CHAP. 16]

THE LAPLACE TRANSFORM METHOD

413

and the ﬁnal value is    s þ 250 i1 ð1Þ ¼ lim½sI1 ðsÞ ¼ lim 6:67 ¼ 10 A s!0 s!0 s þ 166:7 The initial value of i2 is given by   i2 ð0þ Þ ¼ lim ½sI2 ðsÞ ¼ lim 6:667 s!1

s!1

 s ¼ 6:67 A s þ 166:7

and the ﬁnal value is    s i2 ð1Þ ¼ lim½sI2 ðsÞ ¼ lim 6:67 ¼0 s!0 s!0 s þ 166:7 Examination of Fig. 16-13 veriﬁes each of the preceding initial and ﬁnal values. At the instant of closing, the inductance presents an inﬁnite impedance and the currents are i1 ¼ i2 ¼ 100=ð10 þ 5Þ ¼ 6:67 A. Then, in the steady state, the inductance appears as a short circuit; hence, i1 ¼ 10 A, i2 ¼ 0.

16.16 Solve for i1 in Problem 16.14 by determining an equivalent circuit in the s-domain. In the s-domain the 0.02-H inductor has impedance ZðsÞ ¼ 0:02s. Therefore, the equivalent impedance of the network as seen from the source is   ð0:02sÞð5Þ s þ 166:7 ZðsÞ ¼ 10 þ ¼ 15 0:02s þ 5 s þ 250 and the s-domain equivalent circuit is as shown in Fig. 16-14. The current is then     VðsÞ 100 s þ 250 s þ 250 I1 ðsÞ ¼ ¼ ¼ 6:67 ZðsÞ s 15ðs þ 166:7Þ sðs þ 166:7Þ This expression is identical with (37) of Problem 16.14, and so the same time function i1 is obtained.

Fig. 16-14

Fig. 16-15

16.17 In the two-mesh network shown in Fig. 16-15 there is no initial charge on the capacitor. Find the loop currents i1 and i2 which result when the switch is closed at t ¼ 0. The time-domain equations for the circuit are 10i1 þ

1 0:2

ðt i1 d þ 10i2 ¼ 50

50i2 þ 10i1 ¼ 50

0

The corresponding s-domain equations are 1 50 50 I ðsÞ þ 10I2 ðsÞ ¼ 50I2 ðsÞ þ 10I1 ðsÞ ¼ 0:2s 1 s s 5 1 1 I2 ðsÞ ¼  I1 ðsÞ ¼ s þ 0:625 s s þ 0:625

10I1 ðsÞ þ Solving,

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An