Bird J. Electrical Circuit Theory and Technology
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226 Electrical Circuit Theory and Technology
Figure 15.19
(v) The series resonant circuit is often described as an acceptor circuit
since it has its minimum impedance, and thus maximum current, at
the resonant frequency.
(vi) Typical graphs of current I and impedance Z against frequency are
shown in Figure 15.19.
Problem 18. A coil having a resistance of 10 and an inductance
of 125 mH is connected in series with a 60
µF capacitor across a
120 V supply. At what frequency does resonance occur? Find the
current flowing at the resonant frequency.
Resonant frequency, f
r
D
1
2
p
LC
Hz
D
1
2
125
10
3
60
10
6
Hz
D
1
2
125 ð 6
10
8
D
1
2
p
[1256]
10
4
D
10
4
2
p
[1256]
D 58.12 Hz
At resonance, X
L
D X
C
and impedance Z D R
Hence current, I D
V
R
D
120
10
D 12 A
Problem 19. The current at resonance in a series L–C –R circuit is
100
µA. If the applied voltage is 2 mV at a frequency of 200 kHz,
and the circuit inductance is 50
µH, find (a) the circuit resistance,
and (b) the circuit capacitance.
(a) I D 100 µA D 100 ð10
6
A; V D 2mVD 2 ð 10
3
V
At resonance, impedance Z D resistance R
Hence R D
V
I
D
20 ð 10
3
100 ð 10
6
D
2 ð 10
6
100 ð 10
3
D 20 Z
(b) At resonance X
L
D X
C
i.e. 2fL D
1
2fC
Single-phase series a.c. circuits 227
Hence capacitance C D
1
2f
2
L
D
1
2 ð 200 ð10
3
2
50 ð 10
6
F
D
10
6
10
6
4
2
10
10
50
µF
D 0.0127 mFor 12.7nF
15.8 Q-factor
At resonance, if R is small compared with X
L
and X
C
, it is possible for
V
L
and V
C
to have voltages many times greater than the supply voltage
(see Figure 15.12(d)).
Voltage magnification at resonance D
voltage across L (or C)
supply voltage V
This ratio is a measure of the quality of a circuit (as a resonator or tuning
device) and is called the Q-factor.
Hence Q-factor D
V
L
V
D
IX
L
IR
D
X
L
R
D
2pf
r
L
R
Alternatively, Q-factor D
V
C
V
D
IX
C
IR
D
X
C
R
D
1
2pf
r
CR
At resonance f
r
D
1
2
p
LC
i.e. 2f
r
D
1
p
LC
Hence Q-factor D
2f
r
L
R
D
1
p
LC
L
R
D
1
R
L
C
(Q-factor is explained more fully in Chapter 28, page 495)
Problem 20. A coil of inductance 80 mH and negligible resis-
tance is connected in series with a capacitance of 0.25
µF and a
resistor of resistance 12.5 across a 100 V, variable frequency
supply. Determine (a) the resonant frequency, and (b) the current
at resonance. How many times greater than the supply voltage is
the voltage across the reactances at resonance?
(a) Resonant frequencyf
r
D
1
2
80
10
3
0.25
10
6
D
1
2
80.25
10
8
228 Electrical Circuit Theory and Technology
D
10
4
2
p
2
D 1125.4HzD 1.1254 kHz
(b) Current at resonance I D
V
R
D
100
12.5
D 8A
Voltage across inductance, at resonance,
V
L
D IX
L
D I2fL
D 821125.480 ð 10
3
D 4525.5V
(Also, voltage across capacitor,
V
C
D IX
C
D
I
2fC
D
8
21125.40.25 ð 10
6
D 4525.5 V
Voltage magnification at resonance D
V
L
V
or
V
c
V
D
4525.5
100
D 45.255 V
i.e. at resonance, the voltage across the reactances are 45.255 times
greater than the supply voltage. Hence Q-factor of circuit is 45.255.
Problem 21. A series circuit comprises a coil of resistance 2
and inductance 60 mH, and a 30
µF capacitor. Determine the Q-
factor of the circuit at resonance.
At resonance, Q-factor D
1
R
L
C
D
1
2
60 ð 10
3
30 ð 10
6
D
1
2
60 ð 10
6
30 ð 10
3
D
1
2
p
2000 D 22.36
Problem 22. A coil of negligible resistance and inductance
100 mH is connected in series with a capacitance of 2
µF and
a resistance of 10 across a 50 V, variable frequency supply.
Determine (a) the resonant frequency, (b) the current at resonance,
(c) the voltages across the coil and the capacitor at resonance, and
(d) the Q-factor of the circuit.
Single-phase series a.c. circuits 229
(a) Resonant frequency,f
r
D
1
2
p
LC
D
1
2
100
10
3
2
10
6
D
1
2
20
10
8
D
1
2
p
20
10
4
D
10
4
2
p
20
D 355.9Hz
(b) Current at resonance I D
V
R
D
50
10
D 5A
(c) Voltage across coil at resonance,
V
L
D IX
L
D I2f
r
L
D 52 ð 355.9 ð 100 ð 10
3
D 1118 V
Voltage across capacitance at resonance,
V
C
D IX
C
D
I
2f
r
C
D
5
2355.92 ð 10
6
D 1118 V
(d) Q-factor (i.e. voltage magnification at resonance) D
V
L
V
or
V
C
V
D
1118
50
D 22.36
Q-factor may also have been determined by
2f
r
L
R
or
1
2f
r
CR
or
1
R
L
C
Further problems on series resonance and Q-factor may be found in
Section 15.12, problems 21 to 25, page 236.
15.9 Bandwidth and
selectivity
Figure 15.20 shows how current I varies with frequency in an R –L –C
series circuit. At the resonant frequency f
r
, current is a maximum value,
shown as I
r
. Also shown are the points A and B where the current is 0.707
of the maximum value at frequencies f
1
and f
2
. The power delivered to
230 Electrical Circuit Theory and Technology
Figure 15.20
the circuit is I
2
R.AtI D 0.707I
r
, the power is 0.707I
r
2
R D 0.5I
2
r
R, i.e.,
half the power that occurs at frequency f
r
. The points corresponding to
f
1
and f
2
are called the half-power points. The distance between these
points, i.e. (f
2
f
1
), is called the bandwidth.
It may be shown that
Q =
f
r
f
2
− f
1
or
.f
2
− f
1
/ =
f
r
Q
(This formula is proved in Chapter 28, page 495)
Problem 23. A filter in the form of a series L –R –C circuit is
designed to operate at a resonant frequency of 5 kHz. Included
within the filter is a 20 mH inductance and 10 resistance. Deter-
mine the bandwidth of the filter.
Q-factor at resonance is given by
Q
r
D
ω
r
L
R
D
2500020 ð 10
3
10
D 62.83
Since Q
r
D f
r
/f
2
f
1
bandwidth,f
2
f
1
D
f
r
Q
r
D
5000
62.83
D 79.6Hz
Selectivity is the ability of a circuit to respond more readily to signals of
a particular frequency to which it is tuned than to signals of other frequen-
cies. The response becomes progressively weaker as the frequency departs
from the resonant frequency. The higher the Q-factor, the narrower the
bandwidth and the more selective is the circuit. Circuits having high Q-
factors (say, in the order of 100 to 300) are therefore useful in communica-
tions engineering. A high Q-factor in a series power circuit has disadvan-
tages in that it can lead to dangerously high voltages across the insulation
and may result in electrical breakdown.
(For more on bandwidth and selectivity see Chapter 28, page 504)
15.10 Power in a.c.
circuits
In Figures 15.21(a)–(c), the value of power at any instant is given by the
product of the voltage and current at that instant, i.e. the instantaneous
power, p D
vi, as shown by the broken lines.
(a) For a purely resistive a.c. circuit, the average power dissipated, P,
is given by:
P
= VI = I
2
R =
V
2
R
watts (V and I being rms values).
See Figure 15.21(a).
Single-phase series a.c. circuits 231
Figure 15.21
(b) For a purely inductive a.c. circuit, the average power is zero. See
Figure 15.21(b).
(c) For a purely capacitive a.c. circuit, the average power is zero. See
Figure 15.21(c).
Figure 15.22 shows current and voltage waveforms for an R–L circuit
where the current lags the voltage by angle . The waveform for power
(where p D
vi) is shown by the broken line, and its shape, and hence
average power, depends on the value of angle .
For an R –L, R–C or R–L –C series a.c. circuit, the average power P
is given by:
P = VI cosf watts
or
P = I
2
R watts
(V and I being rms values)
The formulae for power are proved in Chapter 26, page 459.
Problem 24. An instantaneous current, i D 250sinωt mA flows
through a pure resistance of 5 k. Find the power dissipated in the
resistor.
Power dissipated, P D I
2
R where I is the rms value of current.
If i D 250sinωt mA, then I
m
D 0.250 A and rms current,
I D 0.707 ð 0.250 A
Hence power P D 0.707 ð 0.250
2
5000 D 156.2 watts
Problem 25. A series circuit of resistance 60 and inductance
75 mH is connected to a 110 V, 60 Hz supply. Calculate the power
dissipated.
Inductive reactance, X
L
D 2fL D 26075 ð 10
3
D 28.27
Impedance, Z D
R
2
C X
2
L
D
[60
2
C 28.27
2
] D 66.33
Current, I D
V
Z
D
110
66.33
D 1.658 A
To calculate power dissipation in an a.c. circuit two formulae may be
used:
(i) P D I
2
R D 1.658
2
60 D 165 W
or (ii) P D VI cos where cos D
R
Z
D
60
66.33
D 0.9046
Hence P D 1101.6580.9046 D 165 W
Figure 15.22
232 Electrical Circuit Theory and Technology
15.11 Power triangle and
power factor
Figure 15.23(a) shows a phasor diagram in which the current I lags the
applied voltage V by angle . The horizontal component of V is V cos
and the vertical component of V is V sin . If each of the voltage phasors
is multiplied by I, Figure 15.23(b) is obtained and is known as the ‘power
triangle’.
Apparent power, S = VI voltamperes (VA)
True or active power, P
= VI cosf watts (W)
Reactive power, Q
= VI sinf reactive
voltamperes (var)
Power factor =
True power P
Apparent power S
For sinusoidal voltages and currents, power factor D
P
S
D
VI cos
VI
,i.e.
p.f. = cosf =
R
Z
(from Figure 15.6)
Figure 15.23
The relationships stated above are also true when current I leads
voltage V. More on the power triangle and power factor is contained
in Chapter 26, page 464.
Problem 26. A pure inductance is connected to a 150 V, 50 Hz
supply, and the apparent power of the circuit is 300 VA. Find the
value of the inductance.
Apparent power S D VI
Hence current I D
S
V
D
300
150
D 2A
Inductive reactance X
L
D
V
I
D
150
2
D 75
Since X
L
D 2fL, inductance L D
X
L
2f
D
75
250
D 0.239 H
Problem 27. A transformer has a rated output of 200 kVA at a
power factor of 0.8. Determine the rated power output and the
corresponding reactive power.
VI D 200 kVA D 200 ð 10
3
;p.f.D 0.8 D cos
Power output, P D VI cos D 200 ð10
3
0.8 D 160 kW
Single-phase series a.c. circuits 233
Reactive power, Q D VI sin
If cos D 0.8, then D arccos 0.8 D 36.87
°
D 36
°
52
0
Hence sin D sin36.87
°
D 0.6
Hence reactive power, Q D 200 ð 10
3
0.6 D 120 kvar
Problem 28. The power taken by an inductive circuit when
connected to a 120 V, 50 Hz supply is 400 W and the current
is 8 A. Calculate (a) the resistance, (b) the impedance, (c) the
reactance, (d) the power factor, and (e) the phase angle between
voltage and current.
(a) Power P D I
2
R. Hence R D
P
I
2
D
400
8
2
D 6.25 Z
(b) Impedance Z D
V
I
D
120
8
D 15 Z
(c) Since Z D
R
2
C X
2
L
, then X
L
D
Z
2
R
2
D
[15
2
6.25
2
]
D 13.64 Z
(d) Power factor D
true power
apparent power
D
VI cos
VI
D
400
1208
D 0.4167
(e) p.f. D cos D 0.4167.Hence phase angle D arccos0.4167
D 65.37
°
D 65
°
22
lagging
Problem 29. A circuit consisting of a resistor in series with a
capacitor takes 100 watts at a power factor of 0.5 from a 100 V,
60 Hz supply. Find (a) the current flowing, (b) the phase angle,
(c) the resistance, (d) the impedance, and (e) the capacitance.
(a) Power factor D
true power
apparent power
i.e. 0.5 D
100
100I
. Hence I D
100
0.5100
D 2A
(b) Power factor D 0.5 D cos. Hence phase angle D arccos0.5
D 60
°
leading
(c) Power P D I
2
R. Hence resistance R D
P
I
2
D
100
2
2
D 25 Z
(d) Impedance Z D
V
I
D
100
2
D 50 Z
234 Electrical Circuit Theory and Technology
(e) Capacitive reactance, X
C
D
Z
2
R
2
D
50
2
25
2
D 43.30
X
C
D
1
2fC
hence capacitance C D
1
2fX
c
D
1
26043.30
F
D 61.26 mF
Further problems on power in a.c. circuits may be found in Section 15.12
following, problems 26 to 36, page 237.
15.12 Further problems
on single-phase series a.c.
circuits
A.c. circuits containing pure inductance and pure capacitance
1 Calculate the reactance of a coil of inductance 0.2 H when it is
connected to (a) a 50 Hz, (b) a 600 Hz and (c) a 40 kHz supply.
[(a) 62.83 (b) 754 (c) 50.27 k]
2 A coil has a reactance of 120 in a circuit with a supply frequency
of 4 kHz. Calculate the inductance of the coil. [4.77 mH]
3 A supply of 240 V, 50 Hz is connected across a pure inductance and
the resulting current is 1.2 A. Calculate the inductance of the coil.
[0.637 H]
4 An e.m.f. of 200 V at a frequency of 2 kHz is applied to a coil of
pure inductance 50 mH. Determine (a) the reactance of the coil, and
(b) the current flowing in the coil. [(a) 628 (b) 0.318 A]
5 Calculate the capacitive reactance of a capacitor of 20
µF when
connected to an a.c. circuit of frequency (a) 20 Hz, (b) 500 Hz,
(c) 4 kHz [(a) 397.9 (b) 15.92 (c) 1.989 ]
6 A capacitor has a reactance of 80 when connected to a 50 Hz
supply. Calculate the value of its capacitance. [39.79
µF]
7 A capacitor has a capacitive reactance of 400 when connected to
a 100 V, 25 Hz supply. Determine its capacitance and the current
taken from the supply. [15.92
µF, 0.25 A]
8 Two similar capacitors are connected in parallel to a 200 V, 1 kHz
supply. Find the value of each capacitor if the circuit current is
0.628 A. [0.25
µF]
R–L a.c. circuits
9 Determine the impedance of a coil which has a resistance of 12
and a reactance of 16 [20 ]
10 A coil of inductance 80 mH and resistance 60 is connected to
a 200 V, 100 Hz supply. Calculate the circuit impedance and the
Single-phase series a.c. circuits 235
current taken from the supply. Find also the phase angle between
the current and the supply voltage.
[78.27 , 2.555 A, 39
°
57
0
lagging]
11 An alternating voltage given by
v D 100sin240t volts is applied
across a coil of resistance 32 and inductance 100 mH. Determine
(a) the circuit impedance, (b) the current flowing, (c) the p.d. across
the resistance, and (d) the p.d. across the inductance.
[(a) 40 (b) 1.77 A (c) 56.64 V (d) 42.48 V]
12 A coil takes a current of 5 A from a 20 V d.c. supply. When
connected to a 200 V, 50 Hz a.c. supply the current is 25 A.
Calculate the (a) resistance, (b) impedance and (c) inductance of
the coil. [(a) 4 (b) 8 (c) 22.05 mH]
13 A coil of inductance 636.6 mH and negligible resistance is connected
in series with a 100 resistor to a 250 V, 50 Hz supply. Calcu-
late (a) the inductive reactance of the coil, (b) the impedance of the
circuit, (c) the current in the circuit, (d) the p.d. across each compo-
nent, and (e) the circuit phase angle.
[(a) 200 (b) 223.6 (c)1.118 A
(d) 223.6V, 111.8 V (e) 63
°
26
0
lagging]
R–C a.c. circuits
14 A voltage of 35 V is applied across a C –R series circuit. If the
voltage across the resistor is 21 V, find the voltage across the
capacitor. [28 V]
15 A resistance of 50 is connected in series with a capacitance of
20
µF. If a supply of 200 V, 100 Hz is connected across the arrange-
ment find (a) the circuit impedance, (b) the current flowing, and
(c) the phase angle between voltage and current.
[(a) 93.98 (b) 2.128 A (c) 57
°
51
0
leading]
16 An alternating voltage
v D 250sin 800 t volts is applied across a
series circuit containing a 30 resistor and 50
µF capacitor. Calcu-
late (a) the circuit impedance, (b) the current flowing, (c) the p.d.
across the resistor, (d) the p.d. across the capacitor, and (e) the phase
angle between voltage and current
[(a) 39.05 (b) 4.527 A (c) 135.8V
(d) 113.2 V (e) 39
°
48
0
]
17 A 400 resistor is connected in series with a 2358 pF capacitor
across a 12 V a.c. supply. Determine the supply frequency if the
current flowing in the circuit is 24 mA. [225 kHz]
R–L–C a.c. circuits
18 A 40
µF capacitor in series with a coil of resistance 8 and induc-
tance 80 mH is connected to a 200 V, 100 Hz supply. Calculate
(a) the circuit impedance, (b) the current flowing, (c) the phase angle