Bird J. Electrical Circuit Theory and Technology

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528 Electrical Circuit Theory and Technology

Figure 29.10

current is a minimum (a) the capacitance of the capacitor, (b) the

dynamic resistance, (c) the supply current, (d) the Q-factor, (e) the

bandwidth, (f) the upper and lower half-power frequencies and

(g) the value of the circuit impedance at the 3 dB frequencies.

[(a) 48.73 nF (b) 6.57 k (c) 3.81 mA (d) 5.03

(e) 497.3 Hz (f) 2761 Hz; 2264 Hz (g) 4.64 k]

3 A 0.1

µF capacitor and a pure inductance of 0.02 H are connected in

parallel across a 12 V variable-frequency supply. Determine (a) the

resonant frequency of the circuit, and (b) the current circulating in

the capacitance and inductance at resonance.

[(a) 3.56 kHz (b) 26.84 mA]

4 A coil of resistance 300  and inductance 100 mH and a 4000 pF

capacitor are connected (i) in series and (ii) in parallel. Find for each

connection (a) the resonant frequency, (b) the Q-factor, and (c) the

impedance at resonance.

[(i) (a) 7958 Hz (b) 16.67 (c) 300 ]

[(ii) (a) 7943 Hz (b) 16.64 (c) 83.33 k]

5 A network comprises a coil of resistance 100  and inductance 0.8 H

and a capacitor having capacitance 30

µF. Determine the resonant

frequency of the network when the capacitor is connected (a) in

series with, and (b) in parallel with the coil.

[(a) 32.5 Hz (b) 25.7 Hz]

Figure 29.11

6 Determine the value of capacitor C shown in Figure 29.10 for which

the resonant frequency of the network is 1 kHz. [2.30

µF]

7 In the parallel network shown in Figure 29.11, inductance L is

40 mH and capacitance C is 5

µF. Determine the resonant frequency

of the circuit if (a) R

D 0 and (b) R

D 40 .

[(a) 355.9 Hz (b) 318.3 Hz]

8 A capacitor of reactance 5  is connected in series with a 10 

resistor. The whole circuit is then connected in parallel with a coil

of inductive reactance 20  and a variable resistor. Determine the

value of this resistance for which the parallel network is resonant.

[10 ]

Figure 29.12

9 Determine, for the parallel network shown in Figure 29.12, the values

of inductance L for which the circuit is resonant at a frequency of

600 Hz. [2.50 mH or 0.45 mH]

10 Find the resonant frequency of the two-branch parallel network

shown in Figure 29.13. [667 Hz]

11 Determine the value of the variable resistance R in Figure 29.14 for

which the parallel network is resonant. [11.87 ]

12 For the parallel network shown in Figure 29.15, determine the

resonant frequency. Find also the value of resistance to be connected

in series with the 10

µF capacitor to change the resonant frequency

to 1 kHz. [928 Hz; 5.27 ]

Parallel resonance and Q-factor 529

Figure 29.13 Figure 29.14 Figure 29.15

13 Determine the overall Q-factor of a parallel arrangement consisting

of a capacitor having a Q-factor of 410 and an inductor having a

Q-factor of 90. [73.8]

14 The value of capacitance in an LR–C parallel network is 49.74 nF. If

the resonant frequency of the circuit is 200 kHz and the bandwidth is

800 Hz, determine for the network (a) the Q-factor, (b) the dynamic

resistance, and (c) the magnitude of the impedance when the supply

frequency is 0.5% smaller than the tuned frequency.

[(a) 250 (b) 4 k (c) 1.486 k]

Assignment 9

This assignment covers the material contained in chapters 27

to 29.

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

the end of each question.

1 In a Schering bridge network PQRS, the arms are made up as follows:

PQ—a standard capacitor C

, QR — a capacitor C

in parallel with

a resistor R

, RS — a resistor R

, SP — the capacitor under test,

represented by a capacitor C

in series with a resistor R

. The detector

is connected between Q and S and the a.c. supply is connected between

P and R.

(a) Sketch the bridge and derive the equations for R

and C

when

the bridge is balanced.

(b) Evaluate R

and C

if, at balance C

D 5 nF, R

D 300 , C

30 nF and R

D 1.5k. (16)

2 A coil of inductance 25 mH and resistance 5  is connected in series

with a variable capacitor C. If the supply frequency is 1 kHz and

the current ﬂowing is 2 A, determine, for series resonance, (a) the

value of capacitance C, (b) the supply p.d., and (c) the p.d. across the

capacitor. (8)

3AnL –R –C series circuit has a peak current of 5 mA ﬂowing in it

when the frequency of the 200 mV supply is 5 kHz. The Q-factor

of the circuit under these conditions is 75. Determine (a) the voltage

across the capacitor, and (b) the values of the circuit resistance, induc-

tance and capacitance. (8)

4 A coil of resistance 15  and inductance 150 mH is connected in

parallel with a 4

µF capacitor across a 50 V variable-frequency supply.

Determine (a) the resonant frequency of the circuit, (b) the dynamic

resistance (c) the current at resonance, and (d) the circuit Q-factor at

resonance. (10)

5 Ω

2 mH

Figure A9.1

5 For the parallel network shown in Figure A9.1, determine the value

of C for which the resonant frequency is 2 kHz. (8)

30 Introduction to

network analysis

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

ž appreciate available methods of analysing networks

ž solve simultaneous equations in two and three unknowns

using determinants

ž analyse a.c. networks using Kirchhoff’s laws

30.1 Introduction

Voltage sources in series-parallel networks cause currents to ﬂow in

each branch of the circuit and corresponding volt-drops occur across

the circuit components. A.c. circuit (or network) analysis involves the

determination of the currents in the branches and/or the voltages across

components.

The laws which determine the currents and voltage drops in a.c.

networks are:

(a) current, I

= V =Z, where Z is the complex impedance and V the

voltage across the impedance;

(b) the laws for impedances in series and parallel, i.e., total

impedance,

D Z

C Z

C ...CZ

for n impedances connected

in series,

and

C ...C

for n impedances

connected in parallel; and

(i) ‘At any point in an electrical circuit the phasor sum of the

currents ﬂowing towards that junction is equal to the phasor

sum of the currents ﬂowing away from the junction.’

(ii) ‘In any closed loop in a network, the phasor sum of the voltage

drops (i.e., the products of current and impedance) taken around

the loop is equal to the phasor sum of the e.m.f.’s acting in that

loop.’

In any circuit the currents and voltages at any point may be determined

by applying Kirchhoff’s laws (as demonstrated in this chapter), or by

532 Electrical Circuit Theory and Technology

extensions of Kirchhoff’s laws, called mesh-current analysis and nodal

analysis (see Chapter 31).

However, for more complicated circuits, a number of circuit theorems

have been developed as alternatives to the use of Kirchhoff’s laws to

solve problems involving both d.c. and a.c. electrical networks. These

include:

(a) the superposition theorem (see Chapter 32)

(b) Th

evenin’s theorem (see Chapter 33)

(d) the maximum power transfer theorems (see Chapter 35).

In addition to these theorems, and often used as a preliminary to

using circuit theorems, star-delta (or T  ) and delta-star (or  T)

transformations provide a method for simplifying certain circuits (see

Chapter 34).

In a.c. circuit analysis involving Kirchhoff’s laws or circuit theorems,

the use of complex numbers is essential.

The above laws and theorems apply to linear circuits, i.e., circuits

containing impedances whose values are independent of the direction and

magnitude of the current ﬂowing in them.

30.2 Solution of

simultaneous equations

using determinants

When Kirchhoff’s laws are applied to electrical circuits, simultaneous

equations result which require solution. If two loops are involved, two

simultaneous equations containing two unknowns need to be solved; if

three loops are involved, three simultaneous equations containing three

unknowns need to be solved and so on. The elimination and substitu-

tion methods of solving simultaneous equations may be used to solve

such equations. However a more convenient method is to use

determinants.

Two unknowns

When solving linear simultaneous equations in two unknowns using deter-

minants:

(i) the equations are initially written in the form:

x C b

y C c

D 0

x C b

y C c

D 0

(ii) the solution is given by:

y

where D



Introduction to network analysis 533

i.e., the determinant of the coefﬁcients left when the x-column is

‘covered up’,



i.e., the determinant of the coefﬁcients left when the y-column is

‘covered up’,

and D D



i.e., the determinant of the coefﬁcients left when the constants-

column is ‘covered up’.

A‘2ð 2’ determinant



is evaluated as ad  bc

Three unknowns

When solving linear simultaneous equations in three unknowns using

determinants:

(i) the equations are initially written in the form:

x C b

y C c

z C d

D 0

x C b

y C c

z C d

D 0

x C b

y C c

z C d

D 0

(ii) the solution is given by:

y

1

where D



i.e., the determinant of the coefﬁcients left when the x-column is

‘covered up’,



i.e., the determinant of the coefﬁcients left when the y-column is

‘covered up’,



i.e., the determinant of the coefﬁcients left when the z-column is

‘covered up’,

534 Electrical Circuit Theory and Technology

and D D



i.e., the determinant of the coefﬁcients left when the constants-

column is ‘covered up’.

To evaluate a 3

× 3 determinant:

(a) The minor of an element o

fa3by3matrix is the value of the

2 by 2 determinant obtained by covering up the row and column

containing that element.

Thus for the matrix





123

456

789





the minor of element 4 is the

determinant



, i.e., 2 ð 9  3 ð 8 D 18  24 D6. Simi-

larly, the minor of element 3 is



, i.e., 4 ð 8  5 ð 7 D

32  35 D3

(b) The sign of the minor depends on its position within the matrix,

the sign pattern being





CC

C

CC





. Thus the signed minor of

element 4 in the above matrix is 



D6 D 6

The signed-minor of an element is called the cofactor of the element.

Thus the cofactor of element 2 is 



D36  42 D 6

the elements and their cofactors of any row or any column of

the corresponding 3 by 3 matrix.

Thus a 3 by 3 determinant



abc

def

ghj



is evaluated as



 b



C c



using the top row,

or  b



C e



 h



using the second column.

There are thus six ways of evaluating a 3 by 3 determinant.

Determinants are used to solve simultaneous equations in some of the

following problems and in Chapter 31.

Introduction to network analysis 535

30.3 Network analysis

using Kirchhoff’s laws

Kirchhoff’s laws may be applied to both d.c. and a.c. circuits. The laws

are introduced in Chapter 13 for d.c. circuits. To demonstrate the method

of analysis, consider the d.c. network shown in Figure 30.1. If the current

ﬂowing in each branch is required, the following three-step procedure

may be used:

(i) Label branch currents and their directions on the circuit diagram.

The directions chosen are arbitrary but, as a starting-point, a useful

guide is to assume that current ﬂows from the positive terminals

of the voltage sources. This is shown in Figure 30.2 where the

three branch currents are expressed in terms of I

and I

only,

since the current through resistance R, by Kirchhoff’s current law,

is (I

C I

)

Figure 30.1

(ii) Divide the circuit into loops— two in this ease (see Figure 30.2)

and then apply Kirchhoff’s voltage law to each loop in turn. From

loop ABEF, and moving in a clockwise direction (the choice of

loop direction is arbitrary), E

D I

r C I

C I

R (note that the

two voltage drops are positive since the loop direction is the same

as the current directions involved in the volt drops). Hence

8 D I

C 5I

C I



or 6I

C 5I

D 8 1

Figure 30.2

From loop BCDE in Figure 30.2, and moving in an anticlockwise

direction, (note that the direction does not have to be the same as

that used for the ﬁrst loop), E

D I

C I

C I

R,

i.e., 3 D 2I

C 5I

C I



or 5I

C 7I

D 3 2

(iii) Solve simultaneous equations (1) and (2) for I

and I

Multiplying equation (1) by 7 gives: 42I

C 35I

D 56 3

Multiplying equation (2) by 5 gives: 25I

C 35I

D 15 4

Equation (3)  equation (4) gives: 17I

D 41

from which, current I

D 41/17 D 2.412 A D 2.41 A, correct to

two decimal places.

From equation (1): 62.412 C 5I

D 8, from which,

current I

8  62.412

D1.294 A

−1.29 A, correct to two decimal places.

The minus sign indicates that current I

ﬂows in the opposite direction

to that shown in Figure 30.2.

536 Electrical Circuit Theory and Technology

The current ﬂowing through resistance R is

I

C I

 D 2.412 C 1.294

D 1.118 A D 1.12 A, correct to two decimal places.

[A third loop may be selected in Figure 30.2, (just as a check), moving

clockwise around the outside of the network. Then E

 E

D I

 I

i.e. 8  3 D I

 2I

. Thus 5 D 2.412 21.294 D 5]

An alternative method of solving equations (1) and (2) is shown below

using determinants. Since

C 5I

 8 D 0 1

C 7I

 3 D 0 2

then



5 8

7 3



I



6 8

5 3



i.e.,

15 C 56

I

18 C 40

42  25

I

from which, I

D 41/17 D 2.41 A and I

D22/17 D −1.29 A, as

obtained previously.

The above procedure is shown for a simple d.c. circuit having two

unknown values of current. The procedure however applies equally well

to a.c. networks and/or to circuits where three unknown currents are

involved. This is illustrated in the following problems.

Problem 1. Use Kirchhoff’s laws to ﬁnd the current ﬂowing in

each branch of the network shown in Figure 30.3.

Figure 30.3

(i) The branch currents and their directions are labelled as shown in

Figure 30.4

(ii) Two loops are chosen. From loop ABEF, and moving clockwise,

25I

C 20I

C I

 D 100

i.e., 45I

C 20I

D 100 1

From loop BCDE, and moving anticlockwise,

10I

C 20I

C I

 D 50

i.e., 20I

C 30I

D j50 2

3 ð equation 1 gives: 135I

C 60I

D 300 3

2 ð equation 2 gives: 40I

C 60I

D j100 4

Introduction to network analysis 537

Equation (3)— equation 4 gives: 95I

D 300  j100,

from which, current I

300  j100

D 3.329

−18.43

A or

.3.158

− j1.052/A

Substituting in equation (1) gives:

453.158  j1.052 C 20I

D 100, from which,

100  453.158  j1.052

D .

−2.106 Y j2.367/A or 3.168

131.66

Thus

C I

D 3.158  j1.052 C 2.106 C j2.367

D .1.052 Y j1.315/ A or 1.684

51.34

Problem 2. Determine the current ﬂowing in the 2  resistor of

the circuit shown in Figure 30.5 using Kirchhoff’s laws. Find also

the power dissipated in the 3  resistance.

Figure 30.5

(i) Currents and their directions are assigned as shown in Figure 30.6.

Figure 30.6

(ii) Three loops are chosen since three unknown currents are required.

The choice of loop directions is arbitrary. From loop ABCDE, and

moving anticlockwise,

C 6I

C 4I

 I

 D 8

i.e., 5I

C 10I

 4I

D 8 1