Page 163 - Schaum's Outline of Theory and Problems of Electric Circuits

FIRST-ORDER CIRCUITS
[CHAP. 7
152
7.16 In the circuit shown in Fig. 7-30, the switch is moved to position 2 at t ¼ 0. Obtain the current i 2
at t ¼ 34:7 ms.
After the switching, the three inductances have the equivalent
10 5ð10Þ
L eq ¼ þ ¼ 5H
6 15
Then ¼ 5=200 ¼ 25 ms, and so, with t in ms,

5
i ¼ 6e t=25 ðAÞ i 2 ¼ i ¼ 2e t=25 ðAÞ
15
and i 2 ð34:7Þ¼ 2e 34:7=25 A ¼ 0:50 A

7.17 In Fig. 7-31, the switch is closed at t ¼ 0. Obtain the current i and capacitor voltage v C , for
t > 0.

Fig. 7-31
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
Knowing the end conditions on v C , we can write
þ
v C ¼½v C ð0 Þ v C ð1ފe t= þ v C ð1Þ ¼ 25ð1 e t=10 Þ ðVÞ
wherein t is measured in ms.
The current in the capacitor is given by

dv C t=10
i C ¼ C ¼ 5e ðAÞ
dt
and the current in the parallel 10-
resistor is
v C t=10
i 10
¼ ¼ 2:5ð1 e Þ ðAÞ
10
Hence, i ¼ i C þ i 10
¼ 2:5ð1 þ e t=10 ÞðAÞ
The problem might also have been solved by assigning mesh currents and solving simultaneous differ-
ential equations.

7.18 The switch in the two-mesh circuit shown in Fig. 7-32 is closed at t ¼ 0. Obtain the currents i 1
and i 2 , for t > 0.