Higher Engineering Mathematics, Sixth Edition

222 Higher Engineering Mathematics

Similarly, for theR−Ccircuit shown in Fig. 20.8(b),

VC lags I by 90◦ (i.e. I leads VC by 90◦)and

VR−jVC=V, from whichR−jXC=Z(whereXC

is the capacitive reactance

1

2 πfC

ohms).

Problem 15. Determine the resistance and

series inductance (or capacitance) foreach of the

following impedances, assuming a frequency of

50Hz:

(a)( 4. 0 +j 7. 0 ) (b)−j 20 

(c) 15∠− 60 ◦

(a) Impedance,Z=( 4. 0 +j 7. 0 )hence,

resistance= 4. 0 and reactance= 7. 00 .

Since the imaginary part is positive, the reactance

is inductive,

i.e.XL= 7. 0 

SinceXL= 2 πfLtheninductance,

L=

XL

2 πf

=

7. 0

2 π( 50 )

= 0 .0223Hor 22 .3mH

(b) Impedance,Z=j20, i.e.Z=( 0 −j 20 )hence

resistance= 0 and reactance= 20 . Since the

imaginary part is negative, the reactance is cap-

acitive, i.e., XC= 20 and since XC=

1

2 πfC

then:

capacitance,C=

1

2 πfXC

=

1

2 π( 50 )( 20 )

F

=

106

2 π( 50 )( 20 )

μF= 159. 2 μF

(c) Impedance,Z

= 15 ∠− 60 ◦=15[cos(− 60 ◦)+jsin(− 60 ◦)]

= 7. 50 −j 12. 99 

Henceresistance= 7. 50 and capacitive reac-

tance,XC= 12. 99 

SinceXC=

1

2 πfC

thencapacitance,

C=

1

2 πfXC

=

106

2 π( 50 )( 12. 99 )

μF

= 245 μF

Problem 16. An alternating voltage of 240V,

50Hz is connected across an impedance of

( 60 −j 100 ). Determine (a) the resistance (b) the

capacitance (c) the magnitude of the impedance and

its phase angle and (d) the current flowing.

(a) ImpedanceZ=( 60 −j 100 ).

Henceresistance= 60 

(b) Capacitive reactance XC= 100  and since

XC=

1

2 πfC

then

capacitance,C=

1

2 πfXC

=

1

2 π( 50 )( 100 )

=

106

2 π( 50 )( 100 )

μF

= 31. 83 μF

(c) Magnitude of impedance,

|Z|=

√

[( 60 )^2 +(− 100 )^2 ]= 116. 6 

Phase angle, argZ=tan−^1

(

− 100

60

)

=− 59. 04 ◦

(d) Current flowing,I=

V

Z

=

240 ∠ 0 ◦

116. 6 ∠− 59. 04 ◦

= 2. 058 ∠ 59. 04 ◦A

The circuit and phasor diagrams are as shown in

Fig. 20.8(b).

Problem 17. For the parallel circuit shown in

Fig. 20.9, determine the value of currentIand its

phase relative to the 240V supply, using complex

numbers.

240 V, 50 Hz

R 3512 V XC 55 V

I

R 2510 V

R 154 V XL 53 V

Figure 20.9