Page 162 - Schaum's Outline of Theory and Problems of Electric Circuits

FIRST-ORDER CIRCUITS
CHAP. 7]
25 3:46 151
t 2 t 1
¼ ¼ ¼ 15:54 ms
ln v 1 ln v 2 ln 20 ln 5
L 2
and so R ¼ ¼ ¼ 128:7
15:54 10 3

7.14 In Fig. 7-28, switch S 1 is closed at t ¼ 0. Switch S 2 is opened at t ¼ 4 ms. Obtain i for t > 0.

Fig. 7-28

As there is always inductance in the circuit, the current is a continuous function at all times. In the
interval 0 t 4 ms, with the 100
shorted out and a time constant ¼ð0:1HÞ=ð50
Þ¼ 2 ms, i starts at
zero and builds toward
100 V
¼ 2A
50
even though it never gets close to that value. Hence, as in Problem 7.12
i ¼ 2ð1 e t=2 Þ ðAÞ ð0 t 4Þ ð22Þ
wherein t is measured in ms. In particular,
2
ið4Þ¼ 2ð1 e Þ¼ 1:729 A
In the interval t 4 ms, i starts at 1.729 A and decays toward 100=150 ¼ 0:667 A, with a time constant
2
0:1=150 ¼ ms. Therefore, with t again in ms,
3
i ¼ð1:729 0:667Þe ðt 4Þ=ð2=3Þ þ 0:667 ¼ 428:4e 3t=2 þ 0:667 ðAÞ ðt 4Þ ð23Þ

7.15 In the circuit of Fig. 7-29, the switch is closed at t ¼ 0, when the 6-mF capacitor has charge
Q 0 ¼ 300 mC. Obtain the expression for the transient voltage v R .
The two parallel capacitors have an equivalent capacitance of 3 mF. Then this capacitance is in series
with the 6 mF, so that the overall equivalent capacitance is 2 mF. Thus, ¼ RC eq ¼ 40 ms.
þ
At t ¼ 0 , KVL gives v R ¼ 300=6 ¼ 50 V; and, as t !1, v R ! 0 (since i ! 0). Therefore,
t= t=40
v R ¼ 50 e ¼ 50e ðVÞ
in which t is measured in ms.

Fig. 7-29 Fig. 7-30