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
- 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
⎤
⎥
⎥
⎦