Bird J. Electrical Circuit Theory and Technology
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658 Electrical Circuit Theory and Technology
(d) From equation (36.16), power dissipated,
P D I
2
R D 13.89
2
12 D 2315 W or 2.315 kW
(Alternatively, equation (36.14) may be used to determine power.)
(e) From equation (36.17),
overall power factor D
2315
173.3513.89
D 0.961
Problem 10. An e.m.f. is represented by
e D 50 C 200 sin ωt C 40 sin
2ωt
2
C 5sin
4ωt C
4
volts,
the fundamental frequency being 50 Hz. The e.m.f. is applied across
a circuit comprising a 100
µF capacitor connected in series with a
50 # resistor. Obtain an expression for the current flowing and
hence determine the rms value of current.
D.c. component
In a d.c. circuit no current will flow through a capacitor. The current wave-
form will not possess a d.c. component even though the e.m.f. waveform
has a 50 V d.c. component. Hence i
0
D 0.
Fundamental
Capacitive reactance,
X
C1
D
1
2fC
D
1
250100 ð 10
6
D 31.83 #
Impedance Z
1
D 50 j31.83# D 59.27
6
32.48
°
#
I
1m
D
V
1m
Z
1
D
200
6
0
°
59.27
6
32.48
°
D 3.374
6
32.48
°
A D 3.374
6
0.567 A
Hence the fundamental current, i
1
D 3.374sinωt C 0.567A
Second harmonic
Capacitive reactance,
X
C2
D
1
22fC
D
31.83
2
D 15.92 #
Impedance Z
2
D 50 j15.92# D 52.47
6
17.66
°
#
I
2m
D
V
2m
Z
2
D
40
6
/2
52.47
6
17.66
°
D 0.762
2
17.66
°
D 0.762
6
72.34
°
A
Complex Waveforms 659
Hence the second harmonic current, i
2
D 0.762sin2ωt 72.34
°
A
D 0.762sin2ωt 1.263A
Fourth harmonic
Capacitive reactance, X
C4
D
1
4
X
C1
D
31.83
4
D 7.958 #
Impedance, Z
4
D 50 j7.958# D 50.63
6
9.04
°
#
I
4m
D
V
4m
Z
4
D
5
6
/4
50.63
6
9.04
°
D 0.099
6
/4 9.04
°
D 0.099
6
54.04
°
A
Hence the fourth harmonic current, i
4
D 0.099sin4ωt C 54.04
°
A
D 0.099sin4ωt C 0.943A
An expression for current flowing is therefore given by
i D i
0
C i
1
C i
2
C i
4
i.e., i= 3.374sin.!t Y 0.567/ Y 0.762sin.2!t − 1.263/
Y 0.099sin.4!t Y 0.943/amperes
From equation (36.5), rms current,
I D
3.374
2
C 0.762
2
C 0.099
2
2
D 2.45 A
Problem 11. A complex voltage is represented by:
D 25 C100sin ωt C 40sin
3ωt C
6
C 20sin
5ωt C
12
volts
where ω D 10
4
rad/s. The voltage is applied to a series circuit
comprising a 5.0 # resistance and a 500
µH inductance.
Determine (a) an expression to represent the current flowing in
the circuit, (b) the rms value of current, correct to two decimal
places, and (c) the power dissipated in the circuit, correct to three
significant figures.
(a) d.c. component
Inductance has no effect on a steady current. Hence the d.c. compo-
nent of the current, i
0
, is given by
i
0
D
0
R
D
25
5.0
D 5.0A
660 Electrical Circuit Theory and Technology
Fundamental
Inductive reactance, X
L1
D ωL D 10
4
500 ð 10
6
D 5 #
Impedance, Z
1
D 5 C j5# D 7.071
6
45
°
#
I
1m
D
V
1m
Z
1
D
100
6
0
°
7.071
6
45
°
D 14.14
6
45
°
A
D 14.14
6
/4 A or 14.14
6
0.785 A
Hence fundamental current, i
1
D 14.14sinωt 0.785A
Third harmonic
Inductive reactance at third harmonic frequency,
X
L3
D 3X
L1
D 15 #
Impedance, Z
3
D 5 C j15# D 15.81
6
71.57
°
#
I
3m
D
V
3m
Z
3
D
40
6
/6
15.81
6
71.57
°
D 2.53
6
41.57
°
A
D 2.53
6
0.726 A
Hence the third harmonic current, i
3
D 2.53sin3ωt 41.57
°
A
D 2.53sin3ωt 0.726A
Fifth harmonic
Inductive reactance at fifth harmonic frequency, X
L5
D 5X
L1
D 25 #
Impedance, Z
5
D 5 C j25# D 25.495
6
78.69
°
#
I
5
D
V
5m
Z
5
D
20
6
/12
25.495
6
78.69
°
D 0.784
6
63.69
°
A
D 0.784
6
1.112 A
Hence the fifth harmonic current, i
5
D 0.784sin5ωt 63.69
°
A
D 0.784sin5ωt 1.112A
Thus current, i D i
0
C i
1
C i
3
C i
5
, i.e.,
i
= 5 Y 14.14sin.!t − 0.785/ Y 2.43sin.3!t − 0.726/
Y 0.784sin.5!t
− 1.112/A
(b) From equation (36.9), rms current,
I D
5.0
2
C
14.14
2
C 2.53
2
C 0.784
2
2
D 11.3348 A D 11.33 A, correct to two decimal places.
Complex Waveforms 661
(c) From equation (36.16), power dissipated,
P D I
2
R D 11.3348
2
5.0 D 642.4W
D 642 W, correct to three significant figures
(Alternatively, from equation (36.15),
power P D 255.0 C
100
p
2
14.14
p
2
cos45
°
C
40
p
2
2.53
p
2
cos71.57
°
C
20
p
2
0.784
p
2
cos78.69
°
D 125 C 499.92 C 16.00 C1.54
D 642.46 W or 642 W,
correct to three significant figures, as above.
Problem 12. The voltage applied to a particular circuit comprising
two components connected in series is given by
D 30 C40sin 10
3
t C 25sin2 ð10
3
t C 15sin4 ð 10
3
tvolts
and the resulting current is given by
i D 0.743sin10
3
t C 1.190 C 0.781sin2 ð10
3
t C 0.896
C 0.636 sin4 ð10
3
t C 0.559A
Determine (a) the average power supplied, (b) the type of compo-
nents present, and (c) the values of the components.
(a) From equation (36.15), the average power P is given by
P D 300 C
40
p
2
0.743
p
2
cos1.190
C
25
p
2
0.781
p
2
cos0.896 C
15
p
2
0.636
p
2
cos0.559
i.e., P D 0 C5.523 C 6.099 C 4.044 D 15.67 W
(b) The expression for the voltage contains a d.c. component of 30 V.
However there is no corresponding term in the expression for current.
This indicates that one of the components is a capacitor (since in
a d.c. circuit a capacitor offers an infinite impedance to a direct
current). Since power is delivered to the circuit the other component
is a resistor.
662 Electrical Circuit Theory and Technology
(c) From equation (36.5), rms current,
I D
0.743
2
C 0.781
2
C 0.636
2
2
D 0.885 A
Average power P D I
2
R, from which,
resistance R D
P
I
2
D
15.67
0.885
2
D 20 Z
At the fundamental frequency, ω D 10
3
rad/s
impedance jZ
1
jD
V
1m
I
1m
D
40
0.743
D 53.84 #
Impedance jZ
1
jD
R
2
C X
2
C1
, from which
X
C1
D
Z
2
1
R
2
D
53.84
2
20
2
D 50 #
Hence 1/ωC D 50, from which
capacitance C D
1
ω50
D
1
10
3
50
D 20 mF
Problem 13. In the circuit shown in Figure 36.17 the supply vol-
tage is given by D 300 sin 314t C 120sin942t C 0.698 volts.
Determine (a) an expression for the supply current, i, (b) the
percentage harmonic content of the supply current, (c) the total
power dissipated, (d) an expression for the p.d. shown as
1
,and
(e) an expression for current i
c
Figure 36.17
(a) Capacitive reactance of the 2.123
µF capacitor at the fundamental
frequency is given by
X
C1
D
1
3142.123 ð 10
6
D 1500 #
At the fundamental frequency the total circuit impedance, Z
1
,is
given by
Z
1
D 560 C
2000j1500
2000 j1500
D 560 C
3 ð 10
6
6
90
°
2500
6
36.87
°
D 560 C 1200
6
53.13
°
D 560 C 720 j960
D 1280 j960# D 1600
6
36.87
°
#
D 1600
6
0.644 #
Since for the nth harmonic the capacitive reactance is 1/nωC,
the capacitive reactance of the third harmonic is
1
3
X
C1
D
1
3
1500 D
500 #. Hence at the third harmonic frequency the total circuit
Complex Waveforms 663
impedance, Z
3
, is given by
Z
1
D 560 C
2000j500
2000 j500
D 560 C
10
6
6
90
°
2061.55
6
14.04
°
D 560 C 485.07
6
75.96
°
D 560 C 117.68 j470.58
D 677.68 j470.58# D 825
6
34.78
°
#
D 825
6
0.607 #
The fundamental current
i
1
D
1
Z
1
D
300
6
0
1600
6
0.644
D 0.188
6
0.644 A
The third harmonic current
i
3
D
3
Z
3
D
120
6
0.698
825
6
0.607
D 0.145
6
1.305 A
Thus, supply current, i
= 0.188sin.314t Y 0.644/
Y 0.145sin.942t Y 1.305/A
(b) Percentage harmonic content of the supply current is given by
0.145
0.188
ð 100% D 77%
(c) From equation (36.14), total active power
P D
300
p
2
0.188
p
2
cos0.644
C
120
p
2
0.145
p
2
cos0.607
i.e., P D 22.55 C7.15 D 29.70 W
(d) Voltage
1
D iR D 560[0.188sin314t C 0.644
C 0.145sin942t C 1.305],
i.e., n
1
D 105.3sin.314t Y 0.644/
Y 81.2sin.942t Y 1.305/ volts
(e) Current, i
c
D i
1
R
R jX
C1
C i
3
R
R jX
C3
by current division
D 0.188
6
0.644
2000
2000 j1500
C 0.145
6
1.305
2000
2000 j500
664 Electrical Circuit Theory and Technology
D 0.188
6
0.644
2000
2500
6
0.644
C 0.145
6
1.305
2000
2061.55
6
0.245
D 0.150
6
1.288 C 0.141
6
1.550
Hence i
c
= 0.150sin314t Y 1.288/ Y 0.141sin.942t Y 1.550/A
Further problems on harmonics in single phase circuits may be found in
Section 36.9, problems 16 to 24, page 673.
36.7 Resonance due to
harmonics
In industrial circuits at power frequencies the typical values of L and
C involved make resonance at the fundamental frequency very unlikely.
(An exception to this is with the capacitor-start induction motor where
the start-winding can achieve unity power factor during run-up.)
However, if the voltage waveform is not a pure sine wave it is quite
possible for the resonant frequency to be near the frequency of one of the
harmonics. In this case the magnitude of the particular harmonic in the
current waveform is greatly increased and may even exceed that of the
fundamental. The effect of this is a great distortion of the resultant current
waveform so that dangerous volt drops may occur across the inductance
and capacitance in the circuit.
When a circuit resonates at one of the harmonic frequencies of the
supply voltage, the effect is called selective or harmonic resonance.
For resonance with the fundamental, the condition is ωL D 1/ωC;for
resonance at, say, the third harmonic, the condition is 3 ωL D 1/3ωC;
for resonance at the nth harmonic, the condition is
n!L = 1=.n!C/
.
Problem 14. A voltage waveform having a fundamental of
maximum value 400 V and a third harmonic of maximum
value 10 V is applied to the circuit shown in Figure 36.18.
Determine (a) the fundamental frequency for resonance with the
third harmonic, and (b) the maximum value of the fundamental
and third harmonic components of current.
(a) Resonance with the third harmonic means that 3ωL D 1/3ωC, i.e.,
ω D
1
9LC
D
1
3
p
0.50.2 ð 10
6
D 1054 rad/s
Figure 36.18
Complex Waveforms 665
from which, fundamental frequency, f D
ω
2
D
1054
2
D 167.7Hz
(b) At the fundamental frequency,
impedance Z
1
D R C j
ωL
1
ωC
D 2 C j
10540.5
1
10540.2 ð 10
6
D 2 j4217#
i.e., Z
1
D 4217
6
89.97
°
#
Maximum value of current at the fundamental frequency,
I
1m
D
V
1m
Z
1
D
400
4217
D 0.095 A
At the third harmonic frequency,
Z
3
D R C j
3ωL
1
3ωC
D R
since resonance occurs at the third harmonic, i.e., Z
3
D 2#
Maximum value of current at the third harmonic frequency,
I
3m
D
V
3m
Z
3
D
10
2
D 5A
(Note that the magnitude of I
3m
compared with I
1m
is 5/0.095, i.e.,
ð 52.6 greater.)
Problem 15. A voltage wave has an amplitude of 800 V at the
fundamental frequency of 50 Hz and its nth harmonic has an ampli-
tude 1.5% of the fundamental. The voltage is applied to a series
circuit containing resistance 5 #, inductance 0.369 H and capaci-
tance 0.122
µF. Resonance occurs at the nth harmonic. Determine
(a) the value of n, (b) the maximum value of current at the nth
harmonic, (c) the p.d. across the capacitor at the nth harmonic and
(d) the maximum value of the fundamental current.
(a) For resonance at the nth harmonic, nωL D 1/nωC, from which
n
2
D
1
ω
2
LC
and n D
1
ω
p
LC
Hence n D
1
250
p
0.3690.122 ð 10
6
D 15
Thus resonance occurs at the 15th harmonic.
666 Electrical Circuit Theory and Technology
(b) At resonance, impedance Z
15
D R D 5 #. Hence the maximum
value of current at the 15th harmonic,
I
15m
D
V
15m
R
D
1.5/100 ð 800
5
D 2.4A
(c) At the 15th harmonic, capacitive reactance,
X
C15
D
1
15ωC
D
1
152500.122 ð 10
6
D 1739 #
Hence the p.d. across the capacitor at the 15th harmonic
D I
15m
X
C15
D 2.41739 D 4.174 kV
(d) At the fundamental frequency, inductive reactance,
X
L1
D ωL D 2500.369 D 115.9 #,
and capacitive reactance,
X
Cl
D
1
ωC
D
1
2500.122 ð 10
6
D 26091 #
Impedance at the fundamental frequency,
jZjD
p
[R
2
C X
C
X
L
2
]
D 25975 #
Maximum value of current at the fundamental frequency,
I
1m
D
V
1m
Z
1
D
800
25975
D 0.031 A or 31 mA
Further problems on harmonic resonance may be found in Section 36.9,
problems 25 to 29, page 676.
36.8 Sources of
harmonics
(i) Harmonics may be produced in the output waveform of an a.c.
generator. This may be due either to ‘tooth-ripple’, caused by the
effect of the slots that accommodate the windings, or to the nonsi-
nusoidal airgap flux distribution.
Great care is taken to ensure a sinusoidal output from genera-
tors in large supply systems; however, non linear loads will cause
harmonics to appear in the load current waveform. Thus harmonics
are produced in devices that have a non linear response to their
inputs. Non linear circuit elements (i.e., those in which the current
flowing through them is not proportional to the applied voltage)
include rectifiers and any large-signal electronic amplifier in which
diodes, transistors, valves or iron-cored inductors are used.
(ii) A rectifier is a device for converting an alternating or an oscil-
lating current into a unidirectional or approximate direct current. A
Complex Waveforms 667
rectifier has a low impedance to current flow in one direction and a
nearly infinite impedance to current flow in the opposite direction.
Thus when an alternating current is applied to a rectifier, current
will flow through it during the positive half-cycles only; the current
is zero during the negative half-cycles. A typical current waveform
is shown in Figure 36.19. This ‘half-wave rectification’ is produced
by using a single diode. The waveform is similar in shape to that
shown in Figure 36.14, page 643, where the d.c. component brought
the negative half-cycle up to the zero current point. The waveform
shown in Figure 36.19 is typical of one containing a fairly large
second harmonic.
Figure 36.19 Typical current waveform containing a fairly large
second harmonic
(iii) Transistors and valves are non linear devices in that sinusoidal
input results in different positive and negative half-cycle amplifica-
tions. This means that the output half-cycles have different ampli-
tudes. Since they have a different shape, even harmonic distortion
is suggested (see Section 36.3).
(iv) Ferromagnetic-cored coils are a source of harmonic generation in
a.c. circuits because of the non-linearity of the B/H curve and
the hysteresis loop, especially if saturation occurs. Let a sinu-
soidal voltage
v D V
m
sinωt be applied to a ferromagnetic-cored
coil (having low resistance relative to inductive reactance) of cross-
section area A square metres and possessing N turns.
If is the flux produced in the core then the instantaneous voltage
is given by
v = N.df=dt/.
If B is the flux density of the core, then, since D BA,
v D N
d
dt
BA D NA
dB
dt
,
since area A is a constant for a particular core.
Separating the variables gives
dB D
1
NA
dt
i.e., B D
1
NA
V
m
sinωt dt D
V
m
ωNA
cosωt