Higher Engineering Mathematics

258 COMPLEX NUMBERS

(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 240 V,

50 Hz is connected across an impedance of

(60−j100). 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−j100).

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 ◦ 2 ′

(d) Current flowing,I=

V

Z

=

240 ∠ 0 ◦

116. 6 ∠− 59 ◦ 2 ′

= 2. 058 ∠ 59 ◦ 2 ′A

The circuit and phasor diagrams are as shown in
Fig. 23.8(b).

Problem 17. For the parallel circuit shown in

Fig. 23.9, determine the value of currentIand

its phase relative to the 240 V supply, using

complex numbers.

R 1 = 4 Ω (^) XL = 3 Ω
R 2 = 10 Ω
R 3 = 12 Ω XC = 5 Ω
l
240 V, 50 Hz
Figure 23.9
CurrentI=
V
Z

. ImpedanceZfor the three-branch
parallel circuit is given by:

1

Z

=

1

Z 1

+

1

Z 2

+

1

Z 3

,

whereZ 1 = 4 +j3,Z 2 =10 andZ 3 = 12 −j 5

Admittance, Y 1 =

1

Z 1

=

1

4 +j 3

=

1

4 +j 3

×

4 −j 3

4 −j 3

=

4 −j 3

42 + 32

= 0. 160 −j 0 .120 siemens

Admittance, Y 2 =

1

Z 2

=

1

10

= 0 .10 siemens

Admittance, Y 3 =

1

Z 3

=

1

12 −j 5

=

1

12 −j 5

×

12 +j 5

12 +j 5

=

12 +j 5

122 + 52

= 0. 0710 +j 0 .0296 siemens

Total admittance, Y=Y 1 +Y 2 +Y 3

=(0. 160 −j 0 .120)+(0.10)

+(0. 0710 +j 0 .0296)

= 0. 331 −j 0. 0904

= 0. 343 ∠− 15 ◦ 17 ′siemens