Bird J. Electrical Circuit Theory and Technology

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236 Electrical Circuit Theory and Technology

between voltage and current, (d) the voltage across the coil, and

(e) the voltage across the capacitor.

[(a) 13.18  (b) 15.17 A (c) 52

(d) 772.1 V (e) 603.6V]

19 Three impedances are connected in series across a 100 V, 2 kHz

supply. The impedances comprise:

(i) an inductance of 0.45 mH and 2  resistance,

(ii) an inductance of 570

µH and 5  resistance, and

(iii) a capacitor of capacitance 10

µF and resistance 3 .

Assuming no mutual inductive effects between the two inductances

calculate (a) the circuit impedance, (b) the circuit current, (c) the

circuit phase angle and (d) the voltage across each impedance. Draw

the phasor diagram.

[(a) 11.12  (b) 8.99 A (c) 25

lagging

(d) 53.92 V, 78.53 V, 76.46 V]

20 For the circuit shown in Figure 15.24 determine the voltages V

and

if the supply frequency is 1 kHz. Draw the phasor diagram and

hence determine the supply voltage V and the circuit phase angle.

D 26.0V,V

D 67.05 V,

V D 50 V, 53

leading]

Figure 15.24

Series resonance and Q-factor

21 Find the resonant frequency of a series a.c. circuit consisting of a coil

of resistance 10  and inductance 50 mH and capacitance 0.05

µF.

Find also the current ﬂowing at resonance if the supply voltage is

100 V. [3.183 kHz, 10 A]

22 The current at resonance in a series L –C–R circuit is 0.2 mA. If the

applied voltage is 250 mV at a frequency of 100 kHz and the circuit

capacitance is 0.04

µF, ﬁnd the circuit resistance and inductance.

[1.25 k, 63.3

µH]

23 A coil of resistance 25  and inductance 100 mH is connected

in series with a capacitance of 0.12

µF across a 200 V, variable

frequency supply. Calculate (a) the resonant frequency, (b) the

current at resonance and (c) the factor by which the voltage across

the reactance is greater than the supply voltage.

[(a) 1.453 kHz (b) 8 A (c) 36.52]

24 Calculate the inductance which must be connected in series with a

1000 pF capacitor to give a resonant frequency of 400 kHz.

[0.158 mH]

25 A series circuit comprises a coil of resistance 20  and inductance

2 mH and a 500 pF capacitor. Determine the Q-factor of the circuit at

resonance. If the supply voltage is 1.5 V, what is the voltage across

the capacitor? [100, 150 V]

Single-phase series a.c. circuits 237

Power in a.c. circuits

26 A voltage

v D 200sin ωt volts is applied across a pure resistance of

1.5 k. Find the power dissipated in the resistor.

[13.33 W]

27 A 50

µF capacitor is connected to a 100 V, 200 Hz supply. Deter-

mine the true power and the apparent power.

[0, 628.3 VA]

28 A motor takes a current of 10 A when supplied from a 250 V a.c.

supply. Assuming a power factor of 0.75 lagging ﬁnd the power

consumed. Find also the cost of running the motor for 1 week contin-

uously if 1 kWh of electricity costs 7.20 p.

[1875 W, £22.68]

29 A motor takes a current of 12 A when supplied from a 240 V a.c.

supply. Assuming a power factor of 0.75 lagging, ﬁnd the power

consumed. [2.16 kW]

30 A substation is supplying 200 kVA and 150 kvar. Calculate the corre-

sponding power and power factor. [132 kW, 0.66]

31 A load takes 50 kW at a power factor of 0.8 lagging. Calculate the

apparent power and the reactive power.

[62.5 kVA, 37.5 kvar]

32 A coil of resistance 400  and inductance 0.20 H is connected to a

75 V, 400 Hz supply. Calculate the power dissipated in the coil.

[5.452 W]

33 An 80  resistor and a 6

µF capacitor are connected in series across

a 150 V, 200 Hz supply. Calculate (a) the circuit impedance, (b) the

current ﬂowing and (c) the power dissipated in the circuit.

[(a) 154.9  (b) 0.968 A (c) 75 W]

34 The power taken by a series circuit containing resistance and

inductance is 240 W when connected to a 200 V, 50 Hz supply.

If the current ﬂowing is 2 A ﬁnd the values of the resistance and

inductance. [60 , 255 mH]

35 A circuit consisting of a resistor in series with an inductance takes

210 W at a power factor of 0.6 from a 50 V, 100 Hz supply. Find

(a) the current ﬂowing, (b) the circuit phase angle, (c) the resistance,

(d) the impedance and (e) the inductance.

[(a) 7 A (b) 53

lagging (c) 4.286 

(d) 7.143  (e) 9.095 mH]

36 A 200 V, 60 Hz supply is applied to a capacitive circuit. The current

ﬂowing is 2 A and the power dissipated is 150 W. Calculate the

values of the resistance and capacitance.

[37.5 , 28.61

µF]

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 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 cos

(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, I

, is in-phase

with the supply voltage V and the current ﬂowing in the inductance, I

lags the supply voltage by 90

. The supply current I is the phasor sum of

and I

and thus the current I lags the applied voltage V by an angle

Single-phase parallel a.c. circuits 239

lying between 0

and 90

(depending on the values of I

and I

), shown

as angle  in the phasor diagram.

From the phasor diagram:

I D



I

C I

, (by Pythagoras’ theorem)

where I

and I

tan D

, sin D

and cos  D

(by trigonometric ratios

Circuit impedance, Z D

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

D 3A

Current ﬂowing in the inductance I

2fL

210002.387 ð 10

3



D 4A

(b) From the phasor diagram, supply current, I D



I

C I





3

C 4



D 5A

D arctan





D 53.13

D 53



lagging

(d) Circuit impedance, Z D

D 12 Z

(e) Power consumed P D VI cos D 605cos 53

 D 180 W

(Alternatively, power consumed P D I

R D 3

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.

240 Electrical Circuit Theory and Technology

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, I

is in-phase with the supply voltage V and the

current ﬂowing in the capacitor, I

, leads V by 90

. The supply current

I is the phasor sum of I

and I

and thus the current I leads the applied

voltage V by an angle lying between 0

and 90

(depending on the values

of I

and I

), shown as angle ˛ in the phasor diagram.

From the phasor diagram:

I D



I

C I

, (by Pythagoras’ theorem)

where I

and I

tan˛ D

, sin˛ D

and cos ˛ D

(by trigonometric ratios)

Circuit impedance Z D

Figure 16.2

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) Current in resistor, I

240

D 3A

Current in capacitor, I



2fC



D 2fCV

D 25030 ð 10

240

D 2.262 A

(b) Supply current, I D



I

C I

 D



3

C 2.262



D 3.757 A

D arctan



2.262



D 37



leading

(d) Circuit impedance, Z D

240

3.757

D 63.88 Z

Single-phase parallel a.c. circuits 241

(e) True or active power dissipated, P D VI cos˛

D 2403.757 cos37

D 720 W

(Alternatively, true power P D I

R D 3

80 D 720 W)

(f) Apparent power, S D VI D 2403.757 D 901.7VA

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),

D I cos53.13

D 20.6

D 1.2A

and I

D I sin53.13

D 20.8

D 1.6A

(Alternatively, I

and I

can be measured from the scaled phasor

diagram.)

Figure 16.3

From the circuit diagram,

from which R D

120

1.2

D 100 Z

and I

D 2fCV, from which, C D

2fV

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, I

lags V by 90

and I

leads V by 90

Theoretically there are three phasor diagrams possible—each depend-

ing on the relative values of I

and I

242 Electrical Circuit Theory and Technology

Figure 16.4

(i) I

(giving a supply current, I D I

 I

lagging V by 90

)

(ii) I

(giving a supply current, I D I

 I

leading V by 90

)

(iii) I

D I

(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

I D phasor difference between I

and I

,andZ D

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

D 2fL D 250120 ð 10

3



D 37.70 

Capacitive reactance, X

2fC

25025 ð 10

6



D 127.3 

Current ﬂowing in inductance, I

100

37.70

D 2.653 A

Current ﬂowing in capacitor, I

100

127.3

D 0.786 A

(b) I

and I

are anti-phase. Hence supply current,

I D I

 I

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

1.867

D 53.56 Z

(d) Power consumed, P D VI cos  D 1001.867cos 90



D 0W

Problem 5. Repeat Problem 4 for the condition when the

frequency is changed to 150 Hz.

Single-phase parallel a.c. circuits 243

(a) Inductive reactance, X

D 2150120 ð 10

3

 D 113.1 

Capacitive reactance, X

215025 ð 10

6



D 42.44 

Current ﬂowing in inductance, I

100

113.1

D 0.884 A

Current ﬂowing in capacitor, I

100

42.44

D 2.356 A

(b) Supply current, I D I

 I

D 2.356  0.884 D 1.472 A leading V

by 90

(see Figure 4(ii))

100

1.472

D 67.93 Z

(d) Power consumed, P D VI cos  D 0W(since  D 90

)

From Problems 4 and 5:

(i) When X

then I

and I lags V by 90

(ii) When X

then I

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 induc-

tance 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 I

of Figure 16.5(d) may be resolved into horizontal and

vertical components. The horizontal component, shown as op is I

cos

and the vertical component, shown as pq is I

sin

. There are three

possible conditions for this circuit:

(i) I

sin

(giving a supply current I leading V by angle

 — as shown in Figure 16.5(e))

(ii) I

sin

(giving I lagging V by angle  —as shown in

Figure 16.5(f))

(iii) I

D I

sin

(this is called parallel resonance, see Section 16.6).

There are two methods of ﬁnding the phasor sum of currents I

and

in Figures 16.5(e) and (f). These are: (i) by a scaled phasor diagram,

244 Electrical Circuit Theory and Technology

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, Z



R

C X



Current, I

and I

Supply current I D phasor sum of I

and I

(by drawing)



fI

cos



C I

sin

¾ I



g (by calculation)

where ¾ means ‘the difference between’.

Circuit impedance Z D

tan

, sin

and cos 

tan D

sin

¾ I

cos

and cos  D

cos

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.

Single-phase parallel a.c. circuits 245

Figure 16.6

The circuit diagram is shown in Figure 16.6(a).

(a) For the coil, inductive reactance X

D 2fL

D 250159.2 ð 10

3



D 50 

Impedance Z



R

C X

 D



40

C 50

 D 64.03 

Current in coil, I

240

64.03

D 3.748 A

Branch phase angle 

D arctan





D arctan1.25

D 51.34

D 51



lagging

(see phasor diagram in Figure 16.6(b))

(b) Capacitive reactance, X

2fC

25030 ð 10

6



D 106.1 

Current in capacitor, I

240

106.1

= 2.262 A leading the supply

voltage by 90

(see phasor diagram of Figure 16.6(b)).

and I

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

and I

may be resolved into their hori-

zontal (or ‘in-phase’) and vertical (or ‘quadrant’) components. The

horizontal component of I

cos51

 D 3.748cos 51

D 2.342 A

The horizontal component of I

is I

cos90

D 0

Thus the total horizontal component, I

D 2.342 A

The vertical component of I

DI

sin51



D3.748sin51

D2.926 A

The vertical component of I

D I

sin90

D 2.262sin90

D 2.262 A