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Alternating Current Theory - J R Lucas
An alternating waveform is a periodic waveform which alternate between positive and negative values. Unlike direct waveforms, they cannot be characterised by one magnitude as their amplitude is continuously varying from instant to instant. Thus various forms of magnitudes are defined for such waveforms.
The advantage of the alternating waveform for electric power is that it can be stepped up or stepped down in potential easily for transmission and utilisation. Alternating waveforms can be of many shapes. The one that is used with electric power is the sinusoidal waveform. This has an equation of the form
v(t) = Vm sin( ω t + φ )
If the period of the waveform is T, then its angular frequency ω corresponds to ωT = 2π.
(a) Instantaneous value: The instantaneous value of a waveform is the value of the waveform at any given instant of time. It is a time variable a(t).
For a sinusoid, Instantaneous value a(t) = Am sin( ω t+ φ )
(b) Peak value: The peak value, or maximum value , of a waveform is the maximum instantaneous value of the waveform. For a sinusoid, Peak value = Am
(c) Mean value: The mean value of a waveform is equal to the mean value over a complete cycle of the waveform. It also corresponds to the direct component of the waveform.
Mean value Amean = ∫
t + T
t
o
o
at dt T
The mean value of a waveform which has equal positive and negative half cycles must thus be always zero.
For a sinusoid, Mean value = ∫
t T
t
m
o
o
A t dt T
sin( ).
ω φ = 0
(d) Average value (rectified): The full-wave rectified average value or average value of a waveform is defined as the mean value of the rectified waveform over a complete cycle.
Average value Aavg =
∫ −^ ∫
hal fcycle negative halfcycle positive
at dt at dt T
For a sinusoid,
Average value =
∫ −^ ∫
T
T m
T
T Am^ tdt A tdt 2
2
0
sin. sin.
ω ω
= Am Am ω T π
The average value is defined in the above manner in electrical engineering as otherwise all alternating waveforms would have zero value and would be indistinguishable.
a(t)
−Am
Amean
v(t)
t
T
Vm T φ/ω
Aavg
vrect
T t
T
(e) Effective value or r.m.s. value:
Neither the peak value, nor the mean value, nor the average value defines the useful value of the waveform with regard to the power or energy. Thus the effective value is defined based on the power equivalent of the quantity.
The effective value is thus defined as the constant value which would cause the same power dissipation as the original quantity over one complete period.
Thus considering current Power dissipation = Ieff^2. R = ∫
t + T
t
o
o
i t Rdt T
giving Ieff = ∫
t + T
t
o
o
i t dt T
or in general Effective value Aeff = ∫
t + T
t
o
o
a t dt T
It is to be noted that the method of calculating the effective value involves the following processes. Taking the root of the mean of the square d waveform over one complete cycle.
Thus it is also designated as the root mean square value, or the r.m.s. value of the waveform.
i.e. r.m.s. value Arms = ∫
t + T
t
o
o
a t dt T
This is defined for both current as well as voltage.
For a sinusoid,
r.m.s. value = ∫
t T
t
m
o
o
A t dt T
sin ( ).
ω φ = 2
A m
Unless otherwise specified, the rms value is the value that is always specified for ac waveforms, whether it be a voltage or a current. For example, 230 V in the mains supply is an rms value of the voltage. Similarly when we talk about a 5 A , 13 A or 15A socket outlet (plug point), we are again talking about the rms value of the rated current of the socket outlet.
For a given waveform, such as the sinusoid , the peak value , average value and the rms value are dependant on each other. The peak factor and the form factor are the two factors that are most commonly defined.
average value
rmsvalue FormFactor =
and for a sinusoidal waveform, Form Factor = 1. 1107 1. 111
π
Vm V m
The form factor is useful such as when the average value has been measured using a rectifier type moving coil meter and the rms value is required to be found. [ Note: You will be studying about these meters later]
Peak Factor = peak value rms value
and for a sinusoidal waveform,
Peak Factor =
The peak factor is useful when defining highly distorted waveforms such as the current waveform of compact fluorescent lamps.
V V m m 2
= 2 = 1 4142.
v^2 (t)
t T
T
t
i
Consider a line OP of length A (^) m which is in the horizontal direction OX at time t=0.
If OP rotates at an angular velocity ω , then in time t its position would correspond to an angle of ω t.
The projection of this rotating phasor OP (a phasor is somewhat similar to a vector, except that it does not have a physical direction in space but a phase angle) on the y-axis would correspond to OP sin ωt or A (^) m sin^ ω^ t and on the x-axis would correspond to A (^) m cos^ ω^ t. Thus the sinusoidal waveform can be thought of being the projection on a particular direction of the complex exponential ejωt^.
If we consider more than one phasor, and each phasor rotates at the same angular frequency, then there is no relative motion between the phasors. Thus if we fix the reference phasor OR in a particular reference direction (without showing its rotation), then all others phasors moving at the same angular frequency would also be fixed at a relative position. Usually this reference direction is chosen as horizontal on the diagram for convenience.
It is also usual to draw the Phasor diagram using the rms value A of the sinusoidal waveform, rather than with the peak value A (^) m. This is shown on an enlarged diagram. Thus unless otherwise specified it is the rms value that is drawn on a phasor diagram.
It should be noted that the values on the phasor diagram are no longer time variables. The phasor A is characterised by its magnitude A and its phase angle φ. These are also the polar co-ordinates of the phasor and is commonly written as A φ. The phasor A can also be characterised by its cartesian co-ordinates Ax and Ay and usually written using complex numbers as A = Ax + j Ay.
Note: In electrical engineering, the letter j is always used for the complex operator − 1 because the letter i is regularly used for electric current.
It is worth noting that (^) A = Ax^2^ + Ay^2 and that (^) tan φ = A A
y x
or φ =
tan −^1
A A
y x
Also, Ax = A cos φ, Ay = A sin φ and A ejφ^ = A cos φ + j A sin φ = Ax + j Ay
Note: If the period of a sinusoidal waveform is T , then the corresponding angle would be ω T. Also, the period of a waveform corresponds to 1 complete cycle or 2 π radians or 360^0.
∴ ω T = 2 π
R
P
φ 0
Am
R
P
φ 0
A
a(t)
t
T
Am sin ( ω t+ φ)
ω
ω
(^0 )
R
X
P
φ
Rotating Phasor diagram
reference direction
φ 0
A Am = 2
Phasor diagram
A
A
Phase difference
Consider the two waveforms Am sin ( ω t+ φ 1 ) and Bm sin ( ω t+ φ 2 ) as shown in the figure. It can be seen that they have different amplitudes and different phase angles with respect to a common reference.
These two waveforms can also be represented by either rotating phasors A (^) m e j( ω t+ φ 1 )^ and B (^) m ej ( ω t+ φ 2 )^ with peak amplitudes A (^) m and B (^) m , or by a normal phasor diagram with complex values A and B with polar co-ordinates A φ 1 and B φ 2 as shown.
Any particular value (such as positive peak, or zero) of a(t) is seen to occur at a time T after the corresponding value of b(t). i.e. the positive peak A (^) m occurs after an angle (φ 2 −φ 1 ) after the positive peak Bm. Similarly the zero of a(t) occurs after an angle (φ 2 −φ 1 ) after the corresponding zero of b(t). In such a case we say that the waveform b(t) leads the waveform a(t) by a phase angle of (φ 2 −φ 1 ). Similarly we could say that the waveform a(t) lags the waveform b(t) by a phase angle of (φ 2 −φ 1 ). [ Note: Only the angle less than 180o^ is used to specify whether a waveform leads or lags another waveform].
We could also define, lead and lag by simply referring to the phasor diagram. Since angles are always measured anticlockwise (convention), we can see from the phasor diagram, that B leads A by an angle of (φ 2 −φ 1 ) anticlockwise or that A lags B by an angle (φ 2 −φ 1 ).
Addition and subtraction of phasors can be done using the same parallelogram and triangle laws as for vectors, generally using complex numbers. Thus the addition of phasor A and phasor B would be
A + B = (A cos φ 1 + j A sin φ 1 ) + (B cos φ 2 + j B sin φ 2 ) = (A cos φ 1 + B cos φ 2 ) + j (A sin φ 1 + B sin φ 2 ) = C (^) x + jC (^) y = C φ c = C
where (^) C = C (^) x^2^ + C (^) y^2 = ( A cos φ (^) 1 + B cos φ 2 ) 2 + ( A sin φ (^) 1 + B sin φ 2 )^2
and (^)
= +
= − − ( cos cos ) tan tan ( sin sin ) 1 2
(^1112) φ φ φ φ φ A B
A B C
C x c y
Example 1
Find the addition and subtraction of the 2 complex numbers given by 10∠ 30 o^ and 25∠ 48 o.
Addition = 10 ∠ 30 o^ + 25 ∠ 48 o^ = 10(0.8660 + j 0.5000) + 25(0.6691 + j 0.7431)
= (8.660 + 16.728) + j (5.000 + 18.577) = 25.388 + j 23.577 = 34.647 ∠ 42.9o
Subtraction = 10 ∠ 30 o^ − 25 ∠ 48 o^ = (8.660 − 16.728) + j (5.000 − 18.577)
= (^) − 8.068 (^) − j 13.577 = 15.793∠239.3o
φ 1 0
A Am = φ 2 - φ 1^2
B B (^) m = 2
φ 1 0
φ 2
B
C
A
φ 1 O φ 2
y(t)
ω t
Am sin ( ω t+ φ 1 )
Bm sin ( ω t+ φ 2 )
ωT
φ 2 − φ 1
φ 2 − φ 1
Am
Bm
ω
It can be seen that the rms magnitude of voltage is related to the rms magnitude of current by the multiplying factor ω L. It also seen that the voltage waveform leads the current waveform by 90o^ or π/2 radians or that the current waveform lags the voltage waveform by 90o^ for an inductor.
Thus it is usual to write the relationship as V = j ω L.I or V = ω L.I 90 ο
The impedance Z of the inductance may thus be defined as j ω L , and V = Z. I corresponds to the generalised form of Ohm’s Law.
Remember also that the power dissipation in a pure inductor is zero, as energy is only stored and as there is no resistive part in it, but that the product V. I is not zero.
(3) Capacitor
for a sinusoid, consider i(t) = Real part of [ Im ej(ωt+θ)^ ] or Im cos (ωt+θ)
v(t) = Real [^1 C
∫ I em^ j^ (^^ ω^ t +θ )^. dt ] =^ Real^ [^^1
C****. j ω
. I (^) m e (jωt+θ)^ ] = Real [^1 j
Vm
e (jωt+θ)^ ] or v(t) =^1 C
∫ I^ m^ cos(^ ω^^ t^ +θ ). dt =^^1
C ω
.Im sin (ωt+θ) =^1 C ω
.Im cos
(ωt+θ−π/2) =Vm cos (ωt+θ−π/2) ∴ Vm =^1 ω C
.I (^) m and Vm/√2 =^1 ω C
.Im/√ 2
It can be seen that the rms magnitude of voltage is related to the rms magnitude of current by the multiplying factor^1 ω C
It also seen that the voltage waveform lags the current waveform by 90o^ or π/2 radians or that the current waveform leads the voltage waveform by 90o^ for a capacitor.
Thus it is usual to write the relationship as V =^1 j ω C
.I or V =^1 ω C
.I 90 ο
The impedance Z of the inductance may thus be defined as^1 j (^) ω C
, and V = Z. I corresponds
to the generalised form of Ohm’s Law.
Remember also that the power dissipation in a pure capacitor is zero, as energy is only stored and as there is no resistive part in it, but that the product V. I is not zero.
Impedance and Admittance in an a.c. circuit
The impedance Z of an a.c. circuit is a complex quantity. It defines the relation between the complex rms voltage and the complex rms current. Admittance Y is the inverse of the impedance Z.
V = Z. I , I = Y. V where Z = R + j X, and Y = G + j B
It is usual to express Z in cartesian form in terms of R and X , and Y in terms of G and B.
The real part of the impedance Z is resistive and is usually denoted by a resistance R , while the imaginary part of the impedance Z is called a reactance and is usually denoted by a reactance X.
v t C
( ) =^1 ∫ i t ( ). dt
O ω t
Vm sin ω t
Im cos ω t
ωT
π/
C
v
i
1 j ω C
V
I
V
I
O
Phasor diagram
V
I
O
Phasor diagram
It can be seen that the pure inductor and the pure capacitor has a reactance only and not a resistive part, while a pure resistor has only resistance and not a reactive part.
Thus Z = R + j 0 for a resistor, Z = 0 + j ω L for an inductor, and Z = 1 j ω C
= 0 − j^1 ω C
for a
capacitor.
The real part of the admittance Y is a conductance and is usually denoted by G , while the imaginary part of the admittance Y is called a susceptance and is denoted by B.
Relationships exist between the components of Z and the components of Y as follows.
G j B Y Z R j X
R j X R X
= −
1 1 2 2
so that (^) G R R X
B X R X
=
= − (^2 2 2) + 2 , and
The reverse process can also be similarly done if necessary.
However, it must be remembered that in a complex circuit, G does not correspond to the inverse of the resistance R but its effective value is influenced by X as well as seen above.
Simple Series Circuits
In the case of single elements R, L and C we found that the angle difference between the voltage and the current was either zero, or ± 90 o. This situation changes when there are more than one component in a circuit.
R-L series circuit
In the series R-L circuit, considering current I as reference
VR = R.I, VL = jωL.I, and V = VR + VL ∴ V = (R + jωL).I so that the total series impedance is Z = R + jωL
The above phasor diagram has been drawn with I as reference. [i.e. I is drawn along the x- axis direction]. The current was selected as reference in this example, because it is common to both the resistance and the inductance and makes the drawing of the circuit diagram easier. In this diagram, the voltage across the resistor VR is in phase with the current, where as the voltage across the inductor VL is 90 o^ leading the current. The total voltage V is then obtained by the phasor addition (similar to vector addition) of V (^) R and VL.
If the total voltage was taken as the reference, the diagram would just rotate as shown. In this diagram, the current is seen to be lagging the voltage by the same angle that in the earlier diagram the voltage was seen to be leading the current. V (^) L has been drawn from the end of VR rather than from the origin for ease of obtaining the resultant V from the triangular law.
In an R-L circuit, the current lags the voltage by an angle less than 90 o^ and the circuit is said to be inductive.
Note that the power dissipation can only occur in the resistance in the circuit and is equal to R. I 2 and that this is not equal to product V. I for the circuit.
R-C series circuit
In the series R-C circuit, VR = R.I, VC = 1 j ω C
. I = − j 1 ω C
and V = VR + VC
R^1
j C
I
V V
V
Phasor diagram
V VC
I VR
φ
Phasor diagram
V
VL
VR
I
φ
R I
V VL
V
φ
V VL
I VR
Phasor diagram
VL
Note that the power dissipation can only occur in the resistance in the circuit and is equal to R. I 2 and that this is not equal to product V. I for the circuit.
Simple Parallel Circuits
R-L parallel Circuit
In the parallel R-L circuit, considering V as reference V = R.IR , V = jωL.IL, and I = IR + IL ∴ (^) I V R
V j L = + ω ∴ total shunt admittance =^1 R j L
ω
R-C parallel Circuit
In the parallel R-C circuit, considering V as reference V = R.IR , IC = jωC.V, and I = IR + IL ∴ (^) I V R
= + V****. j ω C
∴ total shunt admittance =^1 R
R-L-C parallel Circuit
In the parallel R-L-C circuit , considering V as reference V = R.IR , V = jωL.IL, IC = jωC.V and I = IR + IL + IC ∴ (^) I V R
V j L
= + + V j C ω
. ω
∴ total shunt admittance =^1 R j L
ω
As in the case of the series circuit, shunt resonance will occur when^1 ω
ω L
= C giving a
minimum value of shunt admittance.
Note that even in the case of a parallel circuit, power loss can only occur in the resistive elements and that the product V. I is not usually equal to the power loss.
Power and Power Factor
It was noted that in an a.c. circuit, power loss occurs only in resistive parts of the circuit and in general the power loss is not equal to the product V. I and that purely inductive parts and purely capacitive parts of a circuit did not have any power loss. To account for this apparent discrepancy, we define the product V. I as the apparent power S of the circuit.
Apparent power has the unit volt-ampere (VA) and not the watt (W), and watt (W) is used only for the active power P of the circuit (which we earlier called the power )
apparent power S = V. I
Phasor
φ
I IL
IR V
IL
Phasor
φ
I I
L
IR V
IL
R
1 j ω C
I R
V
I C
I
I R^ R
V
I L
I
R
1 j ω C
I R
V
I L
I
I C
j ω L
Phasor
φ
I IL+IC
IR V
IC
IL
Since a difference exists between the apparent power and the active power we define a new term called the reactive power Q for the reactance X.
The instantaneous value of power p(t) is the product of the instantaneous value of voltage v(t) and the instantaneous value of current i(t).
i.e. p(t) = v(t). i(t)
If v(t) = Vm cos ω t and i(t) = Im cos ( ω t − φ ) ,
where the voltage has been taken as reference and the current lags the voltage by a phase angle φ
then p(t) = Vm cos ω t. I (^) m cos ( ω t − φ ) = Vm I (^) m. (^) 21. 2 cos ω t. cos ( ω t − φ )
= 21 Vm I (^) m [cos (2 ω t − φ ) + cos φ]]
It can be seen that the waveform of power p(t) has a sinusoidally varying component and a constant component.
Thus the average value of power (active power) P would be given by the constant value ½ Vm I (^) m cos φ.
active power P = ½ Vm I (^) m cos φ (^) = V^ m Im 2 2
. .cos φ =^ V. I cos^ φ
The term cos φ is defined as the power factor , and is the ratio of the active power to the apparent power.
Note that for a resistor , φ = 0o^ so that P = V. I
and that for an inductor , φ = 90 o^ lagging (i.e. current is lagging the voltage by 90 o) so that P = 0
and that for an capacitor , φ = 90o^ leading (i.e. current is leading the voltage by 90 o) so that P = 0
For combinations of resistor , inductor and capacitor , P lies between V. I and 0
For an inductor or capacitor, V. I exists although P = 0. For these elements the product V. I is defined as the reactive power Q. This occurs when the voltage and the current are quadrature (90 o^ out of phase). Thus reactive power is defined as the product of voltage and current components which are quadrature.
This gives
reactive power Q = V. I sin φ
Vm I (^) m cos t
v(t)
i(t)
t
current lagging voltage by angle φ inphase
Since we are interested in the relationship between e(t) and i (^) r(t), we eliminate the other variables to give
L p. i (^) L(t) = r. i (^) r(t), L p. i(t) = L p .[ i (^) r(t) + i (^) L(t)] = (Lp + r ). i (^) r(t) e(t) = r. ir(t) + v (^) C (t) + R. i(t)
i.e. C p. e(t) = C p. r. i (^) r(t) + i(t) + C p. R. i(t)
L p. C p. e(t) = L p. C p. r. i (^) r(t) + (1 + C p. R). i (^) r(t)
i.e. L.C. p^2. e(t) = [L.C. p^2 .r + 1 + C p. R]. i (^) r(t)
which is a differential equation involving the up to the second derivative of both e(t) and i (^) r(t).
or f 1 (p). e(t) = f 2 (p). i (^) r(t)
or e(t) = Zr(p). i (^) r(t)
Thus we can see that the source emf e(t) and the current i (^) r(t) are related by an impedance transfer function of the differential operator p.
In a similar way, the relationship between any current and any voltage would be related by an admittance transfer function of the differential operator p and between any two voltages and any two currents by a transfer gain of the differential operator p.
It is to be noted, that in the case of sinsusoidal a.c., the differential operator p may be replaced using the relationship p = j ω.
Example 3
Determine the mean value, average value, peak value, rms value, form factor and peak factor for the waveform shown.
Solution
Mean value = ∫ = ∫ − = − = − =
=
T T
t
T E
T
T
T
T
E
T
t t T
E
dt T
t E T
f t dt T (^) 0
2
0
2
0 2
[This result could also have been written by inspection considering areas].
Average value = 6
[ 5 ]
[ 2 ( ( ) )]
3
1 2
1 3
2 2
1 T E
T
E
E T E T
T
× × − × − × = =
Peak value = 2E
rms value =
T
t
T T T
T
t T
E
dt T
t E T
f t dt T (^) 0
3
2
0
2 2 0
=
∫ =^ ∫ − = − × −
E
E
= − [− 1 − 8 ]=
2
form factor = 1. 2 5
E
E
peak factor = 2. 4 5
E
E
2E
0 T 2T 3T
−E
Example 4
A certain 50 Hz , alternating voltage source has an internal emf of 250 V and an internal inductance of 31.83 mH. If the terminal voltage is to be maintained at 230 V , determine the value of the maximum power that can be delivered to a load (R + jX) and the values of R and X under these conditions. Draw also the phasor diagram showing the voltages and currents in the circuit under these conditions.
Solution
at 50 Hz, x s = jωL = j 2π× 50 ×31.83× 10 -3^ = j 10.00 Ω
Across load
Current I = j 10 + R + jX
, | V | = 230 V
active power P = | I |^2 R = R R X
[ + ( + 10 ) ]
2 2
2 , voltage V = (R+jX). I ,
| V |^2 = 230^2 = (R 2 +X^2 ). | I |^2 =
[ ( 10 ) ]
2 2 2
R X
R X
∴ R 2 + (X + 10)^2 = 1.185 R 2 + 1.1815 X^2
i.e. 20 X + 10 2 = 0.1815 (R 2 + X^2 ) or R 2 + X^2 = 5.5104 (10 2 + 20X)
We can differentiate for maximum power keeping the condition in mind.
i.e. P = R X
6. 5104 [ 10 + 20 ]
2
2
dR
dP gives the condition (20X+100).1 – R.20. = 0 dR
dX or X+5 = R dR
dX
also 20 dR
dX
dX )
so that 20(X+5) = 0.363R 2 +0.363X(X+5)
i.e. 20X + 100 = 0.363R 2 +0.363X^2 +1.815X
0.363(R 2 + X^2 ) = 18.185X + 100
but 20 X + 100 = 0.1815 (R 2 + X^2 ) = (9.0925X+50)
i.e. 10.9075X = -
X = -4.584 Ω, giving R = 4.983 Ω
substituting gives Maximum Power = 5750 W,
under the given conditions,
31.83 mH
250 V R+jX
If we differentiate and obtain the condition for maximum power,
0 , = 0 ∂
X
P
and R
P
, we can show that these give the conditions
[R 2 + (X + 10)^2 ].1 – R.2R = 0 and (X + 10) = 0 or X = − 10 Ω and R = 0 Then P = ∞, V = ∞
j10 Ω
250 V 4.983-j4.
