222 Higher Engineering Mathematics
Similarly, for the R−C circuit shown in Fig. 20.8(b),
V C lags I by 90 ◦ (i.e. I leads V C by 90 ◦ ) and
V R − j V C = V , from which R − j X C = Z (where X C
is the capacitive reactance
1
2π fC
ohms).
Problem 15. Determine the resistance and
series inductance (or capacitance) for each of the
following impedances, assuming a frequency of
50 Hz:
(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. X L = 7.0
Since X L = 2πf L then inductance,
L =
X L
2π f
=
7.0
2π(50)
= 0.0223 H or 22.3 mH
(b) Impedance, Z = j 20, i.e. Z = (0 − j 20)) hence
resistance = 0 and reactance = 20 . Since the
imaginary part is negative, the reactance is capacitive, i.e., X C = 20 and since X C =
1
2πf C
then:
capacitance, C =
1
2πf X C
=
1
2π(50)(20)
F
=
10 6
2π(50)(20)
μF = 159.2 μF
(c) Impedance, Z
= 15∠−60
◦
= 15[ cos (−60
◦
) + j sin (−60
◦
)]
= 7.50 − j 12.99
Hence resistance = 7.50 and capacitive reactance, X C = 12.99
Since X C =
1
2πf C
then capacitance,
C =
1
2πf X C
=
10 6
2π(50)(12.99)
μF
= 245 μF
Problem 16. An alternating voltage of 240 V,
50 Hz 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) Impedance Z = (60 − j 100)).
Hence resistance = 60
(b) Capacitive reactance X C = 100 and since
X C =
1
2πf C
then
capacitance, C =
1
2π f X C
=
1
2π(50)(100)
=
10 6
2π(50)(100)
μF
= 31.83 μF
(c) Magnitude of impedance,
|Z | =
[(60) 2 + (−100) 2 ] = 116.6
Phase angle, arg Z = 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 current I and its
phase relative to the 240 V supply, using complex
numbers.
240 V, 50 Hz
R 3 5 12 V
X C 5 5 V
I
R 2 5 10 V
R 1 5 4 V
X L 5 3 V
Figure 20.9
Similarly, for the R−C circuit shown in Fig. 20.8(b),
V C lags I by 90 ◦ (i.e. I leads V C by 90 ◦ ) and
V R − j V C = V , from which R − j X C = Z (where X C
is the capacitive reactance
1
2π fC
ohms).
Problem 15. Determine the resistance and
series inductance (or capacitance) for each of the
following impedances, assuming a frequency of
50 Hz:
(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. X L = 7.0
Since X L = 2πf L then inductance,
L =
X L
2π f
=
7.0
2π(50)
= 0.0223 H or 22.3 mH
(b) Impedance, Z = j 20, i.e. Z = (0 − j 20)) hence
resistance = 0 and reactance = 20 . Since the
imaginary part is negative, the reactance is capacitive, i.e., X C = 20 and since X C =
1
2πf C
then:
capacitance, C =
1
2πf X C
=
1
2π(50)(20)
F
=
10 6
2π(50)(20)
μF = 159.2 μF
(c) Impedance, Z
= 15∠−60
◦
= 15[ cos (−60
◦
) + j sin (−60
◦
)]
= 7.50 − j 12.99
Hence resistance = 7.50 and capacitive reactance, X C = 12.99
Since X C =
1
2πf C
then capacitance,
C =
1
2πf X C
=
10 6
2π(50)(12.99)
μF
= 245 μF
Problem 16. An alternating voltage of 240 V,
50 Hz 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) Impedance Z = (60 − j 100)).
Hence resistance = 60
(b) Capacitive reactance X C = 100 and since
X C =
1
2πf C
then
capacitance, C =
1
2π f X C
=
1
2π(50)(100)
=
10 6
2π(50)(100)
μF
= 31.83 μF
(c) Magnitude of impedance,
|Z | =
[(60) 2 + (−100) 2 ] = 116.6
Phase angle, arg Z = 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 current I and its
phase relative to the 240 V supply, using complex
numbers.
240 V, 50 Hz
R 3 5 12 V
X C 5 5 V
I
R 2 5 10 V
R 1 5 4 V
X L 5 3 V
Figure 20.9