Page 188 - Schaum's Outline of Theory and Problems of Electric Circuits

177
HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
CHAP. 8]
8.2 A series RLC circuit, with R ¼ 50
; L ¼ 0:1H; and C ¼ 50 mF, has a constant voltage V ¼ 100 V
applied at t ¼ 0. Obtain the current transient, assuming zero initial charge on the capacitor.
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R 1
5 2
2
2
2
¼ ¼ 250 s 1 ! 0 ¼ ¼ 2:0 10 s ¼ ! ¼ j370:8 rad=s
0
2L LC
This is an oscillatory case ð < ! 0 Þ, and the general current expression is
i ¼ e 250t ðA 1 cos 370:8 t þ A 2 sin 370:8tÞ
The initial conditions, obtained as in Problem 8.1, are

di
þ
ið0 Þ¼ 0 ¼ 1000 A=s
dt
0 þ
and these determine the values: A 1 ¼ 0, A 2 ¼ 2:70 A. Then
250t
i ¼ e ð2:70 sin 370:8tÞ ðAÞ
8.3 Rework Problem 8.2, if the capacitor has an initial charge Q 0 ¼ 2500 mC.
Everything remains the same as in Problem 8.2 except the second initial condition, which is now

di Q 0 di 100 ð2500=50Þ
0 þ L þ ¼ V or ¼ ¼ 500 A=s
dt C dt 0:1
0 þ 0 þ
The initial values are half those in Problem 8.2, and so, by linearity,
i ¼ e 250t ð1:35 sin 370:8tÞ ðAÞ

8.4 A parallel RLC network, with R ¼ 50:0
, C ¼ 200 mF, and L ¼ 55:6 mH, has an initial charge
Q 0 ¼ 5:0 mC on the capacitor. Obtain the expression for the voltage across the network.

1 1 2 1 4 2
¼ ¼ 50 s ! 0 ¼ ¼ 8:99 10 s
2RC LC
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
2
2
Since ! 0 > , the voltage function is oscillatory and so ! d ¼ ! ¼ 296 rad/s. The general voltage
0
expression is
v ¼ e 50t ðA 1 cos 296t þ A 2 sin 296tÞ
With Q 0 ¼ 5:0 10 3 C, V 0 ¼ 25:0V. At t ¼ 0, v ¼ 25:0 V. Then, A 1 ¼ 25:0.
dv 50t 50t
¼ 50e ðA 1 cos 296t þ A 2 sin 296tÞþ 296e ð A 1 sin 296t þ A 2 cos 296tÞ
dt
At t ¼ 0, dv=dt ¼ V 0 =RC ¼ ! d A 2 A 1 , from which A 2 ¼ 4:22. Thus,
v ¼ e 50t ð25:0 cos 296t 4:22 sin 296tÞðVÞ

8.5 In Fig. 8-19, the switch is closed at t ¼ 0. Obtain the current i and capacitor voltage v , for
C
t > 0.
As far as the natural response of the circuit is concerned, the two resistors are in parallel; hence,
¼ R eq C ¼ð5
Þð2 mFÞ¼ 10 ms
þ

By continuity, v C ð0 Þ¼ v C ð0 Þ¼ 0. Furthermore, as t !1, the capacitor becomes an open circuit, leav-
ing 20
in series with the 50 V. That is,
50
ið1Þ ¼ ¼ 2:5A v C ð1Þ ¼ ð2:5AÞð10
Þ¼ 25 V
20