Page 425

414

THE LAPLACE TRANSFORM METHOD

[CHAP. 16

which invert to i1 ¼ 5e0:625t

ðAÞ

i2 ¼ 1  e0:625t

ðAÞ

16.18 Referring to Problem 16.17, obtain the equivalent impedance of the s-domain network and determine the total current and the branch currents using the current-division rule. The s-domain impedance as seen by the voltage source is ZðsÞ ¼ 10 þ

  40ð1=0:2sÞ 80s þ 50 s þ 5=8 ¼ ¼ 10 40 þ 1=0:2s 8s þ 1 s þ 1=8

ð38Þ

The equivalent circuit is shown in Fig. 16-16; the resulting current is IðsÞ ¼

VðsÞ s þ 1=8 ¼5 ZðsÞ sðs þ 5=8Þ

ð39Þ

Expanding IðsÞ in partial fractions, IðsÞ ¼

1 4 þ s s þ 5=8

from which

i ¼ 1 þ 4e5t=8

ðAÞ

Now the branch currents I1 ðsÞ and I2 ðsÞ can be obtained by the current-division rule. Referring to Fig. 16-17, we have   40 5 and i1 ¼ 5e0:625t ðAÞ ¼ I1 ðsÞ ¼ IðsÞ 40 þ 1=0:2s s þ 5=8   1=0:2s 1 1 and i2 ¼ 1  e0:625t ðAÞ I2 ðsÞ ¼ IðsÞ ¼  40 þ 1=0:2s s s þ 5=8

Fig. 16-16

Fig. 16-17

16.19 In the network of Fig. 16-18 the switch is closed at t ¼ 0 and there is no initial charge on either of the capacitors. Find the resulting current i.

Fig. 16-18

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
Advertisement