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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 I (^) R^2 R watts
Apparent power, S D VI voltamperes (VA)
Reactive power, Q D VI sin reactive voltamperes (var)
Power factor D
true power apparent power
D
P
S
D cos
(These formulae are the same as for series a.c. circuits as used in Chapter 15.)
Figure 16.
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, I (^) R, is in-phase with the supply voltage V and the current flowing in the inductance, I (^) L , lags the supply voltage by 90°. The supply current I is the phasor sum of I (^) R and I (^) L and thus the current I lags the applied voltage V by an angle
lying between 0°^ and 90°^ (depending on the values of I (^) R and I (^) L ), shown as angle in the phasor diagram. From the phasor diagram:
I D
√ I^2 R C I^2 L , (by Pythagoras’ theorem)
where I (^) R D
V
R
and I (^) L D
V
X L
tan D
I L
I R
, sin D
I L
I
and cos D
I R
I
(by trigonometric ratios
Circuit impedance, Z D
V
I
Problem 1. A 20 resistor is connected in parallel with an induc- tance 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) Current flowing in the resistor I (^) R D
V
R
D
D 3 A
Current flowing in the inductance I (^) L D
V
X L
D
V
2 fL
D
2 1000 2. 387 ð 10 ^3
D 4 A
(b) From the phasor diagram, supply current, I D
√ I (^) R^2 C I^2 L
D
√ 32 C 4 2
D 5 A
(c) Circuit phase angle, D arctan
I L
I R
D arctan
( 4 3
) D 53. 13 °
D 53 ° 8 ′^ lagging
(d) Circuit impedance, Z D
V
I
D
D 12 Z
(e) Power consumed P D VI cos D 60 5 cos 53° 80 D 180 W (Alternatively, power consumed P D I (^) R^2 R D 3 2 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.
(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 I (^) R^2 R D 3 2 80 D 720 W)
(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),
I (^) R D I cos 53. 13 °^ D 2 0. 6
D 1. 2 A
and I (^) C D I sin 53. 13 °^ D 2 0. 8
D 1. 6 A
(Alternatively, I (^) R and I (^) C can be measured from the scaled phasor diagram.)
Figure 16.
From the circuit diagram,
I R D
V
R
from which R D
V
I R
D
D 100 Z
and I (^) C D
V
X C
D 2 fCV, from which, C D
I C
2 fV
D
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, I (^) L lags V by 90°^ and I (^) C leads V by 90°. Theoretically there are three phasor diagrams possible — each depend- ing on the relative values of I (^) L and I (^) C:
Figure 16.
(i) I (^) L > I (^) C (giving a supply current, I D I (^) L I (^) C lagging V by 90°) (ii) I (^) C > I (^) L (giving a supply current, I D I (^) C I (^) L leading V by 90°) (iii) I (^) L D I (^) C (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, I (^) L D
V
X L
, I C D
V
X C
I D phasor difference between I (^) L and I (^) C, 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.
The circuit and phasor diagrams are as shown in Figure 16.4.
(a) Inductive reactance, X (^) L D 2 fL D 2 50 120 ð 10 ^3
D 37. 70
Capacitive reactance, X (^) C D
2 fC
D
2 50 25 ð 10 ^6
D 127. 3
Current flowing in inductance, I (^) L D
V
X L
D
D 2. 653 A
Current flowing in capacitor, I (^) C D
V
X C
D
D 0. 786 A
(b) I (^) L and I (^) C are anti-phase. Hence supply current,
I D I (^) L I (^) C D 2. 653 0. 786 D 1. 867 A and the current lags the supply voltage V by 90 °^ (see Figure 16.4(i))
(c) Circuit impedance, Z D
V
I
D
D 53. 56 Z
(d) Power consumed, P D VI cos D 100 1. 867 cos 90°
D 0 W
Problem 5. Repeat Problem 4 for the condition when the frequency is changed to 150 Hz.
Figure 16.
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
√ R^2 C X (^) L 2
Current, I (^) LR D
V
Z LR
and I (^) C D
V
X C
Supply current I D phasor sum of I (^) LR and I (^) C (by drawing)
D
√ f I (^) LR cos 1 2 C I (^) LR sin 1 ¾ I (^) C 2 g (by calculation)
where ¾ means ‘the difference between’.
Circuit impedance Z D
V
I
tan 1 D
VL
VR
D
X L
R
, sin 1 D
X L
Z LR
and cos 1 D
R
Z LR
tan D
I (^) LR sin 1 ¾ I (^) C I (^) LR cos 1
and cos D
I (^) LR 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.
Figure 16.
The circuit diagram is shown in Figure 16.6(a).
(a) For the coil, inductive reactance X (^) L D 2 fL
D 2 50 159. 2 ð 10 ^3
D 50
Impedance Z 1 D
√ R^2 C X^2 L D
√ 40 2 C 50 2 D 64. 03
Current in coil, I (^) LR D
V
Z 1
D
D 3. 748 A
Branch phase angle 1 D arctan
X L
R
D arctan
( 50 40
) D arctan 1. 25
D 51. 34 °^ D 51 ° 20 ′^ lagging (see phasor diagram in Figure 16.6(b))
(b) Capacitive reactance, X (^) C D
2 fC
D
2 50 30 ð 10 ^6
D 106. 1
Current in capacitor, I (^) C D
V
X C
D
= 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 I (^) LR and I (^) C 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 I (^) LR and I (^) C may be resolved into their hori- zontal (or ‘in-phase’) and vertical (or ‘quadrant’) components. The horizontal component of I (^) LR is
I (^) LR cos 51 ° 200 D 3 .748 cos 51° 200 D 2 .342 A
The horizontal component of I (^) C is I (^) C cos 90°^ D 0 Thus the total horizontal component, I (^) H D 2. 342 A
The vertical component of I (^) LR D I (^) LR sin 51 ° 200
D 3 .748 sin 51° 200
D 2 .926 A The vertical component of I (^) C D I (^) C sin 90°
D 2 .262 sin 90°^ D 2 .262 A
(b) Capacitive reactance, X (^) C D
2 fC
D
2 5000 0. 02 ð 10 ^6
D 1592
Capacitor current, I (^) C D
V
X C
D
D 25. 13 mA leading V by 90 °
(c) Currents I (^) LR and I (^) C 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
√ [ I (^) LR cos 51. 5 °^2 C I (^) C I (^) LR sin 51. 5 °^2 ]
D 19 .34 mA
and D arctan
( I (^) C I (^) LR sin 51. 5 ° I (^) LR cos 51. 5 °
) D 74. 50 °
(d) Circuit impedance, Z D
V
I
D
- 34 ð 10 ^3
D 2. 068 kZ
(e) Power consumed, P D VI cos D 40 19. 34 ð 10 ^3 cos 74. 50 °
D 206. 7 mW
(Alternatively, P D I (^) R^2 R D I (^) LR^2 R D 8. 30 ð 10 ^3 2
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 I (^) LR is equal to I (^) C. At this condition the supply current I is in-phase with the supply voltage V.
Resonant frequency
When the quadrature component of I (^) LR is equal to I (^) C then: I (^) C D I (^) LR sin 1 (see Figure 16.9)
Figure 16.
Hence
V
X C
D
( V Z (^) LR
) ( X (^) L Z (^) LR
) , (from Section 16.5)
from which, Z (^) LR^2 D X (^) C X (^) L D 2 f (^) r L
( 1 2 fr C
) D
L
C
Hence [
√ R^2 C X (^) L 2 ] 2 D
L
C
and R^2 C X (^) L 2 D
L
C
Thus 2 fr L 2 D
L
C
R^2 and 2fr L D
√( L C
R^2
)
and fr D
2 L
√ ( L C
R^2
) D
√√ √√
( L L^2 C
R^2
L^2
)
i.e. parallel resonant frequency, f (^) r =
2 p
√√ √ √
( 1 LC
−
R^2
L^2
) Hz
(When R is negligible, then fr D
p LC
, which is the same as for series resonance.)
Current at resonance
Current at resonance, I (^) r D I (^) LR cos 1 (from Figure 16.9)
D
( V Z (^) LR
) ( R Z (^) LR
) (from Section 16.5)
D
VR
Z^2 LR
However from equation (16.1), Z^2 LR D
L
C
hence I (^) r =
VR
L
C
=
VRC
L
The current is at a minimum at resonance.
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, R (^) D (or sometimes, the dynamic impedance).
Figure 16.
The circuit diagram is shown in Figure 16.10.
(a) Parallel resonant frequency, fr D
√( 1 LC
R^2
L^2
)
However, resistance R D 0. Hence,
fr D
√( 1 LC
) D
√[ 1 150 ð 10 ^3 40 ð 10 ^6
]
D
√( 10 7 15 4
)
D
√( 1 6
) D 64. 97 Hz
(b) Current circulating in L and C at resonance,
ICIRC D
V
X C
D
V
( 1 2 fr C
) D 2 fr CV
Hence ICIRC D 2 64. 97 40 ð 10 ^6 50 D 0. 816 A
Alternatively, ICIRC D
V
X L
D
V
2 fr L
D
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, vari- able 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,
fr D
√ ( 1 LC
R^2
L^2
)
D
√ ( 1
- 20 20 ð 10 ^6
)
D
p 250 000 90 000
D
p 160 000 D
D 63. 66 Hz
(b) Dynamic resistance, R (^) D D
L
RC
D
60 20 ð 10 ^6
D 166. 7 Z
(c) Current at resonance, I (^) r D
V
R D
D
D 0. 12 A
(d) Circuit Q-factor at resonance D
2 fr L R
D
D 1. 33
Alternatively, Q-factor at resonance D current magnification (for a parallel circuit) D I (^) c /I (^) r
I (^) c D
V
X (^) c
D
V
( 1 2 fr C
) (^) D 2 fr CV D 2 63. 66 20 ð 10 ^6
D 0. 16 A
Hence Q-factor D
I (^) c I (^) r
D
D 1. 33 , as obtained above
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.
Resonant frequency, fr D
√ ( 1 LC
R^2
L^2
)
Transposing for C gives: 2 fr 2 D
LC
R^2
L^2
2 fr 2 C
R^2
L^2
D
LC
C D
L
{ 2 fr 2 C
R^2
L^2
}
When L D 100 mH, R D 800 and fr D 5000 Hz,
C D
100 ð 10 ^3
{ 2 5000 2 C
100 ð 10 ^3
}
Figure 16.
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 I (^) M 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 differ- ence 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, I (^) C must equal the length ab, such that the phasor sum of I (^) M and I (^) C is I. ab D I (^) M 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 I (^) M 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.
(a) Efficiency D
power output power input
hence
D
power input
Power input D
D 6000 W
Hence, 6000 D VI (^) M cos D 240 I (^) M 0. 625 ,
since cos D p f. D 0. 625
Thus current taken by the motor, I (^) M D
D 40 A
The circuit diagram is shown in Figure 16.13(a). The phase angle between I (^) M 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) When a capacitor C is connected in parallel with the motor a current I (^) C flows which leads V by 90°. The phasor sum of I (^) M and I (^) C 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).
Figure 16.
Figure 16.
The horizontal component of I (^) M (shown as oa) D I (^) M 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
Hence the supply current after p.f. correction, I D
D 26. 32 A
(c) The vertical component of I (^) M (shown as ab) D I (^) M 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 I (^) C (shown as bc) is given by ab ac, i.e. 31. 22 8. 22 D 23 A
(d) Current I (^) C D
V
X (^) c
D
V
( 1 2 fC
) (^) D 2 fCV,
from which, C D
I C
2 fV
D
F D 305 mF
(e) kvar rating of the capacitor D
VI (^) c 1000
D
D 5. 52 kvar
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. Deter- mine, 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.
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) I (^) R D 3 .67 A, I (^) L 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) I (^) R D 0 .625 A, I (^) C 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) I (^) C D 0 .377 A, I (^) L 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) I (^) C D 0 .754 A, I (^) L 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.
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 resis- tance 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 capac- itor, (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 resis- tance, (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]
12 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]
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
