Page 166 - Schaum's Outline of Theory and Problems of Electric Circuits

CHAP. 7]

Fig. 7-34 FIRST-ORDER CIRCUITS Fig. 7-35 155

7.22 The switch in the circuit shown in Fig. 7-35 is moved from 1 to 2 at t ¼ 0. Find v and v , for
R
C
t > 0.

With the switch on 1, the 100-V source results in v C ð0 Þ¼ 100 V; and, by continuity of charge,

þ
v C ð0 Þ¼ v C ð0 Þ. In position 2, with the 50-V source of opposite polarity, v C ð1Þ ¼ 50 V. Thus,
þ
B ¼ v C ð1Þ ¼ 50 V A ¼ v C ð0 Þ B ¼ 150 V
1
¼ RC ¼ s
200
and v C ¼ 150e 200t 50 ðVÞ
Finally, KVL gives v R þ v C þ 50 ¼ 0, or
v R ¼ 150e 200t ðVÞ

7.23 Obtain the energy functions for the circuit of Problem 7.22.
2
1
w C ¼ Cv C ¼ 1:25ð3e 200t 1Þ 2 ðmJÞ
2
ð
t 2
v R 400t
w R ¼ dt ¼ 11:25ð1 e Þ ðmJÞ
0 R
7.24 A series RC circuit, with R ¼ 5k
and C ¼ 20 mF, has two voltage sources in series,
0
v 1 ¼ 25uð tÞðVÞ v 2 ¼ 25uðt t ÞðVÞ
0
Obtain the complete expression for the voltage across the capacitor and make a sketch, if t is
positive.
The capacitor voltage is continuous. For t 0, v 1 results in a capacitor voltage of 25 V.
0
For 0 t t , both sources are zero, so that v C decays exponentially from 25 V towards zero:
0
v C ¼ 25e t=RC ¼ 25e 10t ðVÞ ð0 t t Þ
0
10t
0
In particular, v C ðt Þ¼ 25e (V).
0
0
For t t , v C builds from v C ðt Þ towards the final value 25 V established by v 2 :
0
0
v C ¼½v C ðt Þ v C ð1ފe ðt t Þ=RC þ v C ð1Þ
0
¼ 25½1 ðe 10t 0 1Þe 10t Š ðVÞ ðt t Þ
Thus, for all t,
0 10t 0
0
10t
10t
v C ¼ 25uð tÞþ 25e ½uðtÞ uðt t ފ þ 25½1 ðe 1Þe Šuðt t ÞðVÞ
See Fig. 7-36.