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16
Single-phase parallel
a.c. circuits
At the end of this chapter you should be able to:
ž calculate unknown currents, impedances and circuit phase
angle from phasor diagrams for (a) R–L (b) R–C (c) L –C
(d) LR–C parallel a.c. circuits
ž state the condition for parallel resonance in an LR–C circuit
ž derive the resonant frequency equation for an LR–C parallel
a.c. circuit
ž determine the current and dynamic resistance at resonance in
an LR–C parallel circuit
ž understand and calculate Q-factor in an LR–C parallel circuit
ž understand how power factor may be improved
16.1
Introduction
In parallel circuits, such as those shown in Figures 16.1 and 16.2, the
voltage is common to each branch of the network and is thus taken as
the reference phasor when drawing phasor diagrams.
For any parallel a.c. circuit:
True or active power, P D VI cos watts (W)
or P D IR 2 R watts
Apparent power, S D VI voltamperes (VA)
Reactive power, Q D VI sin reactive voltamperes (var)
Power factor D
P
true power
D D cos apparent power
S
(These formulae are the same as for series a.c. circuits as used in
Chapter 15.)
Figure 16.1
16.2
R –L parallel a.c.
circuit
In the two branch parallel circuit containing resistance R and inductance L
shown in Figure 16.1, the current flowing in the resistance, IR , is in-phase
with the supply voltage V and the current flowing in the inductance, IL ,
lags the supply voltage by 90° . The supply current I is the phasor sum of
IR and IL and thus the current I lags the applied voltage V by an angle
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lying between 0° and 90° (depending on the values of IR and IL ), shown
as angle in the phasor diagram.
From the phasor diagram:
ID
where IR D
tan D
I2R C I2L , (by Pythagoras’ theorem)
V
V
and IL D
R
XL
IR
IL
IL
, sin D
and cos D
(by trigonometric ratios
IR
I
I
Circuit impedance, Z D
V
I
Problem 1. A 20 resistor is connected in parallel with an inductance of 2.387 mH across a 60 V, 1 kHz supply. Calculate (a) the
current in each branch, (b) the supply current, (c) the circuit phase
angle, (d) the circuit impedance, and (e) the power consumed.
The circuit and phasor diagrams are as shown in Figure 16.1.
(a)
V
60
D
D3A
R
20
V
V
D
Current flowing in the inductance IL D
XL
2fL
Current flowing in the resistor
IR D
D
60
2
1000
2.387 ð 103 D4A
(b)
From the phasor diagram, supply current, I D
D
IR 2 C I2L 32 C 42 D5A
(c)
Circuit phase angle, D arctan
IL
4
D arctan
IR
3
V
60
D
D 12 Z
I
5
D 53.13°
D 53° 8 lagging
(d)
Circuit impedance, Z D
(e)
Power consumed P D VI cos D 60
5
cos 53° 80 D 180 W
(Alternatively, power consumed P D IR 2 R D 32 20 D 180 W
Further problems on R–L parallel a.c. circuits may be found in
Section 16.8, problems 1 and 2, page 256.
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16.3 R –C parallel a.c.
circuit
In the two branch parallel circuit containing resistance R and capacitance
C shown in Figure 16.2, IR is in-phase with the supply voltage V and the
current flowing in the capacitor, IC , leads V by 90° . The supply current
I is the phasor sum of IR and IC and thus the current I leads the applied
voltage V by an angle lying between 0° and 90° (depending on the values
of IR and IC ), shown as angle ˛ in the phasor diagram.
From the phasor diagram:
ID
where IR D
tan ˛ D
Figure 16.2
I2R C I2C , (by Pythagoras’ theorem)
V
V
and IC D
R
XC
IR
IC
IC
, sin ˛ D
and cos ˛ D
(by trigonometric ratios)
IR
I
I
Circuit impedance Z D
V
I
Problem 2. A 30 µF capacitor is connected in parallel with an
80 resistor across a 240 V, 50 Hz supply. Calculate (a) the
current in each branch, (b) the supply current, (c) the circuit phase
angle, (d) the circuit impedance, (e) the power dissipated, and
(f) the apparent power.
The circuit and phasor diagrams are as shown in Figure 16.2.
(a)
V
240
D
D3A
R
80
V
V
D Current in capacitor, IC D
1
XC
2fC
D 2fCV
D
Current in resistor, IR
D 2
50
30 ð 106 240
(b)
Supply current, I D
D 2.262 A
IR 2 C IC 2 D
32 C 2.2622 D 3.757 A
(c)
Circuit phase angle, ˛ D arctan
2.262
IC
D arctan
IR
3
D 37° 1 leading
(d)
Circuit impedance, Z D
V
240
D
D 63.88 Z
I
3.757
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(e)
True or active power dissipated, P D VI cos ˛
D 240
3.757 cos 37° 10
D 720 W
(Alternatively, true power P D IR R D 32 80 D 720 W)
2
(f)
Apparent power, S D VI D 240
3.757 D 901.7 VA
Problem 3. A capacitor C is connected in parallel with a resistor
R across a 120 V, 200 Hz supply. The supply current is 2 A at a
power factor of 0.6 leading. Determine the values of C and R.
The circuit diagram is shown in Figure 16.3(a).
Power factor D cos D 0.6 leading, hence D arccos 0.6 D 53.13°
leading.
From the phasor diagram shown in Figure 16.3(b),
IR D I cos 53.13° D 2
0.6
D 1.2 A
and IC D I sin 53.13° D 2
0.8
D 1.6 A
(Alternatively, IR and IC can be measured from the scaled phasor
diagram.)
From the circuit diagram,
Figure 16.3
IR D
and IC D
V
V
120
D
from which R D
D 100 Z
R
IR
1.2
IC
V
D 2fCV, from which, C D
XC
2fV
D
1.6
2
200
120
D 10.61 mF
Further problems on R–C parallel a.c. circuits may be found in
Section 16.8, problems 3 and 4, page 256.
16.4 L–C parallel a.c.
circuit
In the two branch parallel circuit containing inductance L and capacitance
C shown in Figure 16.4, IL lags V by 90° and IC leads V by 90° .
Theoretically there are three phasor diagrams possible — each depending on the relative values of IL and IC :
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(i)
IL > IC (giving a supply current, I D IL IC lagging V by 90° )
(ii)
IC > IL (giving a supply current, I D IC IL leading V by 90° )
(iii)
IL D IC (giving a supply current, I = 0).
The latter condition is not possible in practice due to circuit resistance
inevitably being present (as in the circuit described in Section 16.5).
For the L –C parallel circuit, IL D
V
V
, IC D
XL
XC
I D phasor difference between IL and IC , and Z D
V
I
Problem 4. A pure inductance of 120 mH is connected in parallel
with a 25 µF capacitor and the network is connected to a 100 V,
50 Hz supply. Determine (a) the branch currents, (b) the supply
current and its phase angle, (c) the circuit impedance, and (d) the
power consumed.
Figure 16.4
The circuit and phasor diagrams are as shown in Figure 16.4.
(a)
Inductive reactance, XL D 2fL D 2
50
120 ð 103 D 37.70 Capacitive reactance, XC D
1
1
D
2fC
2
50
25 ð 106 D 127.3 (b)
Current flowing in inductance, IL D
V
100
D
D 2.653 A
XL
37.70
Current flowing in capacitor, IC D
V
100
D
D 0.786 A
XC
127.3
IL and IC are anti-phase. Hence supply current,
I D IL IC D 2.653 0.786 D 1.867 A and the current lags the
supply voltage V by 90° (see Figure 16.4(i))
100
V
D
D 53.56 Z
I
1.867
(c)
Circuit impedance, Z D
(d)
Power consumed, P D VI cos D 100
1.867
cos 90° D0W
Problem 5. Repeat Problem 4 for the condition when the
frequency is changed to 150 Hz.
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(a)
Inductive reactance, XL D 2
150
120 ð 103 D 113.1 Capacitive reactance, XC D
1
D 42.44 2
150
25 ð 106 Current flowing in inductance, IL D
V
100
D
D 0.884 A
XL
113.1
Current flowing in capacitor, IC D
V
100
D
D 2.356 A
XC
42.44
(b)
Supply current, I D IC IL D 2.356 0.884 D 1.472 A leading V
by 90° (see Figure 4(ii))
(c)
Circuit impedance, Z D
(d)
Power consumed, P D VI cos D 0 W (since D 90° )
V
100
D
D 67.93 Z
I
1.472
From Problems 4 and 5:
(i)
When XL < XC then IL > IC and I lags V by 90°
(ii)
When XL > XC then IL < IC and I leads V by 90°
(iii)
In a parallel circuit containing no resistance the power consumed
is zero
Further problems on L –C parallel a.c. circuits may be found in
Section 16.8, problems 5 and 6, page 256.
16.5
LR –C parallel a.c.
circuit
In the two branch circuit containing capacitance C in parallel with inductance L and resistance R in series (such as a coil) shown in Figure 16.5(a),
the phasor diagram for the LR branch alone is shown in Figure 16.5(b)
and the phasor diagram for the C branch is shown alone in Figure 16.5(c).
Rotating each and superimposing on one another gives the complete
phasor diagram shown in Figure 16.5(d).
The current ILR of Figure 16.5(d) may be resolved into horizontal and
vertical components. The horizontal component, shown as op is ILR cos 1
and the vertical component, shown as pq is ILR sin 1 . There are three
possible conditions for this circuit:
(i)
IC > ILR sin 1
(ii)
ILR sin 1 > IC
(iii)
IC D ILR sin 1
(giving a supply current I leading V by angle
— as shown in Figure 16.5(e))
(giving I lagging V by angle — as shown in
Figure 16.5(f))
(this is called parallel resonance, see Section 16.6).
There are two methods of finding the phasor sum of currents ILR and
IC in Figures 16.5(e) and (f). These are: (i) by a scaled phasor diagram,
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Figure 16.5
or (ii) by resolving each current into their ‘in-phase’ (i.e. horizontal) and
‘quadrature’ (i.e. vertical) components, as demonstrated in problems 6
and 7. With reference to the phasor diagrams of Figure 16.5:
Impedance of LR branch, ZLR D
ILR D
Current,
R2 C XL 2 V
V
and IC D
ZLR
XC
Supply current I D phasor sum of ILR and IC (by drawing)
D
f
ILR cos 1 2 C ILR sin 1 ¾ IC 2 g (by calculation)
where ¾ means ‘the difference between’.
V
Circuit impedance Z D
I
tan 1 D
VL
XL
XL
R
D
and cos 1 D
, sin 1 D
VR
R
ZLR
ZLR
tan D
ILR cos 1
ILR sin 1 ¾ IC
and cos D
ILR cos 1
I
Problem 6. A coil of inductance 159.2 mH and resistance 40 is
connected in parallel with a 30 µF capacitor across a 240 V, 50 Hz
supply. Calculate (a) the current in the coil and its phase angle,
(b) the current in the capacitor and its phase angle, (c) the supply
current and its phase angle,(d) the circuit impedance, (e) the power
consumed, (f) the apparent power, and (g) the reactive power. Draw
the phasor diagram.
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The circuit diagram is shown in Figure 16.6(a).
(a)
For the coil, inductive reactance XL D 2fL
D 2
50
159.2 ð 103 Impedance Z1 D
R2 C X2L D
Current in coil, ILR D
D 50 402 C 502 D 64.03 V
240
D
D 3.748 A
Z1
64.03
Branch phase angle 1 D arctan
XL
50
D arctan
R
40
D arctan 1.25
D 51.34° D 51° 20 lagging
(see phasor diagram in Figure 16.6(b))
Figure 16.6
(b)
Capacitive reactance, XC D
1
1
D
2fC
2
50
30 ð 106 D 106.1 Current in capacitor, IC D
V
240
D
XC
106.1
= 2.262 A leading the supply
voltage by 90°
(see phasor diagram of Figure 16.6(b)).
(c)
The supply current I is the phasor sum of ILR and IC This may be
obtained by drawing the phasor diagram to scale and measuring the
current I and its phase angle relative to V. (Current I will always
be the diagonal of the parallelogram formed as in Figure 16.6(b)).
Alternatively the current ILR and IC may be resolved into their horizontal (or ‘in-phase’) and vertical (or ‘quadrant’) components. The
horizontal component of ILR is
ILR cos
51° 200 D 3.748 cos 51° 200 D 2.342 A
The horizontal component of IC is IC cos 90° D 0
Thus the total horizontal component, IH D 2.342 A
The vertical component of ILR D ILR sin
51° 200 D 3.748 sin 51° 200
D 2.926 A
The vertical component of IC D IC sin 90°
D 2.262 sin 90° D 2.262 A
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Thus the total vertical component, IV D 2.926 C 2.262
D −0.664 A
IH and IV are shown in Figure 16.7, from which,
ID
[
2.3422 C 0.6642 ] D 2.434 A
0.664
Angle D arctan
2.342
Figure 16.7
D 15.83° D 15° 500 lagging
Hence the supply current I = 2.434 A lagging V by 15° 50 .
240
V
D
D 98.60 Z
I
2.434
(d)
Circuit impedance, Z D
(e)
Power consumed, P D VI cos D 240
2.434 cos 15° 500
D 562 W
(Alternatively, P D IR 2 R D ILR 2 R (in this case)
D 3.7482 40 D 562 W
(f)
Apparent power, S D VI D 240
2.434 D 584.2 VA
(g)
Reactive power, Q D VI sin D 240
2.434
sin 15° 500 D 159.4 var
Problem 7. A coil of inductance 0.12 H and resistance 3 k is
connected in parallel with a 0.02 µF capacitor and is supplied at
40 V at a frequency of 5 kHz. Determine (a) the current in the coil,
and (b) the current in the capacitor. (c) Draw to scale the phasor
diagram and measure the supply current and its phase angle; check
the answer by calculation. Determine (d) the circuit impedance and
(e) the power consumed.
The circuit diagram is shown in Figure 16.8(a).
(a)
Inductive reactance, XL D 2fL D 2
5000
0.12 D 3770 Impedance of coil, Z1 D
R2 C XL 2 D
[
30002 C 37702 ]
D 4818 Current in coil, ILR D
V
40
D
D 8.30 mA
Z1
4818
Branch phase angle D arctan
Figure 16.8
3770
XL
D arctan
R
3000
D 51.5° lagging
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(b)
Capacitive reactance, XC D
1
1
D
2fC
2
5000
0.02 ð 106 D 1592 Capacitor current, IC D
V
40
D
XC
1592
D 25.13 mA leading V by 90°
(c)
Currents ILR and IC are shown in the phasor diagram of
Figure 16.8(b). The parallelogram is completed as shown and the
supply current is given by the diagonal of the parallelogram. The
current I is measured as 19.3 mA leading voltage V by 74.5°
By calculation, I D
[
ILR cos 51.5° 2 C IC ILR sin 51.5° 2 ]
D 19.34 mA
and D arctan
IC ILR sin 51.5°
ILR cos 51.5°
D 74.50°
V
40
D 2.068 kZ
D
I
19.34 ð 103
(d)
Circuit impedance, Z D
(e)
Power consumed, P D VI cos D 40
19.34 ð 103 cos 74.50° D 206.7 mW
(Alternatively, P D IR 2 R D ILR 2 R D 8.30 ð 103 2 3000
D 206.7 mW)
Further problems on the LR–C parallel a.c. circuit may be found in
Section 16.8, problems 7 and 8, page 256.
16.6 Parallel resonance
and Q-factor
Parallel resonance
Resonance occurs in the two branch network containing capacitance C in
parallel with inductance L and resistance R in series (see Figure 16.5(a))
when the quadrature (i.e. vertical) component of current ILR is equal to
IC . At this condition the supply current I is in-phase with the supply
voltage V.
Resonant frequency
When the quadrature component of ILR is equal to IC then: IC D ILR sin 1
(see Figure 16.9)
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Hence
V
D
XC
V
ZLR
XL
, (from Section 16.5)
ZLR
from which, ZLR 2 D XC XL D 2fr L
Hence [ R2 C XL 2 ]2 D
Thus
1
2fr C
L
L
and R2 C XL 2 D
C
C
L
2fr L D R2 and 2fr L D
C
2
Figure 16.9
and
1
fr D
2L
L
R2
C
D
L
C
L
R2
C
L
1 R2
D
2
2
2
L C L
2
1 1 −R
i.e. parallel resonant frequency, fr =
2p
LC
L2
(When R is negligible, then fr D
(16.1)
Hz
1
p
, which is the same as for
2 LC
series resonance.)
Current at resonance
Current at resonance, Ir D ILR cos 1
D
D
V
ZLR
(from Figure 16.9)
R
ZLR
VR
Z2LR
However from equation (16.1), Z2LR D
L
C
VR
VRC
=
L
L
C
The current is at a minimum at resonance.
hence Ir =
(from Section 16.5)
(16.2)
Dynamic resistance
Since the current at resonance is in-phase with the voltage the impedance
of the circuit acts as a resistance. This resistance is known as the dynamic
resistance, RD (or sometimes, the dynamic impedance).
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From equation (16.2), impedance at resonance D
i.e. dynamic resistance,
RD =
V
L
V
D
D
VRC
Ir
RC
L
L
ohms
RC
Rejector circuit
The parallel resonant circuit is often described as a rejector circuit since
it presents its maximum impedance at the resonant frequency and the
resultant current is a minimum.
Q-factor
Currents higher than the supply current can circulate within the parallel
branches of a parallel resonant circuit, the current leaving the capacitor
and establishing the magnetic field of the inductor, this then collapsing and
recharging the capacitor, and so on. The Q-factor of a parallel resonant
circuit is the ratio of the current circulating in the parallel branches of the
circuit to the supply current, i.e. the current magnification.
Q-factor at resonance D current magnification D
circulating current
supply current
D
IC
ILR sin 1
D
Ir
Ir
D
ILR sin 1
sin 1
D
ILR cos 1
cos 1
D tan 1 D
i.e.
Q-factor at resonance =
XL
R
2pfr L
R
(which is the same as for a series circuit)
Note that in a parallel circuit the Q-factor is a measure of current
magnification, whereas in a series circuit it is a measure of voltage
magnification.
At mains frequencies the Q-factor of a parallel circuit is usually low,
typically less than 10, but in radio-frequency circuits the Q-factor can be
very high.
Problem 8. A pure inductance of 150 mH is connected in parallel
with a 40 µF capacitor across a 50 V, variable frequency supply.
Determine (a) the resonant frequency of the circuit and (b) the
current circulating in the capacitor and inductance at resonance.
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The circuit diagram is shown in Figure 16.10.
(a)
1
Parallel resonant frequency, fr D
2
1
R2
2
LC L
However, resistance R D 0. Hence,
1
fr D
2
Figure 16.10
1
D
2
103
D
2
(b)
1
LC
1
D
2
107
15
4
1
150 ð 103 40 ð 106 1
6
D 64.97 Hz
Current circulating in L and C at resonance,
ICIRC D
V
D
XC
V
D 2fr CV
1
2fr C
Hence ICIRC D 2
64.97
40 ð 106 50 D 0.816 A
Alternatively, ICIRC D
V
V
50
D
D
XL
2fr L
2
64.97
0.15
D 0.817 A
Problem 9. A coil of inductance 0.20 H and resistance 60 is
connected in parallel with a 20 µF capacitor across a 20 V, variable frequency supply. Calculate (a) the resonant frequency, (b) the
dynamic resistance, (c) the current at resonance and (d) the circuit
Q-factor at resonance.
(a)
Parallel resonant frequency,
1
fr D
2
1
D
2
1
R2
2
LC L
1
602
0.20
20 ð 106 0.22
1 p
250 000 90 000
2
1 p
1
160 000 D
400
D
2
2
D
D 63.66 Hz
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(b)
Dynamic resistance, RD D
L
0.20
D
D 166.7 Z
RC
60
20 ð 106 (c)
Current at resonance, Ir D
V
20
D
D 0.12 A
RD
166.7
(d)
Circuit Q-factor at resonance D
2fr L
2
63.66
0.2
D
D 1.33
R
60
Alternatively, Q-factor at resonance D current magnification (for a
parallel circuit) D Ic /Ir
Ic D
V
D Xc
V
D 2fr CV D 2
63.66
20 ð 106 20
1
2fr C
D 0.16 A
Hence Q-factor D
0.16
Ic
D
D 1.33, as obtained above
Ir
0.12
Problem 10. A coil of inductance 100 mH and resistance 800 is connected in parallel with a variable capacitor across a 12 V,
5 kHz supply. Determine for the condition when the supply current
is a minimum: (a) the capacitance of the capacitor, (b) the dynamic
resistance, (c) the supply current, and (d) the Q-factor.
(a)
The supply current is a minimum when the parallel circuit is at
resonance.
1
Resonant frequency, fr D
2
1
R2
2
LC L
Transposing for C gives: 2fr 2 D
1
R2
2
LC L
2fr 2 C
R2
1
D
L2
LC
CD
1
L 2fr
2
R2
C 2
L
When L D 100 mH, R D 800 and fr D 5000 Hz,
CD
100 ð
103
1
2
50002
8002
C
100 ð 103 2
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D
D
(b)
0.1[2 108
1
F
C 0.64108 ]
106
µF D 0.009515 mF or 9.515 nF
0.1
10.51 ð 108 Dynamic resistance, RD D
L
100 ð 103
D
CR
9.515 ð 109 800
D 13.14 kZ
(c)
Supply current at resonance, Ir D
(d)
Q-factor at resonance =
V
12
D
D 0.913 mA
RD
13.14 ð 103
2
5000
100 ð 103 2fr L
D
D 3.93
R
800
Alternatively, Q-factor at resonance D
D
V/Xc
2fr CV
Ic
D
D
Ir
Ir
Ir
2
5000
9.515 ð 109 12
0.913 ð 103
D 3.93
Further problems on parallel resonance and Q-factor may be found in
Section 16.8, problems 9 to 12, page 257.
16.7
Power factor
improvement
For a particular power supplied, a high power factor reduces the current
flowing in a supply system and therefore reduces the cost of cables,
switch-gear, transformers and generators. Supply authorities use tariffs
which encourage electricity consumers to operate at a reasonably high
power factor. Industrial loads such as a.c. motors are essentially inductive (R–L) and may have a low power factor. One method of improving
(or correcting) the power factor of an inductive load is to connect a
static capacitor C in parallel with the load (see Figure 16.11(a)). The
supply current is reduced from ILR to I, the phasor sum of ILR and
IC , and the circuit power factor improves from cos 1 to cos 2 (see
Figure 16.11(b)).
Problem 11. A single-phase motor takes 50 A at a power factor of
0.6 lagging from a 240 V, 50 Hz supply. Determine (a) the current
taken by a capacitor connected in parallel with the motor to correct
the power factor to unity, and (b) the value of the supply current
after power factor correction.
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The circuit diagram is shown in Figure 16.12(a).
(a)
A power factor of 0.6 lagging means that cos D 0.6
i.e. D arccos 0.6 D 53° 80
Hence IM lags V by 53° 80 as shown in Figure 16.12(b).
If the power factor is to be improved to unity then the phase difference between supply current I and voltage V is 0° , i.e. I is in phase
with V as shown in Figure 16.12(c). For this to be so, IC must
equal the length ab, such that the phasor sum of IM and IC is I.
ab D IM sin 53° 80 D 50
0.8 D 40 A
Hence the capacitor current Ic must be 40 A for the power factor
to be unity.
(b)
Supply current I D IM cos 53° 80 D 50
0.6 D 30 A
Problem 12. A motor has an output of 4.8 kW, an efficiency of
80% and a power factor of 0.625 lagging when operated from a
240 V, 50 Hz supply. It is required to improve the power factor to
0.95 lagging by connecting a capacitor in parallel with the motor.
Determine (a) the current taken by the motor, (b) the supply current
after power factor correction, (c) the current taken by the capacitor,
(d) the capacitance of the capacitor, and (e) the kvar rating of the
capacitor.
Figure 16.11
(a)
Efficiency
D
Power input D
power output
80
4800
hence
D
power input
100
power input
4800
D 6000 W
0.8
Hence, 6000 D VIM cos D 240
IM 0.625,
since cos D p f. D 0.625
Thus current taken by the motor, IM D
6000
D 40 A
240
0.625
The circuit diagram is shown in Figure 16.13(a).
The phase angle between IM and V is given by:
D arccos 0.625 D 51.32° D 51° 190 , hence the phasor diagram is as
shown in Figure 16.13(b).
(b)
Figure 16.12
When a capacitor C is connected in parallel with the motor a current
IC flows which leads V by 90° . The phasor sum of IM and IC gives
the supply current I, and has to be such as to change the circuit
power factor to 0.95 lagging, i.e. a phase angle of arccos 0.95 or
18° 120 lagging, as shown in Figure 16.13(c).
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The horizontal component of IM (shown as oa) D IM cos 51° 190
D 40 cos 51° 190
D 25 A
The horizontal component of I (also given by oa) D I cos 18° 120
D 0.95 I
Equating the horizontal components gives: 25 D 0.95 I
25
Hence the supply current after p.f. correction, I D
D 26.32 A
0.95
(c)
The vertical component of IM (shown as ab) D IM sin 51° 190
D 40 sin 51° 190
D 31.22 A
The vertical component of I (shown as ac)
D I sin 18° 120
D 26.32 sin 18° 120
D 8.22 A
The magnitude of the capacitor current IC (shown as bc) is given
by ab ac, i.e. 31.22 8.22 D 23 A
Figure 16.13
(d)
Current IC D
V
D
Xc
from which, C
(e)
D
V
D 2fCV,
1
2fC
IC
23
D
F D 305 mF
2fV
2
50
240
kvar rating of the capacitor D
240
23
VIc
D
D 5.52 kvar
1000
1000
In this problem the supply current has been reduced from 40 A to
26.32 A without altering the current or power taken by the motor. This
means that the size of generating plant and the cross-sectional area of
conductors supplying both the factory and the motor can be less — with
an obvious saving in cost.
Problem 13. A 250 V, 50 Hz single-phase supply feeds the
following loads (i) incandescent lamps taking a current of 10 A
at unity power factor, (ii) fluorescent lamps taking 8 A at a power
factor of 0.7 lagging, (iii) a 3 kVA motor operating at full load and
at a power factor of 0.8 lagging and (iv) a static capacitor. Determine, for the lamps and motor, (a) the total current, (b) the overall
power factor and (c) the total power. (d) Find the value of the static
capacitor to improve the overall power factor to 0.975 lagging.
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A phasor diagram is constructed as shown in Figure 16.14(a), where
8 A is lagging voltage V by arccos 0.7, i.e. 45.57° , and the motor current
is 3000/250, i.e. 12 A lagging V by arccos 0.8, i.e. 36.87°
(a)
The horizontal component of the currents
D 10 cos 0° C 12 cos 36.87° C 8 cos 45.57°
D 10 C 9.6 C 5.6 D 25.2 A
The vertical component of the currents
D 10 sin 0° 12 sin 36.87° 8 sin 45.57°
D 0 7.2 5.713 D 12.91 A
From Figure 16.14(b), total current, IL D
Figure 16.14
[
25.22 C 12.912 ]
D 28.31 A
at a phase angle of D arctan
12.91
, i.e. 27.13° lagging
25.2
(b)
Power factor D cos D cos 27.13° D 0.890 lagging
(c)
Total power, P D VIL cos D 250
28.31
0.890 D 6.3 kW
(d)
To improve the power factor, a capacitor is connected in parallel
with the loads. The capacitor takes a current IC such that the supply
current falls from 28.31 A to I, lagging V by arccos 0.975, i.e.
12.84° . The phasor diagram is shown in Figure 16.15.
oa D 28.31 cos 27.13° D I cos 12.84°
Hence I D
Figure 16.15
28.31 cos 27.13°
D 25.84 A
cos 12.84°
Current IC D bc D ab ac
D 28.31 sin 27.13° 25.84 sin 12.84°
D 12.91 5.742
D 7.168 A
Ic D
V
D Xc
V
D 2fCV
1
2fC
Hence capacitance C D
Ic
7.168
D
F
2fV
2
50
250
= 91.27 mF
Thus to improve the power factor from 0.890 to 0.975 lagging a
91.27 µF capacitor is connected in parallel with the loads.
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16.8 Further problems
on single-phase parallel
a.c. circuits
R –L parallel a.c. circuit
1
A 30 resistor is connected in parallel with a pure inductance of
3 mH across a 110 V, 2 kHz supply. Calculate (a) the current in each
branch, (b) the circuit current, (c) the circuit phase angle, (d) the
circuit impedance, (e) the power consumed, and (f) the circuit power
factor.
[(a) IR D 3.67 A, IL D 2.92 A (b) 4.69 A (c) 38° 300
lagging (d) 23.45 (e) 404 W (f) 0.783 lagging]
2
A 40 resistance is connected in parallel with a coil of inductance
L and negligible resistance across a 200 V, 50 Hz supply and the
supply current is found to be 8 A. Draw a phasor diagram to scale
and determine the inductance of the coil.
[102 mH]
R –C parallel a.c. circuit
3
A 1500 nF capacitor is connected in parallel with a 16 resistor
across a 10 V, 10 kHz supply. Calculate (a) the current in each
branch, (b) the supply current, (c) the circuit phase angle, (d) the
circuit impedance, (e) the power consumed, (f) the apparent power,
and (g) the circuit power factor. Draw the phasor diagram.
[(a) IR D 0.625 A, IC D 0.943 A (b) 1.13 A (c) 56° 280 leading
(d) 8.85 (e) 6.25 W (f) 11.3 VA (g) 0.55 leading]
4
A capacitor C is connected in parallel with a resistance R across a
60 V, 100 Hz supply. The supply current is 0.6 A at a power factor
of 0.8 leading. Calculate the value of R and C.
[R D 125 , C D 9.55 µF]
L–C parallel a.c. circuit
5
An inductance of 80 mH is connected in parallel with a capacitance
of 10 µF across a 60 V, 100 Hz supply. Determine (a) the branch
currents, (b) the supply current, (c) the circuit phase angle, (d) the
circuit impedance and (e) the power consumed.
[(a) IC D 0.377 A, IL D 1.194 A (b) 0.817 A
(c) 90° lagging (d) 73.44 (e) 0 W]
6
Repeat problem 5 for a supply frequency of 200 Hz.
[(a) IC D 0.754 A, IL D 0.597 A (b) 0.157 A
(c) 90° leading (d) 382.2 (e) 0 W]
LR –C parallel a.c. circuit
7
A coil of resistance 60 and inductance 318.4 mH is connected
in parallel with a 15 µF capacitor across a 200 V, 50 Hz supply.
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Calculate (a) the current in the coil, (b) the current in the capacitor,
(c) the supply current and its phase angle, (d) the circuit impedance,
(e) the power consumed, (f) the apparent power and (g) the reactive
power. Draw the phasor diagram.
[(a) 1.715 A (b) 0.943 A (c) 1.028 A at 30° 540 lagging
(d) 194.6 (e) 176.5 W (f) 205.6 VA (g) 105.6 var]
8
A 25 nF capacitor is connected in parallel with a coil of resistance 2 k and inductance 0.20 H across a 100 V, 4 kHz supply.
Determine (a) the current in the coil, (b) the current in the capacitor, (c) the supply current and its phase angle (by drawing a phasor
diagram to scale, and also by calculation), (d) the circuit impedance,
and (e) the power consumed.
[(a) 18.48 mA (b) 62.83 mA (c) 46.17 mA at 81° 290
leading (d) 2.166 k (e) 0.683 W]
Parallel resonance and Q-factor
9
A 0.15 µF capacitor and a pure inductance of 0.01 H are connected in
parallel across a 10 V, variable frequency supply. Determine (a) the
resonant frequency of the circuit, and (b) the current circulating in
the capacitor and inductance.
[(a) 4.11 kHz (b) 38.73 mA]
10
A 30 µF capacitor is connected in parallel with a coil of inductance
50 mH and unknown resistance R across a 120 V, 50 Hz supply. If
the circuit has an overall power factor of 1 find (a) the value of R,
(b) the current in the coil, and (c) the supply current.
[(a) 37.7 (b) 2.94 A (c) 2.714 A]
11
A coil of resistance 25 and inductance 150 mH is connected in
parallel with a 10 µF capacitor across a 60 V, variable frequency
supply. Calculate (a) the resonant frequency, (b) the dynamic resistance, (c) the current at resonance and (d) the Q-factor at resonance.
[(a) 127.2 Hz (b) 600 (c) 0.10 A (d) 4.80]
A coil of resistance 1.5 k and 0.25 H inductance is connected in
parallel with a variable capacitance across a 10 V, 8 kHz supply.
Calculate (a) the capacitance of the capacitor when the supply current
is a minimum, (b) the dynamic resistance, and (c) the supply current.
[(a) 1561 pF (b) 106.8 k (c) 93.66 µA]
12
Power factor improvement
13
A 415 V alternator is supplying a load of 55 kW at a power factor
of 0.65 lagging. Calculate (a) the kVA loading and (b) the current
taken from the alternator. (c) If the power factor is now raised to
unity find the new kVA loading.
[(a) 84.6 kVA (b) 203.9 A (c) 84.6 kVA]
14
A single phase motor takes 30 A at a power factor of 0.65 lagging
from a 240 V, 50 Hz supply. Determine (a) the current taken by the
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capacitor connected in parallel to correct the power factor to unity,
and (b) the value of the supply current after power factor correction.
[(a) 22.80 A (b) 19.5 A]
15
A motor has an output of 6 kW, an efficiency of 75% and a power
factor of 0.64 lagging when operated from a 250 V, 60 Hz supply. It
is required to raise the power factor to 0.925 lagging by connecting a
capacitor in parallel with the motor. Determine (a) the current taken
by the motor, (b) the supply current after power factor correction,
(c) the current taken by the capacitor, (d) the capacitance of the
capacitor and (e) the kvar rating of the capacitor.
[(a) 50 A (b) 34.59 A (c) 25.28 A (d) 268.2 µF (e) 6.32 kvar]
16
A 200 V, 50 Hz single-phase supply feeds the following loads:
(i) fluorescent lamps taking a current of 8 A at a power factor of
0.9 leading, (ii) incandescent lamps taking a current of 6 A at unity
power factor, (iii) a motor taking a current of 12 A at a power factor
of 0.65 lagging. Determine the total current taken from the supply
and the overall power factor. Find also the value of a static capacitor
connected in parallel with the loads to improve the overall power
factor to 0.98 lagging.
[21.74 A, 0.966 lagging, 21.68 µF]
21