Bird J. Electrical Circuit Theory and Technology

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468 Electrical Circuit Theory and Technology

Problem 3. A series circuit possesses resistance R and capaci-

tance C. The circuit dissipates a power of 1.732 kW and has a

power factor of 0.866 leading. If the applied voltage is given by

v D

141.4sin10

t C /9 volts, determine (a) the current ﬂowing

and its phase, (b) the value of resistance R, and (c) the value of

capacitance C.

(a) Since v D 141.4 sin10

t C /9 volts, then 141.4 V represents the

maximum value, from which the rms voltage, V D 141.4/

2 D

100 V, and the phase angle of the voltage DC/9 rad or 20

leading. Hence as a phasor the voltage V is written as 100

Power factor D 0.866 D cos, from which  D arccos 0.866 D 30

Hence the angle between voltage and current is 30

Power P D VI cos . Hence 1732 D 100I cos 30

from which,

current, jIjD

1732

1000.866

D 20 A

Since the power factor is leading, the current phasor leads the

voltage— in this case by 30

. Since the voltage has a phase angle

of 20

current, I D 20

20

C 30

 A D 20

(b) Impedance Z D

100

D 5

30

 or 4.33 j2.5

Hence the resistance, R

= 4.33 Z and the capacitive reactance,

D 2.5 .

Alternatively, the resistance may be determined from active power,

P D I

R. Hence 1732 D 20

R, from which,

resistance R D

1732

20

D 4.33 Z

v D 141.4sin10

t C /9 volts, angular velocity

ω D 10

rad/s. Capacitive reactance, X

D 2.5 , thus

2.5 D

2fC

ωC

from which, capacitance, C D

2.5ω

2.510



F D 40 mF

Problem 4. For the circuit shown in Figure 26.8, determine the

active power developed between points (a) A and B, (b) C and D,

Figure 26.8

Power in a.c. circuits 469

Circuit impedance, Z D 5 C

3 C j4j10

3 C j4 j10

D 5 C

40  j30

3  j6

D 5 C

36.87

6.71

63.43

D 5 C 7.45

26.56

D 5 C 6.66 C j3.33 D 11.66 Cj3.33 or

12.13

15.94



Current I D

100

12.13

15.94

D 8.24

15.94

(a) Active power developed between points A and B D I

R D

8.24

5 D 339.5 W

(b) Active power developed between points C and D is zero, since no

power is developed in a pure capacitor.

D I



C Z



D 8.24

15.94



j10

3  j6



D 8.24

15.94



90

6.71

63.43



D 12.28

42.51

Hence the active power developed between points E and F

D I

R D 12.28

3 D 452.4 W

[Check: Total active power developed D 339.5 C452.4 D 791.9W

or 792 W, correct to three signiﬁcant ﬁgures.

Total active power, P D I

D 8.24

11.66 D 792 W (since

1l.66  is the total circuit equivalent resistance)

or P D VI cos D 1008.24 cos15.94

D 792 W]

Problem 5. The circuit shown in Figure 26.9 dissipates an active

power of 400 W and has a power factor of 0.766 lagging. Determine

(a) the apparent power, (b) the reactive power, (c) the value and

phase of current I, and (d) the value of impedance Z.

Figure 26.9

Since power factor D 0.766 lagging, the circuit phase angle

 D arccos 0.766, i.e.,  D 40

lagging which means that the current I

lags voltage V by 40

(a) Since power, P D VI cos , the magnitude of apparent power,

S D VI D

cos

400

0.766

D 522.2VA

(b) Reactive power Q D VI sin  D 522.2sin 40

 D 335.7 var lagg-

ing. (The reactive power is lagging since the circuit is inductive,

Figure 26.10

470 Electrical Circuit Theory and Technology

Figure 26.11 (a) Circuit

diagram (b) Phasor diagram

which is indicated by the lagging power factor.) The power triangle

is shown in Figure 26.10.

magnitude of current jIjD

522.2

100

D 5.222 A

Since the voltage is at a phase angle of 30

(see Figure 26.9)

and current lags voltage by 40

, the phase angle of current is

 40

D10

. Hence current I = 5.222

−10

(d) Total circuit impedance Z

100

5.222

10

D 19.15

 or 14.67 Cj12.31

Hence impedance Z D Z

 4 D 14.67 C j12.31  4

D .10.67 Y j12.31/Z or 16.29

49.08

Further problems on power in a.c. circuits may be found in Section 26.6,

problems 1 to 12, page 472.

26.5 Power factor

improvement

For a particular active power supplied, a high power factor reduces the

current ﬂowing in a supply system and therefore reduces the cost of cables,

transformers, switchgear and generators, as mentioned in Section 16.7,

page 252. Supply authorities use tariffs which encourage consumers to

operate at a reasonably high power factor. One method of improving the

power factor of an inductive load is to connect a bank of capacitors in

parallel with the load. Capacitors are rated in reactive voltamperes and

the effect of the capacitors is to reduce the reactive power of the system

without changing the active power. Most residential and industrial loads on

a power system are inductive, i.e. they operate at a lagging power factor.

A simpliﬁed circuit diagram is shown in Figure 26.11(a) where a capac-

itor C is connected across an inductive load. Before the capacitor is

connected the circuit current is I

and is shown lagging voltage V by

angle 

in the phasor diagram of Figure 26.11(b). When the capacitor C

is connected it takes a current I

which is shown in the phasor diagram

leading voltage V by 90

. The supply current I in Figure 26.11(a) is now

the phasor sum of currents I

and I

as shown in Figure 26.11(b). The

circuit phase angle, i.e., the angle between V and I, has been reduced from



to 

and the power factor has been improved from cos

to cos 

Figure 26.12(a) shows the power triangle for an inductive circuit with

a lagging power factor of cos 

. In Figure 26.12(b), the angle 

has

been reduced to 

, i.e., the power factor has been improved from

cos

to cos 

by introducing leading reactive voltamperes (shown as

length ab) which is achieved by connecting capacitance in parallel with

Power in a.c. circuits 471

Figure 26.12 Effect of

connecting capacitance in

parallel with the inductive load

the inductive load. The power factor has been improved by reducing the

reactive voltamperes; the active power P has remained unaffected.

Power factor correction results in the apparent power S decreasing

(from 0a to 0b in Figure 26.12(b)) and thus the current decreasing, so

that the power distribution system is used more efﬁciently.

Another method of power factor improvement, besides the use of static

capacitors, is by using synchronous motors; such machines can be made

to operate at leading power factors.

Problem 6. A 300 kVA transformer is at full load with an overall

power factor of 0.70 lagging. The power factor is improved by

adding capacitors in parallel with the transformer until the overall

power factor becomes 0.90 lagging. Determine the rating (in kilo-

vars) of the capacitors required.

At full load, active power, P D VI cos  D 3000.70 D 210 kW.

Circuit phase angle  D arccos 0.70 D 45.57

Reactive power, Q D VI sin D 300sin 45.57

 D 214.2 kvar lagging.

The power triangle is shown as triangle 0ab in Figure 26.13. When the

power factor is 0.90, the circuit phase angle  D arccos 0.90 D 25.84

The capacitor rating needed to improve the power factor to 0.90 is given

by length bd in Figure 26.13.

Figure 26.13

Tan 25.84

D ad/210, from which, ad D 210 tan 25.84

D 101.7 kvar.

Hence the capacitor rating, i.e., bd D ab  ad D 214.2  101.7 D

112.5 kvar leading.

Problem 7. A circuit has an impedance Z D 3 C j4 and a

source p.d. of 50

V at a frequency of 1.5 kHz. Determine

(a) the supply current, (b) the active, apparent and reactive power,

impedance Z to improve the power factor of the circuit to 0.966

lagging, and (d) the value of capacitance needed to improve the

power factor to 0.966 lagging.

(a) Supply current, I D

3 C j4

53.13

D 10

−23.13

(b) Apparent power, S D VI

D 50

10

23.13



D 500

53.13

D 300 C j400VA D P C jQ

Hence active power, P

= 300 W

apparent power, S

= 500 VA and

reactive power, Q

= 400 var lagging.

472 Electrical Circuit Theory and Technology

Figure 26.14

The power triangle is shown in Figure 26.14.

Hence angle  D arccos 0.966 D 15

To improve the power factor from cos 53.13

, i.e. 0.60, to 0.966, the

power triangle will need to change from Ocb (see Figure 26.15) to

0ab, the length ca representing the rating of a capacitor connected in

parallel with the circuit. From Figure 26.15, tan 15

D ab/300, from

which, ab D 300tan15

D 80.38 var.

Hence the rating of the capacitor,caD cb  ab

D 400  80.38

D 319.6 var leading.

Figure 26.15

(d) Current in capacitor, I

319.6

D 6.39 A

Capacitive reactance, X

6.39

D 7.82 

Thus 7.82 D 1/2fC, from which,

required capacitance C D

215007.82

F  13.57 mF

Further problems on power factor improvement may be found in

Section 26.6 following, problems 13 to 16, page 473.

26.6 Further problems

on power in a.c. circuits

Power in a.c. circuits

1 When the voltage applied to a circuit is given by 2 C j5V, the

current ﬂowing is given by 8 C j4A. Determine the power dissi-

pated in the circuit. [36 W]

2 A current of 12 C j5A ﬂows in a circuit when the supply voltage

is 150 C j220V. Determine (a) the active power, (b) the reactive

power, and (c) the apparent power. Draw the power triangle.

[(a) 2.90 kW (b) 1.89 kvar lagging (c) 3.46 kVA]

3 A capacitor of capacitive reactance 40  and a resistance of 30 

are connected in series to a supply voltage of 200

V. Determine

the active power in the circuit. [480 W]

4 The circuit shown in Figure 26.16 takes 81 VA at a power factor of

0.8 lagging. Determine the value of impedance Z.

[4 C j3 or 5

36.87

]

5 A series circuit possesses inductance L and resistance R. The

circuit dissipates a power of 2.898 kW and has a power

factor of 0.966 lagging. If the applied voltage is given by

v D

169.7sin100t  /4 volts, determine (a) the current ﬂowing

Figure 26.16

Power in a.c. circuits 473

and its phase, (b) the value of resistance R, and (c) the value of

inductance L.

[(a) 25

60

A (b) 4.64  (c) 12.4 mH]

6 The p.d. across and the current in a certain circuit are represented

by 190 C j40Vand9 j4A respectively. Determine the active

power and the reactive power, stating whether the latter is leading

or lagging. [1550 W; 1120 var lagging]

7 Two impedances, Z

D 6

 and Z

D 10

 are connected

in series and have a total reactive power of 1650 var lagging. Deter-

mine (a) the average power, (b) the apparent power, and (c) the

power factor. [(a) 2469 W (b) 2970 VA (c) 0.83 lagging]

8 A current i D 7.5sinωt  /4 A ﬂows in a circuit which has

an applied voltage

v D 180sinωt C /12V. Determine (a) the

circuit impedance, (b) the active power, (c) the reactive power, and

(d) the apparent power. Draw the power triangle.

[(a) 24

 (b) 337.5W

Figure 26.17

9 The circuit shown in Figure 26.17 has a power of 480 W and a power

factor of 0.8 leading. Determine (a) the apparent power, (b) the reac-

tive power, and (c) the value of impedance Z.

[(a) 600 VA (b) 360 var leading

50.19

]

10 For the network shown in Figure 26.18, determine (a) the values of

currents I

and I

, (b) the total active power, (c) the reactive power,

and (d) the apparent power.

[(a) I

D 6.20

29.74

A,I

D 19.86

8.92

A (b) 981 W

Figure 26.18

11 A circuit consists of an impedance 5

45

 in parallel with a

resistance of 10 . The supply current is 4 A. Determine for the

circuit (a) the active power, (b) the reactive power, and (c) the power

factor. [(a) 49.34 W (b) 28.90 var leading (c) 0.863 leading]

12 For the network shown in Figure 26.19, determine the active power

developed between points (a) A and B, (b) C and D, (c) E and F

[(a) 254.1 W (b) 0 (c) 65.92 W]

Power factor improvement

13 A 600 kVA transformer is at full load with an overall power factor of

0.64 lagging. The power factor is improved by adding capacitors in

parallel with the transformer until the overall power factor becomes

0.95 lagging. Determine the rating (in kvars) of the capacitors

needed. [334.8 kvar leading]

14 A source p.d. of 130

V at 2 kHz is applied to a circuit having an

impedance of 5 Cj12. Determine (a) the supply current, (b) the

active, apparent and reactive powers, (c) the rating of the capac-

itor to be connected in parallel with the impedance to improve the

Figure 26.19

474 Electrical Circuit Theory and Technology

Figure 26.20

power factor of the circuit to 0.940 lagging, and (d) the value of the

capacitance of the capacitor required.

[(a) 10

27.38

A (b) 500 W, 1300 VA, 1200 var lagging

µF]

15 The network shown in Figure 26.20 has a total active power of

2253 W. Determine (a) the total impedance, (b) the supply current,

factor, (f) the capacitance of the capacitor to be connected in parallel

with the network to improve the power factor to 0.90 lagging, if the

supply frequency is 50 Hz.

[(a) 3.51

58.40

 (b) 35.0 A (c) 4300 VA

(d) 3662 var lagging (e) 0.524 lagging (f) 542.3

µF]

16 The power factor of a certain load is improved to 0.92 lagging with

the addition of a 30 kvar bank of capacitors. If the resulting supply

apparent power is 200 kVA, determine (a) the active power, (b) the

reactive power before power factor correction, and (c) the power

factor before correction.

[(a) 184 kW (b) 108.4 kvar lagging (c) 0.862 lagging]

Assignment 8

This assignment covers the material contained in chapters 23

to 26.

The marks for each question are shown in brackets at the end of

each question.

1 The total impedance Z

of an electrical circuit is given by:

D Z

ð Z

C Z

Determine Z

in polar form, correct to 3 signiﬁcant ﬁgures, when

D 5.5

21

, Z

D 2.6

 and Z

D 4.8

10

2 For the network shown in Figure A8.1, determine

(a) the equivalent impedance of the parallel branches

(b) the total circuit equivalent impedance

(d) the circuit phase angle

(e) currents I

and I

(f) the p.d. across points A and B

(g) the p.d. across points B and C

(h) the active power developed in the inductive branch

(i) the active power developed across the j10  capacitor

(j) the active power developed between points B and C

(k) the total active power developed in the network

(l) the total apparent power developed in the network

(m) the total reactive power developed in the network (30)

8 Ω

j6 Ω

5 Ω

−j3 Ω

170∠0° V

−j10 Ω

Figure A8.1

3 A 400 kVA transformer is at full load with an overall power factor

of 0.72 lagging. The power factor is improved by adding capaci-

tors in parallel with the transformer until the overall power factor

becomes 0.92 lagging. Determine the rating (in kilovars) of the capa-

citors required (10)

27 A.c. bridges

At the end of this chapter you should be able to:

ž derive the balance equations of any a.c. bridge circuit

ž state types of a.c. bridge circuit

ž calculate unknown components when using an a.c. bridge

circuit

27.1 Introduction

A.C. bridges are electrical networks, based upon an extension of

the Wheatstone bridge principle, used for the determination of an

unknown impedance by comparison with known impedances and for

the determination of frequency. In general, they contain four impe-

dance arms, an a.c. power supply and a balance detector which is sensitive

to alternating currents. It is more difﬁcult to achieve balance in an

a.c. bridge than in a d.c. bridge because both the magnitude and the

phase angle of impedances are related to the balance condition. Balance

equations are derived by using complex numbers. A.C. bridges provide

precise methods of measurement of inductance and capacitance, as well

as resistance.

27.2 Balance conditions

for an a.c. bridge

The majority of well known a.c. bridges are classiﬁed as four-arm bridges

and consist of an arrangement of four impedances (in complex form,

Z D R š jX) as shown in Figure 27.1. As with the d.c. Wheatstone

bridge circuit, an a.c. bridge is said to be ‘balanced’ when the current

through the detector is zero (i.e., when no current ﬂows between B and

D of Figure 27.1). If the current through the detector is zero, then the

current I

ﬂowing in impedance Z

must also ﬂow in impedance Z

Also, at balance, the current I

ﬂowing in impedance Z

, must also ﬂow

through Z

Figure 27.1 Four-arm bridge

At balance:

(i) the volt drop between A and B is equal to the volt drop between A

and D,

i.e., V

D V

i.e., I

D I

(both in magnitude and in phase) 27.1

(ii) the volt drop between B and C is equal to the volt drop between D

and C,

i.e., V

D V

A.c. bridges 477

i.e., I

D I

(both in magnitude and in phase) 27.2

Dividing equation (27.1) by equation (27.2) gives

from which

or Z

= Z

27.3

Equation (27.3) shows that at balance the products of the impedances of

opposite arms of the bridge are equal.

If in polar form, Z

, Z

and Z

, then from equation (27.3), 



 D





, which shows that there are two conditions to be

satisﬁed simultaneously for balance in an a.c. bridge, i.e.,

and a

Y a

= a

Y a

When deriving balance equations of a.c. bridges, where at least two of

the impedances are in complex form, it is important to appreciate that for

a complex equation a C jb D c C jd the real parts are equal, i.e. a D c,

and the imaginary parts are equal, i.e., b D d.

Usually one arm of an a.c. bridge circuit contains the unknown

impedance while the other arms contain known ﬁxed or variable

components. Normally only two components of the bridge are variable.

When balancing a bridge circuit, the current in the detector is gradually

reduced to zero by successive adjustments of the two variable components.

At balance, the unknown impedance can be expressed in terms of the ﬁxed

and variable components.

Procedure for determining the balance equations of any a.c. bridge

circuit

(i) Determine for the bridge circuit the impedance in each arm in

complex form and write down the balance equation as in equa-

tion (27.3). Equations are usually easier to manipulate if L and C

are initially expressed as X

and X

, rather than ωL or 1/(ωC).

(ii) Isolate the unknown terms on the left-hand side of the equation in

the form a C jb.

(iii) Manipulate the terms on the right-hand side of the equation into

the form c C jd.

(iv) Equate the real parts of the equation, i.e., a D c, and equate the

imaginary parts of the equation, i.e., b D d.

(v) Substitute ωL for X

and 1/(ωC)forX

where appropriate and

express the ﬁnal equations in their simplest form.