CHAP. 16]

409

THE LAPLACE TRANSFORM METHOD

Applying the method of partial fractions, 104 A B ¼ þ sðs þ 2500Þ s s þ 2500   104  104  A¼ ¼ 4 and B ¼ ¼ 4 s þ 2500 s¼0 s s¼2500

with

Then,

IðsÞ ¼

ð13Þ

4 4 2 4 6   ¼  s s þ 2500 s þ 2500 s s þ 2500

(14)

Taking the inverse Laplace transform of (14), we obtain i ¼ 4  6e2500t (A).

16.8

In the series RL circuit of Fig. 16-7, an exponential voltage v ¼ 50e100t (V) is applied by closing the switch at t ¼ 0. Find the resulting current. The time-domain equation for the given circuit is Ri þ L

di ¼v dt

ð15Þ

In the s-domain, (15) has the form RIðsÞ þ sLIðsÞ  Lið0þ Þ ¼ VðsÞ

ð16Þ

Substituting the circuit constants and the transform of the source, VðsÞ ¼ 50=ðs þ 100Þ, in (16), 10IðsÞ þ sð0:2ÞIðsÞ ¼

5 s þ 100

or

IðsÞ ¼

250 ðs þ 100Þðs þ 50Þ

ð17Þ

By the Heaviside expansion formula, l1 ½IðsÞ ¼ l1



 X Pðan Þ a t PðsÞ ¼ en QðsÞ Q 0 ðan Þ n¼1:2

Here, PðsÞ ¼ 250, QðsÞ ¼ s2 þ 150s þ 5000, Q 0 ðsÞ ¼ 2s þ 150, a1 ¼ 100, and a2 ¼ 50. i ¼ l1 ½IðsÞ ¼

16.9

250 100t 250 50t e e þ ¼ 5e100t þ 5e50t 50 50

Then,

ðAÞ

The series RC circuit of Fig. 16-8 has a sinusoidal voltage source v ¼ 180 sin ð2000t þ Þ (V) and an initial charge on the capacitor Q0 ¼ 1:25 mC with polarity as shown. Determine the current if the switch is closed at a time corresponding to  ¼ 908.

Fig. 16-7

Fig. 16-8

Fig. 16-9

The time-domain equation of the circuit is 40iðtÞ þ

  ðt 1 3 ð1:25  10 Þ þ iðÞ d ¼ 180 cos 2000t 25  106 0

ð18Þ

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An