Higher Engineering Mathematics, Sixth Edition

Complex numbers 223

Current I=

V

Z

. Impedance Z for the three-branch

parallel circuit is given by:

1

Z

=

1

Z 1

+

1

Z 2

+

1

Z 3

,

whereZ 1 = 4 +j 3 ,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. 28 ◦siemens

CurrentI=

V

Z

=VY

=( 240 ∠ 0 ◦)( 0. 343 ∠− 15. 28 ◦)

= 82. 32 ∠− 15. 28 ◦A

Problem 18. Determine the magnitude and

direction of the resultant of the three coplanar

forces given below, when they act at a point.

ForceA, 10N acting at 45◦from the positive

horizontal axis.

ForceB, 87N acting at 120◦from the positive

horizontal axis.

ForceC, 15N acting at 210◦from the positive

horizontal axis.

The space diagram is shown in Fig. 20.10. The forces
may be written as complex numbers.
Thus forceA, fA= 10 ∠ 45 ◦,forceB, fB= 8 ∠ 120 ◦
and forceC,fC= 15 ∠ 210 ◦.

15N

8N 10N

45 

210 

120 

Figure 20.10

The resultant force

=fA+fB+fC

= 10 ∠ 45 ◦+ 8 ∠ 120 ◦+ 15 ∠ 210 ◦

= 10 (cos45◦+jsin45◦)+ 8 (cos120◦

+jsin120◦)+ 15 (cos210◦+jsin210◦)

=( 7. 071 +j 7. 071 )+(− 4. 00 +j 6. 928 )

+(− 12. 99 −j 7. 50 )

=− 9. 919 +j 6. 499

Magnitude of resultant force

=

√

[(− 9. 919 )^2 +( 6. 499 )^2 ]= 11 .86N

Direction of resultant force

=tan−^1

(

6. 499

− 9. 919

)

= 146. 77 ◦

(since− 9. 919 +j 6 .499 lies in the second quadrant).

Now try the following exercise

Exercise 89 Further problemson

applications of complex numbers

1. Determine the resistanceRand series induc-
   tanceL(or capacitanceC) for each of the
   following impedances assuming the frequ-
   ency to be 50Hz.
   (a)( 3 +j 8 ) (b)( 2 −j 3 )
   (c)j 14  (d) 8∠− 60 ◦
   ⎡
   ⎢
   ⎢
   ⎣

(a)R= 3 ,L= 25 .5mH

(b)R= 2 ,C= 1061 μF

(c)R= 0 ,L= 44 .56mH

(d)R= 4 ,C= 459. 4 μF

⎤

⎥

⎥

⎦