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
,
where Z 1 = 4 + j 3, Z 2 = 10 and Z 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
4 2 + 3 2
= 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
12 2 + 5 2
= 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
Current I =
V
Z
= V Y
= (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.
Force A, 10 N acting at 45
◦ from the positive
horizontal axis.
Force B, 87 N acting at 120 ◦ from the positive
horizontal axis.
Force C, 15 N 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 force A, f A = 10∠45 ◦ , force B, f B = 8∠120 ◦
and force C, f C = 15∠210 ◦ .
15 N
8 N
10 N
45Њ
210Њ
120Њ
Figure 20.10
The resultant force
= f A + f B + f C
= 10∠45
◦
+ 8∠120
◦
+ 15∠210
◦
= 10(cos 45
◦
+ j sin 45
◦
) + 8(cos 120
◦
+ j sin 120
◦
) + 15(cos 210
◦
+ j sin 210
◦
)
= (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 ] = 11.86 N
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 problems on
applications of complex numbers
1. Determine the resistance R and series inductance L (or capacitance C) for each of the
following impedances assuming the frequency to be 50 Hz.
(a) (3 + j 8)) (b) (2 − j 3))
(c) j 14
(d) 8∠−60 ◦
⎡
⎢
⎢
⎣
(a) R = 3 , L = 25.5 mH
(b) R = 2 , C = 1061 μF
(c) R = 0, L = 44.56 mH
(d) R = 4 , C = 459.4 μF
⎤
⎥
⎥
⎦
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
,
where Z 1 = 4 + j 3, Z 2 = 10 and Z 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
4 2 + 3 2
= 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
12 2 + 5 2
= 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
Current I =
V
Z
= V Y
= (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.
Force A, 10 N acting at 45
◦ from the positive
horizontal axis.
Force B, 87 N acting at 120 ◦ from the positive
horizontal axis.
Force C, 15 N 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 force A, f A = 10∠45 ◦ , force B, f B = 8∠120 ◦
and force C, f C = 15∠210 ◦ .
15 N
8 N
10 N
45Њ
210Њ
120Њ
Figure 20.10
The resultant force
= f A + f B + f C
= 10∠45
◦
+ 8∠120
◦
+ 15∠210
◦
= 10(cos 45
◦
+ j sin 45
◦
) + 8(cos 120
◦
+ j sin 120
◦
) + 15(cos 210
◦
+ j sin 210
◦
)
= (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 ] = 11.86 N
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 problems on
applications of complex numbers
1. Determine the resistance R and series inductance L (or capacitance C) for each of the
following impedances assuming the frequency to be 50 Hz.
(a) (3 + j 8)) (b) (2 − j 3))
(c) j 14
(d) 8∠−60 ◦
⎡
⎢
⎢
⎣
(a) R = 3 , L = 25.5 mH
(b) R = 2 , C = 1061 μF
(c) R = 0, L = 44.56 mH
(d) R = 4 , C = 459.4 μF
⎤
⎥
⎥
⎦