Sunday, November 22, 2020

AC Power: UNIZOR.COM - Physics4Teens - Electromagnetism - AC Ohm's Law

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

AC Power

Power 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.

As we know from the properties of a direct electrical current, its power is
P = U·I = I²R = U²/R
where U is the voltage around a resistor of resistance R and I is the electric current going through this resistor.
All values in the above expression are constant for direct current.

Alternating current 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.

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
I(t) = (E0 /Z)·sin(ωt+φ) =
= I0·sin(ωt+φ)

where impedance Z=√(XC−XL)²+R²,
tan(φ)=(XC−XL)/R.

Having expressions for generated EMF E(t)=E0·sin(ωt) and electric current in a circuit I(t)=I0·sin(ωt+φ), we can find an instantaneous power
P(t) = E(t)·I(t) =
= E0·I0·sin(ωt)·sin(ωt+φ)


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 effective voltage Eeff = E0 /2 and effective amperage Ieff = I0 /2 of the electric current. The usage of these characteristics allows to calculate the power consumed by a resistor-only circuit during a period [0,T] of time (for T significantly greater than one oscillation of a current) without integrating a variable function P(t)=E(t)·I(t) on interval [0,T], but just performing a multiplication of constants:
W[0,T] = Eeff · Ieff · T

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 Eeff and Ieff 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 effective power of a circuit.

The period of one oscillation with angular speed ω is T=2π/ω.
The instantaneous power consumption by a circuit is
P(t) = E(t)·I(t) =
= E0·sin(ωt)·I0·sin(ωt+φ)

The energy consumed by a circuit during one period of oscillation T=2π/ω equals to
W[0,T] = [0,T]P(t)·dt
where
P(t)=E0·sin(ωt)·I0·sin(ωt+φ)

We can simplify the product of two trigonometric functions to make it easier to integrate:
sin(x)·sin(y) =
= (1/2)·
[cos(x−y)−cos(x+y)]
Using this for x=ωt and y=ωt+φ, we obtain
sin(ωt)·sin(ωt+φ) =
= (1/2)·
[cos(φ)−cos(2ωt+φ)]

To find the power consumption during one period of oscillation T, we have to calculate the following integral
W[0,T] = [0,T]P(t)·dt
where period T=2π/ω and
P(t) = E0·I0·
·(1/2)·
[cos(φ)−cos(2ωt+φ)]

This integral can be expressed as a difference of two integrals
[0,T]E0·I0·(1/2)·cos(φ)·dt
which, considering cos(φ) is a constant for a given circuit, is equal to
E0·I0·(1/2)·cos(φ)·T =
= E0·I0·(1/2)·cos(φ)·2π/ω

and
[0,T]E0·I0·(1/2)·cos(2ωt+φ)·dt
which is equal to zero because integral of a periodical function cos(x) over any argument interval that equals to one or more periods equals to zero.
The same can be proven analytically
[0,T]cos(2ωt+φ)·dt =
= sin(2ωt+φ)/(2ω)|[0,T] =
= sin(2ω·2π/ω+φ)/(2ω) −
− sin(φ)/(2ω) =
=
[sin(4π+φ)−sin(φ)]/(2ω) = 0

Hence, the energy consumed by a circuit during one period of oscillation equals to
W[0,T] = E0·I0·(1/2)·cos(φ)·2π/ω
The average power consumption, that is the average rate of consuming energy that we will call effective power, equals to this amount of energy divided by time, during which it was consumed - one period of oscillation T=2π/ω:
Peff = W[0,T]/T =
= E0·I0·(1/2)·cos(φ)

Since Eeff = E0 /2 and Ieff = I0 /2, the last expression for power equals to
Peff = Eeff·Ieff·cos(φ)
where a phase shift φ depends on resistance and reactances of a circuit as follows
tan(φ) = (XC − XL) /R
XC = 1/(ωC) - capacitive reactance,
XL = ωL - inductive reactance,
R - resistance.
The above formula is derived for RLC-circuit that contains a resistor or resistance R, a capacitor of capacitance C and an inductor of inductance L
Let's analyze different circuits and their effective power consumption rate.

R-Circuit
R-circuit contains only a resistor. Therefore, both reactances XC and XL are zero and phase shift φ is zero as well. Since cos(0)=1, the effective power for this R-circuit is
Peff = Eeff·Ieff
which fully corresponds to a power for a circuit with a direct current running through it.

RC-Circuit
RC-circuit contains a resistor and a capacitor in a series. Reactance XL is zero.
Peff = Eeff·Ieff·cos(φ)
where tan(φ) = XC /R

RL-Circuit
RC-circuit contains a resistor and an inductor in a series. Reactance XC is zero.
Peff = Eeff·Ieff·cos(φ)
where tan(φ) = −XL /R
The negative values of tan(φ) is not important since function cos(φ) is even and cos(φ)=cos(−φ).

Interestingly, if our circuit contains a resistor, but a capacitor and an inductor are in resonance, that is XC=XL, the phase shift will be equal to zero, as if only a resistor is present in a circuit.

L-, C- and LC-Circuits
If no resistor is present in the circuit (assuming the resistance of wiring is zero), the denominator in the expression
tan(φ) = (XC − XL) /R
is equal to zero.
Therefore, the phase shift is φ=π/2=90°, cos(π/2)=0 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 resonance, there is no phase shift, they neutralize each other.

Monday, November 16, 2020

RLC Circuit Ohm Law: UNIZOR.COM - Physics4Teens - Electromagnetism - AC ...

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

RLC Circuit Ohm's Law

This lecture combines the material of the previous two ones dedicated to the Ohm's Law in alternating current circuits.
The first one was analyzing a circuit with a resistor and a capacitor.
The second one analyzed a circuit with a resistor and an inductor.
This lecture analyzes a circuit with all three elements - resistor, inductor and capacitor.

Consider a closed AC circuit with a generator with no internal resistance producing a sinusoidal electromotive force (EMF or voltage)
E(t)=E0·sin(ωt)(volt)
where E0 is a peak EMF produced by an AC generator,
ω is angular speed of EMF oscillations and
t is time.

In this circuit we have a resistor with resistance R, an inductor with inductance L and a capacitor with capacitance C connected in series with EMF generator. This is RLC-circuit.
Since the circuit is closed, the electric current I(t) going through a circuit is the same for all components of a circuit.

Assume that the voltage drop on a resistor (caused by its resistance R) is VR(t), the voltage drop on an inductor (caused by its inductive reactance XL=ω·L) is VL(t) and the voltage drop on a capacitor (caused by its capacitive reactance XC=1/(ω·C)) is VL(t).

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 E(t):
E(t) = VR(t) + VL(t) + VC(t)

As explained in the previous lectures the Ohm's Law localized around a resistor states that
I(t) = VR(t)/R or
VR(t) = I(t)·R

The inductance of an inductor L, voltage drop on this inductor caused by a self-induction effect VL(t), electromagnetic flux going through an inductor Φ(t) and electric current going through it I(t), as explained in the lecture "AC Inductors" of the chapter "Electromagnetism - Alternating Current Induction", are in a relationship
VL = dΦ/dt = L·dI(t)/dt

The capacity of a capacitor C, voltage drop on this capacitor VC(t) and electric charge accumulated on its plates QC(t), according to a definition of a capacitance C, are in a relationship
C = QC(t)/VC(t) or
VC(t) = QC(t)/C

We have expressed both voltage drops VR(t) and VL(t) in terms of an electric current I(t):
VR(t) = I(t)·R
VL(t) = L·I'(t)
Since an electric charge QC(t) 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
I(t) = QC'(t) and
I'(t) = QC"(t)

Now all three voltage drops can be expressed as functions of QC(t) and, equating their sun to a generated EMF E(t), we can the following differential equation
E0·sin(ωt) = QC'(t)·R +
+ L·QC"(t) + QC(t)/C =
= L·QC"(t) + R·QC'(t) +
+ (1/C)·QC(t)


As we know, capacitive and inductive reactance are functionally equivalent to resistance, and they all have the same unit of measurement - ohm. Therefore, it's convenient, instead of capacitance C and inductance L, to use corresponding reactance XC=1/(ω·C) and XL=ω·L.
So, we will substitute
C = 1/(ω·XC)
L = XL

Substitute QC(t)=y(t) for brevity. Then our equation looks like
(XL/ω)·y"(t) + R·y'(t) +
+ ω·XC·y(t) = E0·sin(ωt)


Solving this equation for y(t)=QC(t) and differentiating it by time t, we will obtain the expression for an electric current I(t) in this circuit as a function of all given parameters and time.

First of all, let's find a particular solution of this differential equation.
The form of the right side of this equation prompts to look for a solution in trigonometric form
y(t) = F·sin(ωt) + G·cos(ωt)
Then
y'(t) =
= ω·
[F·cos(ωt)−G·sin(ωt)]
y"(t) =
= −ω²·
[F·sin(ωt)+G·cos(ωt)] =
= −ω²·y(t)


Using this trigonometric representation of potential solution y(t), the left side of our equation is
(XL/ω)·y"(t) + R·y'(t) +
+ ω·XC·y(t) =
= −(XL/ω)·ω²·y(t) + R·y'(t) +
+ ω·XC·y(t) =
= −ω·XL·y(t) + R·y'(t) +
+ ω·XC·y(t) =
= R·y'(t) + ω·(XC−XL)·y(t) =
= R·ω·
[F·cos(ωt)−G·sin(ωt)] +
+ ω·(XC−XL
[F·sin(ωt)+
+G·cos(ωt)
] =
= ω·
[(XC−XL)·F−R·G]·sin(ωt)+
+ω·
[R·F+(XC−XL)·G]·cos(ωt)

Since this is supposed to be equal to E0·sin(ωt), we have the following system of two linear equations with two variables F and G
ω·[(XC−XL)·F−R·G] = E0
ω·[R·F+(XC−XL)·G] = 0
or
(XC−XL)·F−R·G = E0/ω
R·F+(XC−XL)·G = 0

Determinant of the matrix that defines this system is
(XC−XL)²+R².
It's always positive and usually is denoted as
Z² = (XC−XL)²+R²
The value Z is called the impedance of a circuit and, as we will see, plays the role of a resistance for an entire RLC-circuit.

This system of two linear equations with two variables has a solution:
F = (E0/ω)·(XC−XL)/
G = −(E0/ω)·R/

Consider two expressions that participate in the above solutions F and G:
(XC−XL)/ and R/
Since Z² = (XC−XL)²+R², we can always find an angle φ such that
(XC−XL)/Z = sin(φ) and
R/Z = cos(φ)

Using this, we express the solutions to the above system as
F = (E0/(ω·Z))·sin(φ)
G = −(E0/(ω·Z))·cos(φ)

Now the solution of the differential equation for y(t)=QC(t) that we were looking for in a format
y(t) = F·sin(ωt) + G·cos(ωt)
looks like
y(t)=(E0/(ω·Z))·sin(φ)sin(ωt) −
− (E0/(ω·Z))·cos(φ)cos(ωt) =
= −(E0/(ω·Z))·cos(ωt+φ)


As we noted before, differentiation of this function gives the electric current in the circuit I(t):
I(t) = y'(t) =
= (E0 /(ω·Z))·ω·sin(ωt+φ) =
= (E0 /Z)·sin(ωt+φ) =
= I0·sin(ωt+φ)

where I0=E0 /Z is an equivalent of the Ohm's Law for an RLC-circuit.

Similar relationship exists between effective voltage and effective current
Ieff = I0 /2 =
= E0 /(√2·Z) = Eeff /Z


Here impedance Z, defined by resistance R, inductive reactance XL and capacitive reactance XC as
Z = √(XC−XL)²+R²
plays a role of a resistance in the RLC-circuit.

There is a phase shift φ of the electric current oscillations relative to EMF. It is also defined by the same characteristics of an RLC-circuit:
(XC−XL)/Z = sin(φ)
R/Z = cos(φ)
Hence
tan(φ) = (XC−XL)/R
This makes phase shift positive or negative depending on the values of capacitive and inductive reactance.
If XC is greater than XL, the phase shift is positive, if the reverse is true, the phase shift is negative.
If the values of reactance are the same, that is XC=XL, there is no phase shift. From definition of reactance, it happens when
1/(ω·C) = ω·L or
1/(L·C) = ω² or
L·C = 1/ω²
This relationship between inductance, capacitance and angular speed of EMF oscillation is call resonance.

Expressions for electric current
I(t) = (E0 /Z)·sin(ωt+φ) =
= I0·sin(ωt+φ)

in RLC-circuit, where impedance Z=√(XC−XL)²+R² and tan(φ)=(XC−XL)/R, 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.

Tuesday, November 10, 2020

Ohm's Law for RL_Circuit: UNIZOR.COM - Physics4Teens - Electromagnetism ...

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

RL Circuit Ohm's Law

Consider a closed AC circuit with a generator with no internal resistance producing a sinusoidal electromotive force (EMF or voltage)
E(t)=E0·sin(ωt)(volt)
where E0 is a peak EMF produced by an AC generator,
ω is angular speed of EMF oscillations and
t is time.
In this circuit we have a resistor with resistance R and an inductor with inductance L connected in series with EMF generator. This is RL-circuit.
Since the circuit is closed, the electric current I(t) going through a circuit is the same for all components of a circuit.

Assume that the voltage drop on a resistor caused by its resistance is VR(t) and the voltage drop on an inductor caused by self-induction effect is VL(t). Since a resistor and an inductor are connected in a series, the sum of these voltage drops should be equal to a generated EMF E(t):
E(t) = VR(t) + VL(t)

As explained in the previous lecture for an R-circuit, the Ohm's Law localized around a resistor states that
I(t) = VR(t)/R or
VR(t) = I(t)·R

The inductance of an inductor L, voltage drop on this inductor caused by a self-induction effect VL(t), electromagnetic flux going through an inductor Φ(t) and electric current going through it I(t), as explained in the lecture "AC Inductors" of the chapter "Electromagnetism - Alternating Current Induction", are in a relationship
VL = dΦ/dt = L·dI(t)/dt

We have expressed both voltage drops VR(t) and VL(t) in terms of an electric current I(t):
VR(t) = I(t)·R
VL(t) = L·I'(t)

Now we can substitute them into a formula for their sum being equal to a generated EMF
E(t) = VR(t) + VL(t) =
= I(t)·R + I'(t)·L


Since E(t)=E0·sin(ωt) we should solve the following differential equation to obtain function I(t)
E0·sin(ωt) = I(t)·R + I'(t)·L

Let's divide this equation by L and use simplified notation for brevity:
y(t) = I(t)
a = R/L
b = E0/L
Then our equation looks simpler
y'(t)+a·y(t) = b·sin(ωt)

This exact differential equation was solved in the previous lecture dedicated to RC-circuit, notes for that lecture contain detail analysis of its solution
y(t) = −b·cos(ωt+ψ)/√(a²+ω²) + K
where phase shift ψ=arctan(a/ω) and K is a constant that can be determined by initial condition.

Using original notation,
I(t) = −(E0/L)·cos(ωt+ψ)/
/(R/L)²+ω² + K =
= −E0·cos(ωt+ψ)/
/R²+(ω·L)² + K =
= −E0·cos(ωt+ψ)/R²+XL² + K

where XL=ω·L is inductive reactance of an inductor - a characteristic of an inductor functionally equivalent to a resistance for a resistor and measured in the same units - ohm and
tan(ψ) = a/ω = R/(L·ω) = R/XL

Let's apply some Trigonometry to simplify the above formula.
−cos(ωt+ψ) =
= −sin((π/2)−ωt−ψ) =
= −sin((π/2)−ψ−ωt) =
= sin(ωt−((π/2)−ψ)) =
= sin(ωt−φ)

where φ=(π/2)−ψ and, therefore, tan(φ)=1/tan(ψ)=XL/R.

Using phase shift φ, the equation for the current I(t) looks like
I(t) = E0·sin(ωt−φ)/R²+XL² + K
where tan(φ)=XL/R.

The value of a constant K 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 K=0

The final version of the current in this RL-circuit is
I(t) = E0·sin(ωt−φ)/R²+XL² =
= I0·sin(ωt−φ)

where I0 = E0/R²+XL²

The last issue is to analyze the effective current in this RL-circuit. Since for sinusoidal oscillations the effective current is by √2 less than peak amperage, the effective current is
Ieff = I0 /2 =
= E0 /(√XL²+R²·√2) =
= Eeff /XL²+R²

This is the Ohm's Law for effective voltage and amperage in RC-circuit.

It's important to notice that in the absence of a resistor (that is, R=0) the formula for I(t) transforms into
I(t) = I0·sin(ωt−φ)
where peak amperage I0=E0/XL and phase shift φ=arctan(∞)=π/2, which corresponds to results obtained in a lecture "AC Inductors" of the "Alternating Current Induction" chapter of this course dedicated to only an inductor in the AC circuit, taking the current as given and deriving the EMF.

Saturday, November 7, 2020

Ohm's Law for R- and RC-Circuits: UNIZOR.COM - Physics4Teens - Electromagnetism - AC O...

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

R and RC Circuit Ohm's Law

In this chapter we will examine different aspects of the Ohm's Law as they occur in different alternating current (AC) circuits.
Four different types of AC circuits will be considered in this and subsequent lectures:
(a) R-circuit that contains only a resistor;
(b) RC-circuit that contains a resistor and a capacitor;
(c) RL-circuit that contains a resistor and an inductor;
(d) RLC-circuit that contains a resistor, an inductor and a capacitor.
The first two are subject to this lecture.


R-circuit

Consider a closed AC circuit with a generator with no internal resistance producing a sinusoidal electromotive force (EMF or voltage)
E(t)=E0·sin(ωt)(volt)
where E0 is a peak EMF produced by an AC generator,
ω is angular speed of EMF oscillations
and t is time.
In this circuit there is only a resistor with resistance R(ohm).

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 R, the electric current at any moment of time will obey the Ohm's Law for direct current.

Therefore the electric current I(t) in this R-circuit is
I(t) = E(t)/R = E0·sin(ωt)/R =
= (E0/R)·sin(ωt) = I0·sin(ωt)

where I0 = E0/R is a peak current.
Now we can express the Ohm's Law for peak voltage and peak amperage in the R-circuit as
I0 = E0 /R

Since in many cases we are interested in effective voltage and effective electric current, and knowing that
Eeff = E0 /2 and Ieff = I0 /2,
the Ohm's Law for effective voltage and effective amperage in the R-circuit can be expressed in a form
Ieff = Eeff /R


RC-circuit

Let's add a capacitor with capacity C (farad) to R-circuit, connecting it in series with a resistor. This is RC-circuit.
As we know, alternating current goes through a capacitor, so we have a closed circuit of resistor R and capacitor C connected in a series with EMF generator that produces sinusoidal voltage E(t)=E0·sin(ωt).
The electric current I(t) going through a circuit is the same for all components of a circuit.

Assume that the voltage drop on a resistor is VR(t) and the voltage drop on a capacitor is VC(t). Since a resistor and a capacitor are connected in a series, the sum of these voltage drops should be equal to a generated EMF E(t):
E(t) = VR(t) + VC(t)

As explained above for an R-circuit, the Ohm's Law localized around a resistor states that
I(t) = VR(t)/R or
VR(t) = I(t)·R

The capacity of a capacitor C, voltage drop on this capacitor VC(t) and electric charge accumulated on its plates QC(t), according to a definition of a capacity, are in a relationship
C = QC(t)/VC(t) or
VC(t) = QC(t)/C

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:
I(t) = Q'C(t)

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
VR(t) = Q'C(t)·R

We have expressed both voltage drops VR(t) and VC(t) in terms of an electric charge QC(t) accumulated on the plates of a capacitor:
VR(t) = Q'C(t)·R
VC(t) = QC(t)/C

Now we can substitute them into a formula for their sum being equal to a generated EMF
E(t) = VR(t) + VC(t) =
= Q'C(t)·R + QC(t)/C


Since E(t)=E0·sin(ωt) we should solve the following differential equation to obtain function QC(t)
E0·sin(ωt) =
= Q'C(t)·R + QC(t)/C


Let's divide this equation by R and use simplified notation for brevity:
y(t) = QC(t)
a = 1/(R·C)
b = E0/R
Then our equation looks simpler
y'(t)+a·y(t) = b·sin(ωt)

Let's discuss how to solve (integrate) this differential equation.
Recall the formula for a derivative of a product of two functions
(x(t)·y(t))' =
= x(t)·y'(t) + x'(t)·y(t)


Let's find such function x(t) that, if we use it as a multiplier of our differential equation, the left side will look like a complete derivative of x(t)·y(t).

Multiplied by this function x(t), our equation looks like this:
x(t)·y'(t)+a·x(t)·y(t) =
= b·x(t)·sin(ωt)

Let's find function x(t) such that a·x(t) is x'(t). Then the left side of the equation will be equal to (x(t)·y(t))'.

Obvious choice for such function is ea·t. It's an intelligent guess, but it can be determined analytically.
Indeed,
a·x(t) = x'(t)
a·x(t) = dx/dt
dt = dx/x(t)
d(a·t) = dln(x(t))
a·t = ln(x(t)) + K1
x(t) = K2·ea·t
Here K1 and K2 are any constants, so let's use K2=1.
Then x(t)=ea·t.

Now the equation takes the following form
ea·t·y'(t)+a·ea·t·y(t) =
= b·ea·t·sin(ωt)

which is equivalent to
[ea·t·y(t)]' = b·ea·t·sin(ωt)

The above equation should be integrated to get ea·t·y(t) and, finally y(t).

The indefinite integral of the left side is ea·t·y(t).

The indefinite integral of the right side can be found straight forward using Euler formula
cos(φ)+i·sin(φ) = ei·φ
and following from it expressions for trigonometric functions
cos(φ) = (ei·φ+e−i·φ)/2
sin(φ) = (ei·φ−e−i·φ)/(2i)

The result of integration of the right side (without multiplier b) is
ea·t·sin(ωt)·dt =
=
[ea·t/(a²+ω²)]·
·
[a·sin(ωt)−ω·cos(ωt)] + K3
(K3 is any constant)

Therefore,
ea·t·y(t) = b·[ea·t/(a²+ω²)]·
·
[a·sin(ωt)−ω·cos(ωt)] + K3

Reducing by ea·t both sides, the expression for y(t)=QC(t) looks like
y(t) = [b/(a²+ω²)]·
·
[a·sin(ωt)−ω·cos(ωt)] + K4

Consider two constants a/√(a²+ω²) and ω/√(a²+ω²). It is convenient to represent them as sin(φ) and cos(φ) correspondingly for some angle φ=arctan(a/ω).
Then
a·sin(ωt)−ω·cos(ωt) =
= √(a²+ω²)·
·
[sin(φ)·sin(ωt)−cos(φ)·cos(ωt)]

The last expression equals to
−√(a²+ω²)·cos(ωt+φ)
and our equation for y(t) looks like
y(t) = −b·cos(ωt+φ)/√(a²+ω²) +
+ K4


Since y(t) is the electric charge QC(t) accumulated on the plates of a capacitor, its derivative is an electric current in the circuit:
I(t) = b·ω·sin(ωt+φ)/√(a²+ω²)

Let's restore this equation to original constants
a = 1/(R·C)
b = E0/R
The factor at sin(ωt+φ) in the equation above then is
b·ω/√(a²+ω²) =
= E0·ω/(R·√1/(R·C)²+ω²) =
= E0/(√1/(ωC)²+R²) =
= E0/(√XC²+R²)

where XC=1/(ωC) was defined in one of the previous lectures as reactance of a capacitor (or capacitive reactance).

The angle φ=arctan(a/ω) in terms of original constants is
φ=arctan(1/(R·C·ω)) =
= arctan (XC / R)

This angle is called phase shift of the current from the voltage.

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
I(t) = E0·sin(ωt+φ)/√XC²+R² =
= I0·sin(ωt+φ)

where I0=E0/√XC²+R² is the peak amperage of the electric current.
This is the Ohm's Law for RC-circuit.

The expression XC²+R² plays in this case the same role as a resistance in case of direct current circuits.

It is important that there is a phase shift φ of the oscillations of an electric current relatively to oscillation of generated EMF.

The last issue is to analyze the effective current in this RC-circuit. Since for sinusoidal oscillations the effective current is by √2 less than peak amperage, the effective current is
Ieff = I0 /2 =
= E0 /(√XC²+R²·√2) =
= Eeff /XC²+R²

This is the Ohm's Law for effective voltage and amperage in RC-circuit.

It's important to notice that in the absence of a resistor (that is, R=0) the formula for I(t) transforms into
I(t) = I0·sin(ωt+φ)
where peak amperage I0=E0/XC and phase shift φ=arctan(∞)=π/2, which corresponds to results obtained in a lecture "AC Capacitors" of the "Alternating Current Induction" chapter of this course dedicated to only a capacitor in the AC circuit.

Monday, October 26, 2020

Problems on LC Circuits: UNIZOR.COM - Physics4Teens - Electromagnetism -...

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

Problems on LC Circuit

Problem A

Consider a circuit that contains an AC generator, an inductor of inductance L and a capacitor of capacity C in a series.


Generated EMF has frequency f=50Hz and effective voltage Eeff=220V (volt).
The effective AC current is Ieff=5A (ampere).
The capacitance of a capacitor is C=10μF (microfarad).
Find inductance L (henry) of an inductor?

Solution

Let's start with an expression for the AC current in the LC circuit in terms of generated EMF and reactance of the capacitor and inductor.

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=2πf is an angular velocity of a rotor in radians per second, where f is a frequency of generated EMF in number of cycles per second.

Alternating electric current in the circuit
I(t) = I0·cos(ωt)
where
I0 = E0/(XC−XL) is peak amperage,
XC = 1/(ω·C) is capacitive reactance,
XL = ω·L is inductive reactance

Effective voltage and effective amperage are less than, correspondingly, peak voltage and peak amperage by 2.
Therefore, from the expression for AC current follows
I(t) = I0·cos(ωt) =
= E0·cos(ωt)/(XC−XL)

Ieff = I0/√2 =
= E0/
[2(XC−XL)] =
= Eeff/(XC−XL)


The unknown in this expression is XL that can be found:
XC−XL = Eeff/Ieff
XL = XC−Eeff/Ieff

Since XC=1/(ω·C) and XL=ω·L L·ω = 1/(ω·C) − Eeff/Ieff
L = 1/(ω²·C) − Eeff/(ω·Ieff)

Substituting
ω=2πf=2π·50Hz=314(1/sec)
C=10μF=10−5F
Eeff=220V
Ieff=5A
into above expression for L, we obtain (rounded to 0.001)
L= 1/(314²·10−5)−220/(314·5)=
= 0.874
H (henry)



Problem B

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 R and a capacitor with capacitance C connected parallel to a resistor.


The signal (electrical current) coming through a resistor will have higher frequencies weakened by a presence of a capacitor.
Explain why.

Explanation

Reactance of a capacitor XC, which functionally similar to a resistance of a resistor, is inversely proportional to a frequency of the voltage on its ends
XC = 1/(ω·C) = 1/(2πf·C),
where
ω=2πf is the angular velocity of oscillations,
f is a frequency of oscillations,
C is the capacitance of a capacitor.
Therefore, the greater frequency - the less reactance of a capacitor.
Since reactance 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.

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.

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 E, but also at any moment t, when generated EMF oscillates (even irregularly) as a function of time E(t).
Therefore, a capacitor will lower reactance will have higher current going through it, thus weakening the higher frequency oscillations of the current going through a resistor.

Problem C

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 R and an inductor with inductance L connected parallel to a resistor.


The signal (electrical current) coming through a resistor will have lower frequencies weakened by a presence of a inductor.
Explain why.

Explanation

Reactance of an inductor XL, which functionally similar to a resistance of a resistor, is proportional to a frequency of the voltage on its ends
XL = ω·L = 2πf·L,
where
ω=2πf is the angular velocity of oscillations,
f is a frequency of oscillations,
L is the inductance of an inductor.
Therefore, the lower frequency - the less reactance of an inductor.
Since reactance 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.

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.

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 E, but also at any moment t, when generated EMF oscillates (even irregularly) as a function of time E(t).
Therefore, an inductor will lower reactance will have higher current going through it, thus weakening the lower frequency oscillations of the current going through a resistor.

SUMMARY
Capacitors and inductors can be used to filter certain "useful" frequencies from the noisy signals.

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)






Answer



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.