## Wednesday, October 7, 2020

### Series LC Circuit: UNIZOR.COM - Physics4Teens - Electromagnetism - Alter...

Notes to a video lecture on http://www.unizor.com

Series LC Circuit

Consider a circuit that contains an AC generator, an inductor of inductance L and a capacitor of capacity C in a series. The current IL(t) going through an inductor is the same as the current IC(t) going through a capacitor. So, we will use an expression I(t) for both.

The electromotive force (EMF) generated by an AC generator depends only
on its own properties and can be described as a function of time t

E(t) = E0·sin(ωt)

where

E0 is a peak voltage on the terminals of a generator,

ω is an angular velocity of a rotor in radians per second.

Inductance L of an inductor and capacity C of a capacitor produce voltage drops VL(t) and VC(t) correspondingly.

As we know, the voltage drop on an inductor is causes by self-induction
and depends on the rate of change (that is, the first derivative by
time) of a magnetic flux Φ(t) going through it

VL(t) = dΦ(t)/dt

Magnetic flux, in turn, depends on a current going through a wire of an inductor I(t) and the inductor's inductance L

Φ(t) = L·I(t)

Therefore, the voltage drop on an inductor equals to

VL(t) = L·dI(t)/dt

As we know, the amount of electricity Q(t) accumulated in a capacitor is proportional to voltage VC(t) applied to its plates (that is, voltage drop on a capacitor). The constant capacity of a capacitor C
is the coefficient of proportionality (see lecture "Electric Fields" -
"Capacitors" in this course) that depends on a type of a capacitor

C = Q(t)/VC(t)

Therefore,

Q(t)=C·VC(t)

Knowing the amount of electricity Q(t) accumulated in a capacitor as a function of time t, we can determine the electric current I(t) in a circuit, which is a rate of change (that is, the first derivative by time) of the amount of electricity

I(t) = dQ(t)/dt = C·dVC(t)/dt

The sum of voltage drops on an inductor and a capacitor is supposed to be equal to EMF produced by an AC generator E(t)=E0·sin(ωt), which is the final equation in our system:

VL(t) = L·dI(t)/dt

I(t) = C·dVC(t)/dt

E0·sin(ωt) = VL(t) + VC(t)

To solve this system of three equations, including two differential ones, let's resolve the third equation for VC(t) and substitute it in the second one.

VC(t) = E0·sin(ωt) − VL(t)

I(t)=C·d[E0·sin(ωt)−VL(t)]/dt

In the last equation we can differentiate each component and, using symbol ' for a derivative, obtain

I(t)=CωE0·cos(ωt)−C·V'L(t)

Together with the first equation from the original system of three
equations above we have reduced the system to two equations (again, we
use symbol ' for brevity to signify differentiation)

VL(t) = L·I'(t)

I(t)=CωE0·cos(ωt)−C·V'L(t)

Substituting VL(t) from the first of these equations into the second, we obtain one equation for I(t), which happens to be a differential equation of the second order (we will use symbol " to signify a second derivative of I(t) for brevity)

I(t)=CωE0·cos(ωt)−CL·I"(t)

or in a more traditional for differential equation form

a·I"(t) + b·I(t) = E0·cos(ωt)

where

a = L/ω

b = 1/(Cω)

Without getting too deep into a theory of differential equations, notice that the one and only known function in this equation that depends on time t is cos(ωt). It's second derivative also contains cos(ωt). So, if I(t) is proportional to cos(ωt), its second derivative I"(t) will also be proportional to cos(ωt) and we can find the coefficient of proportionality to satisfy the equation.

Let's try to find such coefficient K that function I(t)=K·cos(ωt) satisfies our equation.

I'(t) = −ωK·sin(ωt)

I"(t) = −ω²K·cos(ωt)

Now our differential equation for I(t) is

−a·ω²K·cos(ωt) + b·K·cos(ωt) = E0·cos(ωt)

From this we can easily find a coefficient K

K = E0/(b−aω²)

Since a=L/ω and b=1/(Cω)

K = E0/{[1/(Cω)] − Lω}

In the previous lectures we have introduced concepts of capacitive reactance XC=1/(Cω) and inductive reactance XL=Lω. Using these variables, the expression for coefficient K is
K = E0/(XC−XL)

Therefore,

I(t) = E0·cos(ωt)/(XC−XL)

or

I(t) = I0·cos(ωt)

where

I0 = E0/(XC−XL)

The last equation brings us to a concept of a reactance of the LC circuit

XC−XL

that is similar to resistance of regular resistors.

Using a concept of reactance, the last equation resembles the Ohm's Law.

Let's determine the voltage drops on a capacitor VC(t) and an inductor VL(t) using the expression for the current I(t).

Since Q(t)=C·VC(t) and I(t)=Q'(t), we can find VC(t) by integrating I(t)/C.

VC(t) = [0,t]I(t)·dt/C =

= I0·sin(ωt)/(C·ω) =

= XC·E0·sin(ωt)/(XC−XL)

VL(t) = L·I'(t) =

= −L·E0·sin(ωt)·ω/(XC−XL) =

= −XL·E0·sin(ωt)/(XC−XL)

Let's check that the sum of voltage drops in the circuit VL(t) and VC(t) is equal to the original EMF generated by a source of electricity.

Indeed,

VL(t) + VC(t) =

=(XC−XL)·E0·sin(ωt)/(XC−XL)=

= E0·sin(ωt)

Summary

EDF generated by a source of electricity

E(t) = E0·sin(ωt)

where

E0 is a peak voltage on the terminals of a generator,

ω is an angular velocity of a rotor in radians per second.

Alternating electric current in the circuit

I(t) = I0·cos(ωt)

where

I0 = E0/(XC−XL)

XC = 1/(ω·C)

XL = ω·L

Voltage drop on a capacitor

VC(t) = XC·E0·sin(ωt)/(XC−XL)

Voltage drop on an inductor

VL(t) = −XL·E0·sin(ωt)/(XC−XL)

Phase Shift

Notice that cos(x)=sin(x+π/2). Graph of function y=sin(x+π/2) is shifted to the left by π/2 relative to graph of y=sin(x).

Therefore, oscillations of the current I(t) in the LC circuit are ahead of oscillation of the EMF generated by a source of electricity E(t) by a phase shift of π/2.

Oscillations of the voltage drop on a capacitor VC(t) in the LC circuit are synchronized (in phase) with generated EMF.

Notice that −sin(x)=sin(x+π). Graph of function y=sin(x+π) is shifted to the left by π relative to graph of y=sin(x).

Therefore, oscillations of the voltage drop on an inductor VL(t) in the LC circuit are ahead of oscillation of the EMF generated by a source of electricity E(t) by π (this is called in antiphase).

## Friday, October 2, 2020

### AC Inductors: UNIZOR.COM - Physics4Teens - Electromagnetism

Notes to a video lecture on http://www.unizor.com

Alternating Current and Inductors

For the purpose of this lecture it's important to be familiar with the concept of a self-induction explained in "Electromagnetism - Self-Induction" chapter of this course.

In this lecture we will discuss the AC circuit that contains an inductor - a wire wound in a reel or a solenoid, thus making multiple loops, schematically presented on the following picture. Both direct and alternating current go through an inductor, but, while
direct current goes with very little resistance through a wire, whether
it's in a shape of a loop or not, alternating current meets some
additional resistance when this wire is wound into a loop.

Consider the following experiment. 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.

While the rod is not inside a solenoid, the lamp lights with normal
intensity. But let's gradually insert an iron core into a solenoid. As
the core goes deeper into a solenoid, the lamp produces less and less
light, as if some kind of resistance is increasing in the circuit.

This experiment demonstrates that inductors in the AC circuit produce
effect similar to resistors, and the more "inductive" the inductor - the
more resistance can be observed in a circuit.

The theory behind this is explained in this lecture.

The cause of this resistance is self-induction. This concept was
explained earlier in this course and its essence is that variable
magnetic field flux, going through a wire loop, creates electromotive
force (EMF) directed against the original EMF that drives electric
current through a loop.

Any current that goes along a wire creates a magnetic field around this
wire. Since the current in our wire loop is alternating, the magnetic
field that goes through this loop is variable. According to the
Faraday's Law, the variable magnetic field going through a wire loop
generates EMF equal to a rate of change of the magnetic field flux and
directed opposite to the EMF that drives the current through a wire,
thus resisting it.

Magnetic flux Φ(t) going through inductor, as a function of time t, is proportional to an electric current I(t) going through its wire

Φ(t) = L·I(t)

where L is a coefficient of proportionality that depends
on physical properties of the inductor (number of loop in a reel, type
of its core etc.) called inductance of the inductor.

If the current is alternating as

I(t) = Imax·sin(ωt)

the flux will be

Φ(t) = L·Imax·sin(ωt)

According to Faraday's Law, self-induction EMF Ei
is equal in magnitude to a rate of change of magnetic flux and opposite
in sign (see chapter "Electromagnetism - "Self-Induction" in this
course)

Ei(t) = −dΦ/dt =

= −L·
dI(t)/dt =

= −L·ω·Imax·cos(ωt) =

= −L·ω·Imax·sin(ωt+π/2) =

= −Eimax·sin(ωt+π/2)

where

Eimax = L·ω·Imax

The unit of measurement of inductance is henry (H) with 1H being an inductance of an inductor that generates 1V electromotive force, if the rate of change of current is 1A/sec.

That is,

henry = volt·sec/ampere = ohm·sec

An expression XL=L·ω in the above formula for Ei is called inductive reactance. It plays the same role for an inductor as resistance for resistors.

The units of the inductive reactance is Ohm (Ω) because

henry/sec = ohm·sec/sec = ohm.

Using this concept of inductive reactance XL of an inductor, the time dependent induced EMF is

Ei(t) = −XL·Imax·sin(ωt+π/2) = −Eimax·sin(ωt+π/2)

and

Eimax = XL·Imax,

which for inductors in AC circuit is an analogue of the Ohm's Law for resistors.

What's most important in the formula

Ei(t) = −Eimax·sin(ωt+π/2)

and the most important property of an inductor in an AC circuit is
that, while the electric current in a circuit oscillates with angular
speed ω, the voltage drop on an inductor oscillates with the same angular speed ω as the current, but its period is shifted in time by π/2 relative to the current.

## Tuesday, September 29, 2020

### UNIZOR.COM - Physics4Teens - Electromagnetism - Ohm's Law - Problems 4

Notes to a video lecture on http://www.unizor.com

Direct Current - Ohm's Law - Problems 4

Problem A

Given a circuit presented on a picture below. Initially, a red switch is in position A to fully charge a capacitor of capacity C from a battery producing a direct current with voltage V.

When a capacitor is fully charged, a switch is moved to position B, disconnecting a capacitor from a battery and forming a new circuit that includes only a fully charged capacitor and a resistor of resistance R.

When a switch is in position B, a
capacitor starts discharging its charge through a resistor. It's charge
will gradually diminish to zero, when all excess electrons on its one
plate will flow through a resistor to a plate with deficiency of
electrons.

During this process of discharge the electric current in a circuit that
contains a capacitor and a resistor will change from some maximum value
in the beginning of this process to zero, when the discharge is
completed.

Find the charge on a capacitor Q(t) and an electric current flowing trough a resistor I(t) as functions of time t.

Solution

Assume, our switch is in position A, and we are at the charging stage, when the battery of voltage V is charging a capacitor of capacity C.

The capacity of a capacitor is defined as the constant ratio of a charge
accumulated by a capacitor to a voltage applied to its plate (see
"Capacitors" lecture in the "Electromagnetism - Electric Field"
chapter):

C = Q/V

Therefore, the full charge of a capacitor at the end of the first stage of charging is V·C.

Then we flip a switch into position B, starting the second stage - discharging of a capacitor through a resistor.

At the beginning of this second stage a capacitor is fully charged. So, at time t=0 its charge is

Q(0) = V·C

The charge on a capacitor at any time produces a voltage between its plates

V(t) = Q(t)/C

This voltage produces a current flowing through a resistor I(t) that, according to the Ohm's Law, should be equal to

I(t) = V(t)/R

From the two equations above we conclude

Q(t)/C = I(t)·R

This is our first equation that connects two time-dependent (that is,
functions of time) variables - an electric current in a circuit I(t) and a charge on a capacitor Q(t).

The second functional equation is, basically, a definition of an
electric current as the rate of electric charge flowing in a circuit
(that is, amperage is how much electricity in coulombs flows through a circuit per unit of time - a second).

Mathematically speaking, an electric current is the first derivative of
an electric charge by time, taken with a sign that depends on the
direction of the change of the charge (plus if the charge is increasing and minus if decreasing):

I(t) = −dQ(t)/dt

Considering the charge Q(t) is decreasing and, therefore,
its derivative is negative, while we would like the electric current to a
be a positive number, we have to use a minus sign in this equation.

This is our second functional equation (that happens to be differential)
connecting two functions - an electric current in a circuit I(t) and a charge on a capacitor Q(t).

Now we have two functional equations, one of them is differential, and an initial condition:

Q(t)/C = I(t)·R

I(t) = −dQ(t)/dt

Q(0) = V·C

It's up to our mathematical skills to solve this system of equations.

First, we substitute I(t) from the second equation into the first, getting a differential equation for Q(t)

Q(t)/C = −R·dQ(t)/dt

This can be converted into

dQ(t)/Q(t) = −dt/(R·C)

or

d[ln(Q(t))] = d[−t/(R·C)]

If differentials of two functions are equal, the functions themselves
are just separated by a constant that can be determined using the
initial condition. Let denote that constant as K.

ln(Q(t)) = −t/(R·C) + K

or, applying an exponent to both sides of this equation,

Q(t) = eK·e−t/(R·C)

It's time to use the initial condition Q(0)=V·C to determine the multiplier eK.

For t=0 the right side of an expression for Q(t) equals to eK. Therefore, this multiplier equals to V·C.

Therefore, the final expression for a charge on a capacitor as a function of time Q(t) is

Q(t) = V·C·e−t/(R·C)

So, a charge on a capacitor is exponentially diminishing.

From the expression of Q(t) we can find the expression on an electric current going through a resistor, using the equation

I(t) = −dQ(t)/dt

from which follows

I(t) = −d[V·C·e−t/(R·C)]/dt =

= −V·C·
d
[e−t/(R·C)]/dt =

= −(V·C)·(−1/(R·C))·e−t/(R·C) =

= (V/R)·e−t/(R·C)

Q(t) = (V·C)·e−t/(R·C)

The multiplier V·C is the initial full charge of a capacitor.

I(t) = (V/R)·e−t/(R·C)

The multiplier V/R is the current that would flow through a resistor, if there were no capacitor. This follows from the Ohm's Law.

## Tuesday, September 8, 2020

### 3-Phase AC Problem: UNIZOR.COM - Physics4Teens - Electromagnetism - Alte...

Notes to a video lecture on http://www.unizor.com

Problems on AC Induction

Problem A

Three-phase generator has four wires coming out from it connected to a "star" with three phase wires carrying sinusoidal EMF shifted by 120°=2π/3 from each other and one neutral wire.

The angular speed of a generator's rotor is ω.

Assume that the peak difference in electric potential between each phase wire and a neutral one is E.

Describe the difference in electric potential between each pair of phase wires as a function of time.

Solution

The difference in electric potential between phase wires and a neutral one can be described as

Phase 1: E1(t)=E·sin(ωt)

Phase 2: E2(t)=E·sin(ωt−2π/3)

Phase 3: E3(t)=E·sin(ωt+2π/3)

The difference in electric potential between phase 1 wire and phase 2 wire can be represented as

E1,2(t) = E1(t) − E2(t)

Similarly, the difference in electric potential between two other pairs of phase wires is

E2,3(t) = E2(t) − E3(t)

E3,1(t) = E3(t) − E1(t)

Let's calculate all these voltages.

E1,2(t) = E·sin(ωt) − E·sin(ωt−2π/3) =

= E·sin((ωt−π/3)+π/3) − E·sin((ωt−π/3)−π/3) =

[substitute φ=ωt−π/3]

= E·sin(φ+π/3)−E·sin(φ−π/3) =

= E·
[sin(φ+π/3)−sin(φ−π/3)]

Let's simplify the trigonometric expression.

sin(φ+π/3) − sin(φ−π/3) =

= sin(φ)·cos(π/3) +

+ cos(φ)·sin(π/3) −

− sin(φ)·cos(π/3) +

+ cos(φ)·sin(π/3) =

= 2cos(φ)·sin(π/3) =

= √3cos(φ) =

= √3cos(ωt−π/3)

Therefore,

E1,2(t) = E√3cos(ωt−π/3)

As we see, the electric potential between phase 1 and phase 2 wires is also sinusoidal (since cos(x)=sin(x+π/2), cos(ωt−π/3) equals to sin(ωt+π/6)),
but shifted in time, and its peak voltage is greater than the peak
voltage between a phase wire and a neutral one by a factor of √3.

Similar factor difference of √3 is between effective voltages of these pairs of wires.

Analogous calculations for the other pairs of phase wires produce the following.

E2,3(t) = E·sin(ωt−2π/3) − E·sin(ωt+2π/3) =

[substitute φ=ωt]

= sin(φ)·cos(2π/3) −

− cos(φ)·sin(2π/3) −

− sin(φ)·cos(2π/3) −

− cos(φ)·sin(2π/3) =

= −2cos(φ)·sin(2π/3) =

= −√3cos(φ) =

= −√3cos(ωt)

Therefore,

E2,3(t) = −E√3cos(ωt)

Also the same factor difference of √3 relative to phase/neutral voltage.

Finally, the third phase/phase voltage calculations produce the following.

E3,1(t) = E·sin(ωt+2π/3) − E·sin(ωt) =

= E·sin((ωt+π/3)+π/3) − E·sin((ωt+π/3)−π/3) =

[substitute φ=ωt+π/3]

= E·sin(φ+π/3)−E·sin(φ−π/3) =

= E·
[sin(φ+π/3)−sin(φ−π/3)]

Let's simplify the trigonometric expression.

sin(φ+π/3) − sin(φ−π/3) =

= sin(φ)·cos(π/3) +

+ cos(φ)·sin(π/3) −

− sin(φ)·cos(π/3) +

+ cos(φ)·sin(π/3) =

= 2cos(φ)·sin(π/3) =

= √3cos(φ) =

= √3cos(ωt+π/3)

Therefore,

E3,1(t) = E√3cos(ωt+π/3)

Again, the same factor difference of √3 relative to phase/neutral voltage.

CONCLUSION

The voltage between any two phase wires is in magnitude greater than phase to neutral voltage by √3. Both are sinusoidal, but one is shifted in time relatively to another.

The phase to phase effective voltage (which is by √2 less then peak voltage) also is greater than phase to neutral by the same √3.

EXAMPLES

If phase to neutral effective voltage is 127V, the phase to phase effective voltage is 220V.

If phase to neutral effective voltage is 220V, the phase to phase effective voltage is 380V.

## Thursday, August 27, 2020

### Three-phase AC: UNIZOR.COM - Physics4Teens - Electromagnetism - Alternat...

Notes to a video lecture on http://www.unizor.com

Three Phases AC

The Basic Principle

of 3-Phase AC Generation

Recall the process of generating alternating current (AC) using a pair
of permanent magnets and a wire frame (a coil) rotating around the axis
perpendicular to magnetic field lines.

The pictures below represent the schematic design of such a system and a
graph of an electromotive force (EMF) generated between the ends a1 and a2 of the wire frame

Ea1a2 = Emax·sin(ωt)

where

Emax is the maximum absolute value of EMF,

ω is the angular speed of rotation of the wire frame,

t is time.  An obvious improvement to this design is to use the power of rotation
more efficiently by having two wire frames on the same axis positioned
perpendicularly to each other. In this case we can have two independent
sources of EMF with the only difference of one of them to be shifted in
time relatively to another by 1/2 of the time of rotation.

This shift is related to a simple fact that at the moment one wire
frame, aligned along the magnetic line, crosses these magnetic lines
with the highest rate, while another wire frame, that is perpendicular
to magnetic field lines, moves along these lines without actual
crossing. Then the roles are changed, as the coils rotate.

The EMF between the ends b1 and b2 of the second wire frame will then be

Eb1b2 = Emax·sin(ωt−π/2)  Why stop at two wire frames? Let's have three coils positioned at 120°
relative to each other. Now we will have three independent sources of
EMF shifted in time from each other by 1/3 of the time of rotation (phase
shift) - the time needed by one wire frame to take the position between
the magnet poles, previously taken by another wire frame.

Three different EMF, therefore, will be equal to

Ea1a2 = Emax·sin(ωt)

Eb1b2 = Emax·sin(ωt−2π/3)

Ec1c2 = Emax·sin(ωt−4π/3) = Emax·sin(ωt+2π/3)  Practical Implementation

The design of a three phase generator, as depicted above, is just the
first try of an idea. If the magnet is fixed and three wire frames
(coils) are rotating between its poles, it presents a problem to connect
these coils to transmit the generated electricity to consumers, we need
sliding contacts, brushes and other impractical devices.

In real life generators the three coils make up a stator - a fixed part
of a generator, while the magnet is rotating inside a circle of coils by
external power (like steam, water, wind etc.), generating the
alternating current in the coils, which allows to make an electric
connection to coils fixed.

At the first glance, three coils have three pairs of connections with
sinusoidal EMF generated in each pair and, to transfer AC electricity
from all coils to consumers, we seem to need three pairs of wires, two
from each coil - six wires altogether. This, however, can be improved by
using the following technique.

Let's connect ends a2, b2 and c2
of three coils together (see a picture below, it's a black wire at the
bottom connected to a black circle going around all coils) and see what
kind of resulting voltage will be observed on each end of the coils.
This is called a star connection of the generator's coils. We know that the electric potential (EMF) on each of the above contacts
has a sinusoidal magnitude with a time shift by 1/3 of a period relative
to each other. When we connect these three contacts, the potential at
the joint will be

E0 = Emax·sin(ωt) + Emax·sin(ωt−2π/3) + Emax·sin(ωt+2π/3) = Emax·X

where

X = sin(ωt) + sin(ωt−2π/3) + sin(ωt+2π/3) =

= sin(ωt) + sin(ωt)·cos(2π/3) − cos(ωt)·sin(2π/3) + sin(ωt)·cos(2π/3) + cos(ωt)·sin(2π/3) =

= sin(ωt) + sin(ωt)·(−1/2) − cos(ωt)·(√3/2) + sin(ωt)·(−1/2) +
cos(ωt)·(√3/2) = 0

Therefore, E0 = 0

There will be no difference in electrical potential between the joint and the ground.

This fact enables to transmit all three phases of generated electricity along four wires - one from a1 (phase 1) contact, one from b1 (phase 2) contact, one from c1 (phase 3) contact and one neutral from a joint connection to a2, b2 and c2.

The neutral wire is usually grounded since its electric potential is equal to zero.

With this arrangement we still have an advantage of having three
independent phases of alternating current, but we need only four wires
to transmit it - three phase wires and one neutral.

Connecting any device to any phase and a neutral wires, we will get a closed circuit with AC running in it.

Energy Consideration

Obviously, putting two or three coils in a stator of a generator doubles
or triples the energy output carried by outgoing wires. The Law of
Energy Conservation must work, so where is the energy is coming from?

Recall the electromagnetic induction experiment described in the lecture
"Faraday's Law" in the "Electromagnetic Induction" chapter of this
course with a wire moving in the uniform magnetic field. Since we physically move wire's electrons in one direction
perpendicularly to magnetic field lines, the Lorentz force pushes them
perpendicularly to both, the direction of the movement of a wire and the
direction of the magnetic field lines, that is, along the wire, thereby
creating an electric current between wire ends.

Now electrons are moving with a wire in one direction and along the wire in another.

The first movement maintains the electric current in the wire, but the
second, again, is a subject of the Lorentz force that pushes the
electrons perpendicularly to their direction, that is opposite to the
original direction of a wire movement.

This force resists the movement of a wire in its original direction. We
have to perform work against this force to move the wire.

Similar considerations are true in a case of a circular movement of a
wire frame in a magnetic field or, if wire coils are in a stator, the
force is needed to rotate the magnet in a rotor. That is, we have to
spend energy to generated the electricity, the rotor's rotation is
possible only if we apply the force against the Lorentz forces resisting
this rotation. The magnetic field generated by the electric current in a
wire coil of a stator resists the rotation of a magnet in a rotor.

If we have more than one coil in a stator, each one resists the rotation
of a rotor, so we have to spend proportionally more effort to rotate
the rotor.

The Law of Energy Conservation works. The more coils we have in a stator
- the more electricity is generated, but the more resistance to a
rotor's rotation needs to be overcome.

Three Phase AC Motor

The lecture "AC Motors" of this chapter described the necessity of having a rotating magnetic field to make an AC motor.

To achieve such a rotating field we had to resort to artificially create
a second AC current with a phase shift by 90° using a capacitor or a
transformer.

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 phase wire and a neutral one.

Powerful industrial level AC motors (like in a water pump that works in a
tall building to pump water to the roof tank) needs more power, and we
can use all three phases to create a rotating magnetic field.

So, all four wires coming from the AC generator, three phase wires and one neutral
one go into an AC motor, whose principal construction very much
resembles the one described in the previous lecture. The only difference
is, we already have three wires with AC phase shifted by 120°
relatively to each other. So, we have to position three wire coils in a
stator at 120° angles to each other, connect one end of each coil to a
corresponding phase wire and another end - to a common neutral
wire, and the rotating field is ready. Then it will work pretty much as
it was described in the "AC Motors" lecture, but smoother because three
phases make a smoother rotation of a magnetic field than two phases.

At the end I would like to say again, that it was Nicola Tesla's genius
that created all the basic principles, based on which all the current AC
motors are working now. His contributions to our industrial development
are grossly underappreciated. Calling an electric car model "Tesla" is a
late but well deserved tribute to his creativity.

## Sunday, August 23, 2020

### AC MotorsUNIZOR.COM - Physics4Teens - Electromagnetism - Alternating Cur...

Notes to a video lecture on http://www.unizor.com

Alternating Current Motors

Alternating current (AC) motors are used where the sources of electricity produce alternating current - in our homes, at industrial facilities etc.

For example, AC motors work in refrigerators, water pumps, air conditioners, large fans and many other devices.

The idea of AC motors belongs to Nicola Tesla, who invented it at the end of 19th century.

And it was a brilliant idea!

Before talking about AC motors, let's recall how DC motors are working.
Their basic design was presented in a lecture Physics 4 Teens -
Electromagnetism - Magnetism of Electric Current - DC Motors.

The important feature of DC motor, which was the starting idea behind
its design, was the rotational momentum exerted by a magnetic field of a
permanent magnet onto a wire frame with the direct electric current
going through it due to Lorentz force. Without a commutator or
any electronic switches the rotating frame would rotate until its plane
will reach a perpendicular position relative to magnetic lines, which
will be its equilibrium position.

Let's start with this idea of an AC motor and try to make it work.

Firstly, as in the case of DC motors, let's replace the permanent magnet
in a stator with two coils around an iron cores - electromagnets that
create magnetic field. Notice, however, that the current in these
electromagnets is alternating, which means that the magnetic field
between these electromagnets is alternating as well in magnitude and
direction.

Secondly, we will place a permanent magnet on an axis, so it freely rotates between the poles of the electromagnets

What will happen if we switch on the alternating current in the electromagnets?

Well, nothing noticeable. The problem is, the polarity of electromagnets
will start switching with frequency of the AC - 50 or 60 oscillations
per second in usual commercial wiring, and a permanent magnet in-between
will be forced in two opposite directions with the same frequency and
it will not start rotating.

But let's manually start rotating the magnet in any direction strong
enough to force it to rotate with significant angular speed.

At different positions the variable external magnetic field of
electromagnetic stator interferes with rotating permanent magnet rotor,
sometimes slowing it, sometimes speeding its rotation.

In a short while the rotation of the permanent magnet rotor will
synchronize with variable external field of electromagnetic stator and
rotation will be maintained with the same angular speed as the frequency
of AC in the coils of electromagnets.

Let's analyze this rotation in steps after the synchronization is achieved.

Let a pole facing the rotor of one electromagnet be X and an opposite pole of another electromagnet be Y.

As AC changes its direction and magnitude, pole X is gradually changing
from North (X=N) to zero (X=0), to South (X=S), again to zero (X=0),
again to North (X=N) etc.

At the same time pole Y of an opposite electromagnet is gradually
changing from South (Y=S) to zero (Y=0), to North (Y=N), again to zero
(Y=0), again to South (X=N) etc.

Synchronously rotating permanent magnet of a rotor should with its North
pole approach X=S and, simultaneously, with its South pole approach
Y=N. As a rotor approaches with its poles the poles of a stator, X and Y
should gradually weaken and, when the permanent magnet is fully aligned
along XY line, the magnitudes of the magnetic fields of electromagnets
should diminish to zero (X=0, Y=0).

Permanent magnet rotor will pass this point by inertia and the polarity of the electromagnets switches, so now X=N and Y=S. That causes rotation to continue until the permanent magnet of a rotor again takes a position along XY line.

By that time the electromagnets will be at X=0 and Y=0, rotor will
continue rotation by inertia, then AC switches the polarity of
electromagnets again and rotation continues in a similar manner
indefinitely.

Our first version of an AC motor is functional, but it has two obvious disadvantages.

One disadvantage is the usage of permanent magnet, they are very
expensive. We could not avoid it in a DC motor, but in an AC case we
might think that variable magnetic field might help to use the induction
effect to avoid it.

Another disadvantage is that switching the AC on does not really start
the rotation, we manually started it, and only then, if we gave a strong
push, it started to rotate and maintained this rotation.

Let's use a wire loop instead of a permanent magnet rotor. But, to act
as a permanent magnet, it needs an electric current running through the
wire, and we don't want any extra sources of electricity to connect to
it, it's complicated, needs a commutator, like in the original DC
motors.

Instead, we will count on the induction effect created by an alternating
current in a stator, which creates an alternating magnetic field,
which, in turn, induces the electric current in the wire loop of a
rotor.

The induced electric current in a wire loop of our rotor appears when a
wire crosses the magnetic field lines. Magnetic field produced by a
stator is directed always along a center line XY and is changing in
magnitude from a maximum in one direction (from X to Y) to zero, to a
maximum in another direction (from Y to X), again to zero etc.

The problem is, if the rotor is standing still, its wire does not cross
magnetic field lines. Therefore, no electric current will be produced in
it, and there will be no rotation. Rotor needs an initial push
sufficient to synchronize its rotation with the frequency of alternating
magnetic field to continue the rotation. Our second disadvantage is
still unresolved.

Let's imagine that, besides two main electromagnets with poles X and Y,
we have another pair of auxiliary electromagnets in a stator, that are
positioned perpendicularly to line XY. Let their poles be A and B and
(very important!) the current in them, also sinusoidal, is shifted in
time relatively to a current in main electromagnets by 90°.

So, when the magnetic field in one direction is maximum along line XY,
it's zero along line AB; then it gradually decreases along XY and
increases along AB until it's zero along XY and maximum along AB. This
cycle repeats itself indefinitely.

Graphically, it can be represented as follows. Graphs on the left and on the right of a picture represent the electric
current in each pair of electromagnets - XY and AB. Notice, they are
shifted by π/2=90°. The middle part of a picture represents the
direction of the magnetic field created by both pairs of electromagnets
of a stator.

As the time goes, the magnetic field direction is rotating. The rotating
magnetic field around a closed wire loop of a rotor causes the rotor's
wire to cross the magnetic field lines of a stator, which, in turn,
causes induction of electric current in a rotor's closed wire loop,
which will act now as a permanent magnet, which is supposed to follow
the rotation of the magnetic field around it.

This causes the rotation of the rotor. The rotor's actual physical
rotation will follow the stator's magnetic field rotation, created
without any physical movement.

The rotor cannot go after the stator's magnetic field rotation exactly
synchronously because then there will be no crossing of magnetic field
lines. So, the rotor, after it reached the same speed as the stator's
magnetic field rotation, will lose its rotational momentum and slow
down. Then the rotor's wire, rotating slower than the magnetic field
around it, will cross the magnetic field lines, there will be induction
current in it, it's magnetic properties will be restored and it will try
to catch up with the stator's magnetic field rotation. This process
will continue as long as AC is supplied.

An obvious improvement is to use multiple wire loops as a rotor with
common axis of rotation. That will cause more uniform rotation.

What's remaining in our project of designing the AC motor is to create
another alternating current for the second pair of electromagnets in a
stator with a time delay of π/2=90° relatively to the main AC.

This problem is resolved and described in the previous lecture about AC
capacitors. If we introduce another circuit fed from the same AC source,
but with a capacitor in it, this circuit will have the alternating
current shifted in time exactly as we want. This current will go through
the second pair of electromagnets and both pairs will create a
revolving magnetic field.

The interaction between magnetic fields of a stator and a rotor is quite
complex, when the rotor rotates. The magnetic flux going through a
rotor wire loop depends on a variable magnetic field of a stator and
variable area of a rotor wire loop in a direction of a magnetic field
lines of a stator. The exact calculations of this process are beyond the
scope of this course.

The main idea, however, is clear - to create an AC motor we have to create a revolving magnetic field - the great idea of Nicola Tesla. That can be accomplished by using known methods described above.

## Wednesday, August 19, 2020

### DC Motors: UNIZOR.COM - Physics4Teens - Electromagnetism - Magnetism of ...

Notes to a video lecture on http://www.unizor.com

Direct Current Motors

Direct current (DC) motors are used where the sources of electricity produce direct current, like batteries.

For example, DC motors work in car engine starter, computers, toys, drones and many other devices.

We will concentrate on principles of their work without going into many
details. Basically, we will describe these motors as they were first
thought of by their inventors. Obviously, initial ideas were further
improved by many people and improvements are still introduced after more
than 200 years after their invention, but the basic ideas are still
there.

The beginning of a development of a DC motor is associated with a simple
experiment we described before - the one that demonstrated the Lorentz
force on a current in a magnetic field.

If the direction of a current I is perpendicular to a direction of magnetic field lines B,
the magnetic field "pushes" the conductor that carries this current in a
direction perpendicular to both magnetic field lines and a current by
force F. The first modification of this experiment was to use a rectangular wire
frame instead of a straight line conductor and let it spin around an
axis. The force exerted by a magnetic field will act on both sides of a
wire frame that are perpendicular to magnetic lines in opposite
directions, creating a rotational momentum, so the frame with rotate. The force F of a magnetic field "pushes" segment AB up and segment CD, where the electric current goes in an opposite to segment AB direction, is "pushed" down by this force. This creates a rotational momentum.

The problem is, the rotation will stop when the wire frame plane will be
perpendicular to a direction of a magnetic field, because the forces
exerted by a magnetic field on both sides of a wire frame AB and CD
will no longer create a rotating momentum, they will act against each
other within the wire frame plane and after a short oscillation caused
by a rotational inertia the rotation stops.

On the picture above, when the plane of a wire loop reaches the position perpendicular to the magnetic field force, the segment AB will be on the top and the magnetic field force F will "push" this segment up. Segment CD will be on the bottom and the magnetic field force F
will "push" this segment down. Both forces are acting within the same
plane with the axis of rotation and nullify each other. No rotational
momentum is created.

Our next improvement is related to overcoming this problem.

What happens if exactly at the moment when our wire frame plane is
perpendicular to magnetic field, when the forces exerted by a magnetic
field no longer create a rotational moment, we switch off the electric
current in a wire and a very short moment later we switch it on, but in
the opposite direction of the electric current?

First of all, the rotation will continue by inertia during the time
electricity is off. Then, we turn the electric current on but in
opposite direction. That means, the direction of the magnetic field
force will change to an opposite. Segment AB will be "pushed" down, segment CD
will be "pushed" up. Since the wire frame has passed the point of
perpendicularity to magnetic field by rotational inertia during electric
current off time, the newly formed magnetic field forces will create a
rotational moment and the direction of rotation will be the same as
before.

Our frame will continue the rotation in the same direction until segment AB will be on the bottom and segment CD - on the top.

At that moment we will do the same switching off the electric current to
let our frame pass the perpendicular position towards the magnetic
field force, and a very short moment later we switch electric current on
in an opposite direction. Now segment AB will be "pushed" up again, segment CD will be "pushed" down, which will allow to complete the cycle of rotation and the rotation will continue in the same direction.

All we need is to explain how to switch the direction of an electric current.

Here is a drawing of this simple device called commutator. Direct current comes through brushes to a commutator and to a revolving
frame connected to it. As this construction revolves, the brushes lose
the contact with a commutator for a very short period, then again touch
the commutator, but with opposite poles.

That's how we change the direction of the current that forces the frame to constantly rotate in the same direction.

The inner rotating part of this type of a DC motor is called rotor, the outer stationary part with a magnet is called stator.

The detail implementation of this design of a DC motor is outside the scope of this course. Our purpose is to convey the idea.

The weak part of a DC motor design with brushes is that these brushes
wear out with time and are the source of sparks, which might be
prohibitive in some environments. Plus, they are noisy.

With development of electronics inventors came up with a better design
that does not involve brushes at all, rotation is accomplished without
any mechanical switches.

First step on the design of this brushless DC motor is to realize that
we have to invert the functionality of permanent magnet and a wire
frame. The need for a commutator with mechanical brushes was related to the fact that electric circuit was rotating.

Let's invert roles and put two parallel wire frames on opposite ends of a circle as a stator and a permanent magnet on an axis in the middle as a rotor.

Recall that a wire loop with direct current running through it acts like a magnet Putting two such loops on opposite ends of a stator creates a magnetic
field and a freely rotating permanent magnet in between will have to
turn to align itself with the field of these two wire loops If at the moment our permanent magnet aligns with magnetic field of both
wire loops we change the direction of electric current in the loops to
an opposite, thus changing the polarity of the magnetic field they
generate, the permanent magnet in between will have to continue the
rotation until it will align again. Repeating this cycle creates the
rotation of a permanent magnet.

This rotation will not be smooth. The force of magnetic field attraction
will create a stronger moment of rotation when the direction of the
permanent magnet is perpendicular to a direction of the magnetic field
of the loops. This can be improved by certain additional details
described below.

The changing of the direction of the current in the wire loops now is
much easier than in the previous design because the wires are not
moving. Simple electronic switch working off some kind of a marker on a
rotating permanent magnet can signal its position and trigger the switch
of direction of the current.

Basically, the idea is finished here. But some very important improvements should be mentioned.

1. Instead of a single wire loop we can use a copper coil around an iron
core to make the magnetic field of this electromagnet stronger.

2. Instead of a pair of such electromagnets with switching the direction
after the permanent magnet in the middle turns by 180°, we can use two
pairs and properly engage another pair after 90° turn, switching off the
previous pair, thus creating a rotating magnetic field that results in a
smoother movement of a permanent magnet rotor. Even better, we can use
three pairs of electromagnets and engage another pair after each 60°,
switching off the previous two pairs. This will allow even smoother
rotation. In some DC motors they use even six pairs of electromagnets,
which results in a very uniform rotation with practically constant
angular speed.

The electronic switches that engage and disengage the coils can be designed for any type of coil arrangement.

3. Instead of a bar permanent magnet inside the circle of electromagnets
we can use a ring permanent magnet outside it. It creates a better
response to a revolving magnetic field of a stator, the rotation will be
smoother because of inertia of a rotating ring.

With all the above improvements the DC motor used in hard disk of computers and other devices looks like the one below This DC motor has 9 electromagnets sequentially engaged after each 40° turn of a rotor.

Rotor is a ring-shaped permanent magnet that rotates around the electromagnets.

A marker on the rotor sends a signal to an electronic switch to indicate
the position of a rotor, which is used by electronics on the attached
board to properly engage the electromagnets of a stator.

The most important detail of this design of a DC motor is the rotating
magnetic field achieved through purely electronic means without any
moving parts. This results in a smooth rotation of a rotor made of a
ring permanent magnet around a stator.