CHAP. 7]

155

FIRST-ORDER CIRCUITS

Fig. 7-34

7.22

Fig. 7-35

The switch in the circuit shown in Fig. 7-35 is moved from 1 to 2 at t ¼ 0. t > 0.

Find vC and vR , for

With the switch on 1, the 100-V source results in vC ð0 Þ ¼ 100 V; and, by continuity of charge, vC ð0 Þ ¼ vC ð0 Þ. In position 2, with the 50-V source of opposite polarity, vC ð1Þ ¼ 50 V. Thus, þ

B ¼ vC ð1Þ ¼ 50 V

A ¼ vC ð0þ Þ  B ¼ 150 V

 ¼ RC ¼

1 s 200

vC ¼ 150e200t  50 ðVÞ

and Finally, KVL gives vR þ vC þ 50 ¼ 0, or

vR ¼ 150e200t

7.23

ðVÞ

Obtain the energy functions for the circuit of Problem 7.22. wC ¼ 12 Cv2C ¼ 1:25ð3e200t  1Þ2 ðmJÞ ðt 2 vR dt ¼ 11:25ð1  e400t Þ ðmJÞ wR ¼ 0 R

7.24

A series RC circuit, with R ¼ 5 k and C ¼ 20 mF, has two voltage sources in series, v1 ¼ 25uðtÞ ðVÞ

v2 ¼ 25uðt  t 0 Þ ðVÞ

Obtain the complete expression for the voltage across the capacitor and make a sketch, if t 0 is positive. The capacitor voltage is continuous. For t  0, v1 results in a capacitor voltage of 25 V. For 0  t  t 0 , both sources are zero, so that vC decays exponentially from 25 V towards zero: vC ¼ 25et=RC ¼ 25e10t

ð0  t  t 0 Þ

ðVÞ

0

In particular, vC ðt 0 Þ ¼ 25e10t (V). For t  t 0 , vC builds from vC ðt 0 Þ towards the ﬁnal value 25 V established by v2 : 0

vC ¼ ½vC ðt 0 Þ  vC ð1Þeðtt Þ=RC þ vC ð1Þ 0

¼ 25½1  ðe10t  1Þe10t 

ðt  t 0 Þ

ðVÞ

Thus, for all t, 0

vC ¼ 25uðtÞ þ 25e10t ½uðtÞ  uðt  t 0 Þ þ 25½1  ðe10t  1Þe10t uðt  t 0 Þ ðVÞ See Fig. 7-36.

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An