic <br />field no longer create a rotational moment, we switch off the electric <br />current in a wire and a very short moment later we switch it on, but in <br />the opposite direction of the electric current?<br /><br />First of all, the rotation will continue by inertia during the time <br />electricity is off. Then, we turn the electric current on but in <br />opposite direction. That means, the direction of the magnetic field <br />force will change to an opposite. Segment <i><b>AB</b></i> will be "pushed" down, segment <i><b>CD</b></i><br /> will be "pushed" up. Since the wire frame has passed the point of <br />perpendicularity to magnetic field by rotational inertia during electric<br /> current off time, the newly formed magnetic field forces will create a <br />rotational moment and the direction of rotation will be the same as <br />before.<br /><br /><br /><br />Our frame will continue the rotation in the same direction until segment <i><b>AB</b></i> will be on the bottom and segment <i><b>CD</b></i> - on the top.<br /><br />At that moment we will do the same switching off the electric current to<br /> let our frame pass the perpendicular position towards the magnetic <br />field force, and a very short moment later we switch electric current on<br /> in an opposite direction. Now segment <i><b>AB</b></i> will be "pushed" up again, segment <i><b>CD</b></i> will be "pushed" down, which will allow to complete the cycle of rotation and the rotation will continue in the same direction.<br /><br /><br /><br />All we need is to explain how to switch the direction of an electric current.<br /><br />Here is a drawing of this simple device called <i>commutator</i>.<br /><br /><img src="http://www.unizor.com/Pictures/DCcommutator.jpg" style="height: 111px; width: 200px;" /><br /><br />Direct current comes through brushes to a commutator and to a revolving <br />frame connected to it. As this construction revolves, the brushes lose <br />the contact with a commutator for a very short period, then again touch <br />the commutator, but with opposite poles.<br /><br />That's how we change the direction of the current that forces the frame to constantly rotate in the same direction.<br /><br /><br /><br />The inner rotating part of this type of a DC motor is called <i>rotor</i>, the outer stationary part with a magnet is called <i>stator</i>.<br /><br />The detail implementation of this design of a DC motor is outside the scope of this course. Our purpose is to convey the idea.<br /><br /><br /><br />The weak part of a DC motor design with brushes is that these brushes <br />wear out with time and are the source of sparks, which might be <br />prohibitive in some environments. Plus, they are noisy.<br /><br />With development of electronics inventors came up with a better design <br />that does not involve brushes at all, rotation is accomplished without <br />any mechanical switches.<br /><br /><br /><br />First step on the design of this brushless DC motor is to realize that <br />we have to invert the functionality of permanent magnet and a wire <br />frame. The need for a <i>commutator</i> with mechanical brushes was related to the fact that electric circuit was rotating.<br /><br />Let's invert roles and put two parallel wire frames on opposite ends of a circle as a <i>stator</i> and a permanent magnet on an axis in the middle as a <i>rotor</i>.<br /><br />Recall that a wire loop with direct current running through it acts like a magnet<br /><br /><img src="http://www.unizor.com/Pictures/LoopMagnet.jpg" style="height: 200px; width: 200px;" /><br /><br />Putting two such loops on opposite ends of a stator creates a magnetic <br />field and a freely rotating permanent magnet in between will have to <br />turn to align itself with the field of these two wire loops<br /><br /><img src="http://www.unizor.com/Pictures/DCmotorBrushless.png" style="height: 200px; width: 200px;" /><br /><br />If at the moment our permanent magnet aligns with magnetic field of both<br /> wire loops we change the direction of electric current in the loops to <br />an opposite, thus changing the polarity of the magnetic field they <br />generate, the permanent magnet in between will have to continue the <br />rotation until it will align again. Repeating this cycle creates the <br />rotation of a permanent magnet.<br /><br /><br /><br />This rotation will not be smooth. The force of magnetic field attraction<br /> will create a stronger moment of rotation when the direction of the <br />permanent magnet is perpendicular to a direction of the magnetic field <br />of the loops. This can be improved by certain additional details <br />described below.<br /><br /><br /><br />The changing of the direction of the current in the wire loops now is <br />much easier than in the previous design because the wires are not <br />moving. Simple electronic switch working off some kind of a marker on a <br />rotating permanent magnet can signal its position and trigger the switch<br /> of direction of the current.<br /><br /><br /><br />Basically, the idea is finished here. But some very important improvements should be mentioned.<br /><br /><br /><br />1. Instead of a single wire loop we can use a copper coil around an iron<br /> core to make the magnetic field of this electromagnet stronger.<br /><br /><br /><br />2. Instead of a pair of such electromagnets with switching the direction<br /> after the permanent magnet in the middle turns by 180°, we can use two <br />pairs and properly engage another pair after 90° turn, switching off the<br /> previous pair, thus creating a rotating magnetic field that results in a<br /> smoother movement of a permanent magnet rotor. Even better, we can use <br />three pairs of electromagnets and engage another pair after each 60°, <br />switching off the previous two pairs. This will allow even smoother <br />rotation. In some DC motors they use even six pairs of electromagnets, <br />which results in a very uniform rotation with practically constant <br />angular speed.<br /><br />The electronic switches that engage and disengage the coils can be designed for any type of coil arrangement.<br /><br /><br /><br />3. Instead of a bar permanent magnet inside the circle of electromagnets<br /> we can use a ring permanent magnet outside it. It creates a better <br />response to a revolving magnetic field of a stator, the rotation will be<br /> smoother because of inertia of a rotating ring.<br /><br /><br /><br />With all the above improvements the DC motor used in hard disk of computers and other devices looks like the one below<br /><br /><img src="http://www.unizor.com/Pictures/DCmotorReal.jpg" style="height: 480px; width: 210px;" /><br /><br />This DC motor has 9 electromagnets sequentially engaged after each 40° turn of a rotor.<br /><br />Rotor is a ring-shaped permanent magnet that rotates around the electromagnets.<br /><br />A marker on the rotor sends a signal to an electronic switch to indicate<br /> the position of a rotor, which is used by electronics on the attached <br />board to properly engage the electromagnets of a stator.<br /><br /><br /><br />The most important detail of this design of a DC motor is the rotating <br />magnetic field achieved through purely electronic means without any <br />moving parts. This results in a smooth rotation of a rotor made of a <br />ring permanent magnet around a stator.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-82782206556858281582020-08-15T13:26:00.001-07:002020-08-15T13:26:02.359-07:00AC Capacitors: UNIZOR.COM - Physics4Teens - Electromagnetism - AC Induction<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/-q4WLYIEdqw" width="480"></iframe><br /><br /><br /><br /><u><i>Notes to a video lecture on http://www.unizor.com</i></u><br /><br /><br /><br /><u>Alternating Current and Capacitors</u><br /><br /><br /><br />Let's start with a clear illustration of what we will discuss in this lecture.<br /><br /><img src="http://www.unizor.com/Pictures/ACcapacitor.png" style="height: 300px; width: 200px;" /><br /><br />The main idea is that alternating current (AC) goes through a capacitor, while direct current (DC) does not.<br /><br /><br /><br />To explain this fact, consider how electricity is produced in both cases DC and AC.<br /><br /><br /><br />For DC case we will use a battery that uses some chemical reaction to <br />separate some outer layer ("free") electrons from their atoms and moves <br />them toward a battery's negative pole, where excess of electrons will be<br /> present, leaving deficiency of electrons on the positive pole.<br /><br /><br /><br />If we connect the poles of this battery to two plates of a capacitor, <br />the excess of electrons from the negative pole of a battery will travel <br />to one plate, while the deficiency of electrons will be on the other <br />plates. It will continue until we reach a saturation of electrons on one<br /> plate and their absence on another. The electromotive power of the <br />battery could not push electrons anymore to one plate, because their <br />repulsive force will be too strong. And that's it. Electrons will not <br />jump from one plate of a capacitor to another (unless the EMF of a <br />battery is so strong that it breaks the isolation between the plates of a<br /> capacitor, which defeats the purpose), the circuit is not closed, the <br />current will not go through.<br /><br /><br /><br />In AC case the generated electromotive force (EMF) is variable in magnitude and direction.<br /><br /><br /><br />It starts with separating "free" electrons from their atoms, gradually <br />increasing the concentration of electrons on one pole and on the <br />capacitor's plate connected to it.<br /><br />At the same time deficiency of electrons is observed on the other pole and a capacitor's plate connected to it.<br /><br /><br /><br />That lasts for some short period of time (half a period), when the <br />generated EMF gets weaker and goes down to zero. This causes the excess <br />electrons from one capacitor plate to go back to a generator and <br />compensate the deficiency of electrons on the other plate. This ends <br />half a period of EMF oscillation.<br /><br /><br /><br />On the next half a period the EMF changes the sign, electrons will be <br />accumulated on the plate, where previously we observed their deficiency.<br /> On the plate where we had excess of electrons during the previous half a<br /> period we will have their deficiency.<br /><br /><br /><br />So, during the full cycle of two half-periods electrons move to one <br />plate of a capacitor, then back to a generator and to another plate of a<br /> capacitor. Then the cycle repeats itself, electrons move within a <br />circuit with a capacitor back and forth, thus facilitating the <br />alternating current in this circuit.<br /><br /><br /><br />Let's compare the alternating current in a case of a closed circuit without a capacitor with a circuit that has a capacitor.<br /><br />We start in both cases with a device that generates alternating sinusoidal EMF<br /><br /><i><b>U(t) = U<sub>max</sub>·sin(ωt)</b></i><br /><br />where<br /><br /><i><b>U(t)</b></i> is the generated EMF as a function of time <i><b>t</b></i>,<br /><br /><i><b>U<sub>max</sub></b></i> is the peak value of EMF,<br /><br /><i><b>ω</b></i> is the angular speed of rotation of a device generating the alternating EMF.<br /><br /><br /><br />In a closed circuit with an alternating EMF and no capacitors, according to the Ohm's Law, the electric current <i><b>I(t)</b></i>, as a function of time <i><b>t</b></i>, will be proportional to an EMF and will alternate synchronously with it<br /><br /><i><b>I(t) = I<sub>max</sub>·sin(ωt)</b></i><br /><br />where<br /><br /><i><b>I<sub>max</sub></b></i> is the peak value of electric current in a circuit.<br /><br /><br /><br />In case of a capacitor being a part of a circuit the situation is more complex.<br /><br />As we know, the amount of electricity <i><b>Q(t)</b></i> accumulated in the capacitor is proportional to voltage <i><b>U(t)</b></i> applied to its plates, and <i>capacity</i> of a capacitor <i><b>C</b></i><br /> (see lecture "Electric Fields" - "Capacitors" in this course) is the <br />constant proportionality factor that depends on a type of a capacitor<br /><br /><i><b>C = Q(t)/U(t)</b></i><br /><br /><br /><br />Therefore,<br /><br /><i><b>Q(t) = C·U(t) = C·U<sub>max</sub>·sin(ωt)</b></i><br /><br />Knowing the amount of electricity <i><b>Q(t)</b></i> accumulated in a capacitor as a function of time <i><b>t</b></i>, we can determine the electric current <i><b>I(t)</b></i> in a circuit, which is a rate of change (that is, derivative by time) of the amount of electricity<br /><br /><i><b>I(t) = </b>d<b>Q(t)/</b>d<b>t =<br /><br />= C·U<sub>max</sub>·ω·cos(ωt) =<br /><br />= I<sub>max</sub>·cos(ωt) =<br /><br />= I<sub>max</sub>·sin(ωt+π/2)</b></i><br /><br />where<br /><br /><i><b>I<sub>max</sub> = C·U<sub>max</sub>·ω</b></i><br /><br /><br /><br />Incidentally, expression which is called <i>reactance</i>, plays for a capacitor similar role to that of a resistance because, using this <i>reactance</i>, the formula for a peak value of the electric current in a circuit with a capacitor resembles the Ohm's Law<br /><br /><i><b>I<sub>max</sub> = U<sub>max</sub>/X<sub>C</sub></b></i><br /><br /><br /><br />What's most important in the formula <i><b>I(t)=I<sub>max</sub>·sin(ωt+π/2)</b></i><br /> and the most important property of a capacitor in an AC circuit is <br />that, while an EMF in a circuit oscillates with angular speed <i><b>ω</b></i>, the <b>electric current oscillates with the same angular speed <i><b>ω</b></i> as EMF, but its period is shifted in time by <i>π/2</i> relative to EMF</b>.<br /><br /><br /><br />This is a very important capacitor's property used in AC motors, which will be a subject of the next lecture.<br /><br /><nobr><i><b>X<sub>C</sub> = 1/(ω·C)</b></i>,</nobr>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-40804915874119424402020-08-13T09:44:00.001-07:002020-08-13T09:44:03.855-07:00AC Induction: UNIZOR.COM - Physics4Teens - Electromagnetism<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Alternating Current Induction</u><br /><br /><br /><br />The Faraday's Law of electromagnetic induction states that a <b>variable <i>magnetic flux Φ(t)</i> going through a wire loop generates in this wire loop an <i>electromotive force EMF=U(t)</i> equal in magnitude to a rate of change of the magnetic flux</b>:<br /><br /><i><b>U(t) = −</b>d<b>Φ(t)/</b>d<b>t</b></i><br /><br />where the meaning of the minus sign was explained in the lecture about <i>self-induction</i> previously in this course.<br /><br /><br /><br />There are different ways to generate a variable magnetic flux. For <br />example, we can physically move a permanent magnet through or around a <br />wire loop, which will generate EMF in the wire.<br /><br />But, having <i>alternating current (AC)</i> at our disposal and knowing <br />that any electric current creates a magnetic field around it, we can <br />create a variable magnetic field just by running AC through a wire.<br /><br /><br /><br />Let's start from a simple experiment just to demonstrate the principle of AC induction.<br /><br /><img src="http://www.unizor.com/Pictures/ACinduction.png" style="height: 225px; width: 200px;" /><br /><br />Here the bottom wire loop (we will call it <i>primary</i>) is connected to a source of an <i>alternating current</i>, while the top wire loop (<i>secondary</i>)<br /> has no source of electricity, but is connected to a device to measure <br />the effective electric current in it - an AC ammeter (or amp-meter, or <br />ampermeter).<br /><br /><br /><br />Since the AC in the primary loop is variable (more precisely, sinusoidal), the <i>intensity</i> of the magnetic field generated by it is also variable.<br /><br /><br /><br />Therefore, the <i>magnetic flux</i> going through the secondary loop is <br />variable (also sinusoidal) with the generated EMF (first derivative of <br />magnetic flux) not equal to zero and also sinusoidal since the <br />derivative of <i>sin(x)</i> is <i>cos(x)</i> and derivative of <i>cos(x)</i> is <i>−sin(x)</i>, as we know.<br /><br /><br /><br />Variable (sinusoidal) EMF in the secondary loop causes the variable <br />(sinusoidal) electric current in it, which we can observe on the AC <br />ammeter.<br /><br />We have just demonstrated the principle of <i>AC induction</i> - <br />transformation of the alternating electric current from one circuit <br />(primary loop) to another (secondary loop) without physical connection <br />between them.<br /><br /><br />Let's switch to more practical aspects of AC induction.<br /><br /><br /><br />Recall from the earlier presented lecture "Magnetism of Electric Current in a Loop" that the intensity <i><b>B</b></i> of a magnetic field at the center of a wire loop in a vacuum equals to<br /><br /><i><b>B = μ<sub>0</sub>·I/(2R)</b></i><br /><br />where<br /><br /><i><b>μ<sub>0</sub></b></i> is the <i>permeability</i> of free space,<br /><br /><i><b>I</b></i> is the electric current going through a wire loop (maybe variable as in the case of AC),<br /><br /><i><b>R</b></i> is the radius of a wire loop.<br /><br /><br /><br />If, instead of a vacuum, we have some other material inside a loop, the <br />intensity of a magnetic field will be different. The formula will be <br />almost the same, just the <i>permeability</i> of free space <i><b>μ<sub>0</sub></b></i> should be replaced with <i>permeability</i> of the corresponding material <i><b>μ</b></i>.<br /><br /><br /><br />Experimentally obtained data tell us that one of the greatest <i>permeability</i><br /> is that of so-called ferromagnetic materials - those that contain iron.<br /> Actually, pure iron has the permeability of about 200,000 greater than <br />vacuum. So, if we have a ferromagnetic material inside the primary wire <br />loop in our experiment, the magnetic energy will have a choice - to go <br />through highly resistant space around this ferromagnetic material or to <br />go through it with much less resistance.<br /><br /><br /><br />The general principle of field propagation is that the most energy goes <br />through least resistance. So, we can expect that ferromagnetic material <br />inside the wire loop will concentrate the magnetic field generated by a <br />primary wire loop and much less of the field energy will be dissipated <br />around it.<br /><br /><br /><br />Our first improvement to the experiment above is to use a cylinder made of <i>ferromagnetic</i> material going through both our loops.<br /><br /><img src="http://www.unizor.com/Pictures/ACinductionFerro.png" style="height: 225px; width: 200px;" /><br /><br />Ferromagnetic material has temporary magnetic properties and quickly <br />reacts to external magnetic field by rearranging the orientation of its <br />atoms. The variable magnetic field created by AC inside the primary loop<br /> will force the atoms inside this cylinder to react by orienting along <br />the vector of intensity of the magnetic field, thus increasing this <br />intensity.<br /><br />So, the purpose of this ferromagnetic cylinder going through both loops <br />is to increase the intensity of the magnetic field going through the <br />secondary wire loop without spending any more energy. The ferromagnetic <br />cylinder just concentrates and directs more of the energy of the <br />magnetic field to go through the secondary wire loop and less will be <br />dissipated outside the secondary loop.<br /><br /><br /><br />Still, the energy of magnetic field generated by the primary wire loop <br />and directed through the ferromagnetic cylinder through both loops will <br />be dissipated from the ends of this cylinder.<br /><br />To make our induction even more efficient, we can close the magnetic <br />field on the ends of a ferromagnetic cylinder as on this picture:<br /><br /><img src="http://www.unizor.com/Pictures/ACinductionClosed.png" style="height: 150px; width: 200px;" /><br /><br />Now the magnetic field is not dissipated outside the ferromagnetic <br />material, and almost all energy generated by the primary wire loop with <br />AC running through it is used to create the magnetic flux inside the <br />secondary wire loop.<br /><br /><br /><br />Our next practical improvement is related to the fact that the vector of<br /> intensity of a magnetic field is an additive function. That is, the <br />intensity of a combined magnetic field of two or more sources equals to a<br /> vector sum of intensities of all component fields.<br /><br /><br /><br />We can use multiple loops in the primary coil, which is a source of <br />magnetic field, instead of a single primary wire loop in our experiment.<br /><br />Let's wind <i>N<sub>P</sub></i> primary loops of wire in a coil around <br />our ferromagnetic core. Connected to the same source of AC and having <br />the same sinusoidal current going through them, each loop will create a <br />magnetic field and all these <i>N<sub>P</sub></i> fields will have to be added together to get the resulting magnetic field intensity.<br /><br /><br /><br />So, the generated magnetic flux of such a coil will be <i>N<sub>P</sub></i> times greater than the one by a single loop. Its inductive effect on the secondary wire loop will be <i>N<sub>P</sub></i> times stronger.<br /><br /><br /><br />As the next practical improvement of our experiment, let's use a coil of <i>N<sub>S</sub></i> wire turns instead of a single secondary loop.<br /><br />Each turn of this secondary winding will have the same magnetic flux <br />flowing through it and will induce an electromotive force (EMF) <br />proportional to a rate of change of magnetic flux. All these EMF in each<br /> turn of a secondary wire are sequential to each other and should be <br />added together to get the resulting EMF in the secondary circuit.<br /><br />So, the EMF induced in a secondary coil of <i>N<sub>S</sub></i> wire turns will be <i>N<sub>S</sub></i> stronger then the EMF induced in a single secondary loop.<br /><br /><br /><br />So, changing the number of turns in the primary wire coil <i>N<sub>P</sub></i>,<br /> we proportionally change the variable magnetic flux, which, in turn, <br />proportionally change the induced EMF in the secondary coil.<br /><br />Analogously, changing the number of turns in the secondary wire coil <i>N<sub>S</sub></i>, we proportionally change the induced EMF in this wire coil.<br /><br />These two facts lead us to use the device we have constructed to <br />transform the EMF of alternating current for whatever purpose we need.<br /><br /><br /><br />Indeed, if <i>N<sub>P</sub></i>=<i>N<sub>S</sub></i> and our device is <br />ideal in a sense that the energy does not dissipate and the coils are <br />not heated up because of internal resistance, the same magnetic flux <br />goes through the same coils, which means the EMF in both primary and <br />secondary coils is the same.<br /><br /><br /><br />If we want to induce the secondary EMF twice as stronger as the primary <br />one, we can double the number of turns in the secondary coil.<br /><br />If we want to induce the secondary EMF twice as weaker as the primary <br />one, we can reduce the number of turns in the secondary coil by half.<br /><br /><br /><br />This can be expressed as<br /><br /><i><b>U<sub>P</sub>(t)/U<sub>S</sub>(t) = N<sub>P</sub>/N<sub>S</sub></b></i><br /><br />where<br /><br /><i>U<sub>P</sub>(t)</i> is the EMF in the primary coil,<br /><br /><i>U<sub>S</sub>(t)</i> is the EMF in the secondary coil,<br /><br /><i>N<sub>P</sub></i> is the number of wire turns in the primary coil,<br /><br /><i>N<sub>S</sub></i> is the number of wire turns in the secondary coil.<br /><br /><br /><br />But changing the EMF has its consequences.<br /><br />The Conservation of Energy Law is still supposed to be held. The power <br />produced by AC in the primary coil in an ideal transformer is completely<br /> transferred to the secondary coil, which means<br /><br /><i><b>P(t) = U<sub>P</sub>(t)·I<sub>P</sub>(t) = U<sub>S</sub>(t)·I<sub>S</sub>(t)</b></i><br /><br />where<br /><br /><i>P(t)</i> is the power produced by primary and transferred to secondary coil, variable for AC,<br /><br /><i>U<sub>P</sub>(t)</i> is the EMF in the primary coil,<br /><br /><i>I<sub>P</sub>(t)</i> is the electric current in the primary coil,<br /><br /><i>U<sub>S</sub>(t)</i> is the EMF in the secondary coil,<br /><br /><i>I<sub>S</sub>(t)</i> is the electric current in the secondary coil.<br /><br /><br /><br />From the equation above follows that increasing the EMF in the secondary<br /> coil causes proportional decreasing in the electric current in it and <br />vice versa:<br /><br /><i><b>U<sub>P</sub>(t)/U<sub>S</sub>(t) = I<sub>S</sub>(t)/I<sub>P</sub>(t)</b></i><br /><br /><br /><br />The formulas above are the basic principle of working of transformers - <br />devices used to increase the AC voltage, while proportionally decrease <br />the electric current and, therefore, decrease losses related to <br />transmission of electricity along long wires and also used to decrease <br />this voltage, while proportionally increase the electric current at the <br />location of homes and electric devices that need the lower voltage in <br />the outlets with higher amperage.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-26067326828103625712020-08-07T09:42:00.001-07:002020-08-07T09:42:46.858-07:00AC - Effective Voltage: UNIZOR.COM - Physics4Teens - Electromagnetism<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Effective Voltage<br />and Current (RMS)</u><br /><br /><br /><br />We all know the voltage in our outlets at home. In some countries it's <br />220V, in others 110V or some other. But we also know that electric <br />current distributed to consumers from power plants is <i>alternating (AC)</i>.<br /><br />The actual value of <i>voltage</i> and <i>amperage</i> are sinusoidal, changing with time as:<br /><br /><i><b>U(t) = U</b><sub>max</sub><b>·sin(ωt)</b></i><br /><br /><i><b>I(t) = U(t)/R = I</b><sub>max</sub><b>·sin(ωt)</b></i><br /><br />where<br /><br /><i><b>ω</b></i> is some parameter related to the generation of electricity at the power plant,<br /><br /><i><b>t</b></i> is time,<br /><br /><i><b>U</b><sub>max</sub></i> is the voltage <i>amplitude</i>,<br /><br /><i><b>I</b><sub>max</sub><b> = U</b><sub>max</sub><b>/R</b></i> is the amperage <i>amplitude</i>,<br /><br /><i><b>R</b></i> is resistance of the circuit where <i>alternating current</i> is running through.<br /><br /><br /><br />If <i>voltage</i> is variable, what does it mean that in the outlet it is, for example, 220V?<br /><br /><br /><br />The answer to this question is in comparing energy the electric current is carrying in case of a variable <i>voltage</i> with energy of a <i>direct current</i>.<br /><br /><br /><br />The AC <i>voltage</i> is sinusoidal, so we can calculate the energy it carries through a circuit during the time of its <i>period</i> <i><b>T=2π/ω</b></i> and calculate the corresponding DC <i>voltage</i> that carries the same amount of energy during the same time through the same circuit.<br /><br />That corresponding value of the DC <i>voltage</i> is, <u>by definition</u>, the <i>effective voltage</i> in the AC circuit, which is usually called just <i>voltage</i> for alternating current.<br /><br /><br /><br />First, let's evaluate how much energy the AC with voltage <i><b>U(t)=U</b><sub>max</sub><b>·sin(ωt)</b></i> carries through some circuit of resistance <i><b>R</b></i> during the time of its period <i><b>T=2π/ω</b></i>.<br /><br /><br /><br />Consider an infinitesimal time interval from <i><b>t</b></i> to <i><b>t+</b>d<b>t</b></i>.<br /> During this interval we can consider the voltage and amperage to be <br />constant and, therefore, use the expression of the energy flowing <br />through a circuit of resistance <i><b>R</b></i> for direct current:<br /><br /><i>d<b>W(t) = U(t)·I(t)·</b>d<b>t = U²(t)·</b>d<b>t<span style="font-size: medium;">/</span>R</b></i><br /><br />(see "Electric Heat" topic of this course and "Ohm's Law" for direct current)<br /><br /><br /><br />To calculate the energy going through a circuit during any period <i><b>T=2π/ω</b></i>, we have to integrate this expression for <i>d<b>W</b></i> on an interval from <i><b>0</b></i> to <i><b>T</b></i>.<br /><br /><i><b>W</b></i><sub>[0,T]</sub> = <b><span style="font-size: large;">∫</span></b><sub>[0,T]</sub><i>d<b>W(t) = </b></i><br /><br /><b>= <span style="font-size: large;">∫</span></b><sub>[0,T]</sub><i><b>U²(t)·</b>d<b>t<span style="font-size: medium;">/</span>R</b></i> =<br /><br /><b>= <span style="font-size: large;">∫</span></b><sub>[0,T]</sub><i><b>U²</b><sub>max</sub><b>·sin²(ωt)·</b>d<b>t<span style="font-size: medium;">/</span>R</b></i> =<br /><br />= (<i><b>U²</b><sub>max</sub><b><span style="font-size: medium;">/</span>R</b></i>)·<b><span style="font-size: large;">∫</span></b><sub>[0,T]</sub><i><b>sin²(ωt)·</b>d<b>t</b></i><br /><br /><br /><br />To simplify the integration we will use the known trigonometric identity<br /><br /><i><b>cos(2x) = cos²(x) − sin²(x) = 1 − 2sin²(x)</b></i><br /><br />from which follows<br /><br /><i><b>sin²(x) = </b></i>[<i><b>1 − cos(2x)</b></i>]<i><b> <span style="font-size: medium;">/</span>2</b></i><br /><br /><br /><br />Next we will do the substitution<br /><br /><i><b>x = ω·t</b></i><br /><br /><i><b>t = x/ω</b></i><br /><br /><i>d<b>t = </b>d<b>x/ω</b></i><br /><br /><i><b>t</b></i> ∈ [0,T] <span style="font-size: medium;">⇒</span> <i><b>x</b></i> ∈ [0,ωT]=[0,2π]<br /><br /><br /><br />The integral above, expressed in terms of <i><b>x</b></i>, is<br /><br /><b><span style="font-size: large;">∫</span></b><sub>[0,2π]</sub><i><b>sin²(x)·</b>d<b>x/ω =<br /><br />= </b></i><b><span style="font-size: large;">∫</span></b><sub>[0,2π]</sub>[<i><b>1−cos(2x)</b></i>]<i><b>·</b>d<b>x/(2ω) =<br /><br />= </b></i>[<i><b>2π−</b></i><span style="font-size: large;">∫</span><sub>[0,2π]</sub><i><b>cos(2x)·</b>d<b>x</b></i>]<i><b>/(2ω) =<br /><br />= π/ω</b></i><br /><br />Therefore,<br /><br /><i><b>W</b></i><sub>[0,T]</sub><i><b> = π·U²</b><sub>max</sub><b><span style="font-size: medium;">/</span>(ω·R)</b></i><br /><br /><br /><br />At the same time we have defined the <i>effective</i> voltage as the one that delivers the same amount of energy as <i><b>W</b></i><sub>[0,T]</sub> to the same circuit of resistance <i><b>R</b></i> during the same time <i><b>T=2π/ω</b></i> if the current is constant and direct.<br /><br />That is<br /><br /><i><b>U²</b><sub>eff</sub><b>·2π/(R·ω) = W</b></i><sub>[0,T]</sub><i><b> = π·U²</b><sub>max</sub><b><span style="font-size: medium;">/</span>(ω·R)</b></i><br /><br /><br /><br />Cancelling and simplifying the above equality, we conclude<br /><br /><i><b>2·U²</b><sub>eff</sub><b> = U²</b><sub>max</sub></i><br /><br />or<br /><br /><i><b>√<span style="text-decoration: overline;">2</span>·U</b><sub>eff</sub><b> = U</b><sub>max</sub></i><br /><br />For <i>effective</i> voltage 110V the peak voltage (amplitude) is 156V,<br /><br />for <i>effective</i> voltage 120V the peak voltage (amplitude) is 170V,<br /><br />for <i>effective</i> voltage 220V the peak voltage (amplitude) is 311V.<br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0