Page 163

152

7.16

FIRST-ORDER CIRCUITS

[CHAP. 7

In the circuit shown in Fig. 7-30, the switch is moved to position 2 at t ¼ 0. at t ¼ 34:7 ms.

Obtain the current i2

After the switching, the three inductances have the equivalent Leq ¼

10 5ð10Þ þ ¼ 5H 6 15

Then  ¼ 5=200 ¼ 25 ms, and so, with t in ms, i ¼ 6et=25

  5 i ¼ 2et=25 15

ðAÞ

i2 ð34:7Þ ¼ 2e34:7=25 A ¼ 0:50 A

and

7.17

i2 ¼

ðAÞ

In Fig. 7-31, the switch is closed at t ¼ 0. t > 0.

Obtain the current i and capacitor voltage vC , for

Fig. 7-31 As far as the natural response of the circuit is concerned, the two resistors are in parallel; hence,  ¼ Req C ¼ ð5 Þð2 mFÞ ¼ 10 ms By continuity, vC ð0þ Þ ¼ vC ð0 Þ ¼ 0. Furthermore, as t ! 1, the capacitor becomes an open circuit, leaving 20  in series with the 50 V. That is, ið1Þ ¼

50 ¼ 2:5 A 20

vC ð1Þ ¼ ð2:5 AÞð10 Þ ¼ 25 V

Knowing the end conditions on vC , we can write vC ¼ ½vC ð0þ Þ  vC ð1Þet= þ vC ð1Þ ¼ 25ð1  et=10 Þ

ðVÞ

wherein t is measured in ms. The current in the capacitor is given by iC ¼ C

dvC ¼ 5et=10 dt

ðAÞ

and the current in the parallel 10- resistor is i10 ¼ Hence,

vC ¼ 2:5ð1  et=10 Þ 10 

ðAÞ

i ¼ iC þ i10  ¼ 2:5ð1 þ et=10 Þ ðAÞ

The problem might also have been solved by assigning mesh currents and solving simultaneous differential equations.

7.18

The switch in the two-mesh circuit shown in Fig. 7-32 is closed at t ¼ 0. and i2 , for t > 0.

Obtain the currents i1

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An