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412

THE LAPLACE TRANSFORM METHOD

5I1 ðsÞ þ

Q0 1 þ I ðsÞ þ 5I2 ðsÞ ¼ VðsÞ 2s 2s 1

[CHAP. 16

10I2 ðsÞ þ 2sI2 ðsÞ  2i2 ð0þ Þ þ 5I1 ðsÞ ¼ VðsÞ

ð33Þ

When this set of s-domain equations is written in matrix form,      5 þ ð1=2sÞ 5 VðsÞ  ðQ0 =2sÞ I1 ðsÞ ¼ þ VðsÞ þ 2i2 ð0 Þ 5 10 þ 2s I2 ðsÞ the required s-domain circuit can be determined by examination of the ZðsÞ, IðsÞ, and VðsÞ matrices (see Fig. 16-12).

Fig. 16-12

Fig. 16-13

16.14 In the two-mesh network of Fig. 16-13, find the currents which result when the switch is closed. The time-domain equations for the network are di1 di  0:02 2 ¼ 100 dt dt di2 di1 0:02 þ 5i2  0:02 ¼0 dt dt

10i1 þ 0:02

ð34Þ

Taking the Laplace transform of set (34), ð10 þ 0:02sÞI1 ðsÞ  0:02sI2 ðsÞ ¼ 100=s

ð5 þ 0:02sÞI2 ðsÞ  0:02sI1 ðsÞ ¼ 0

ð35Þ

From the second equation in set (35) we find  I2 ðsÞ ¼ I1 ðsÞ

 s s þ 250

ð36Þ

which when substituted into the first equation gives   s þ 250 10 3:33 I1 ðsÞ ¼ 6:67 ¼  sðs þ 166:7Þ s s þ 166:7

ð37Þ

Inverting (37), i1 ¼ 10  3:33e166:7t Finally, substitute (37) into (36) and obtain   1 I2 ðsÞ ¼ 6:67 s þ 166:7

whence

ðAÞ

i2 ¼ 6:67e166:7t

ðAÞ

16.15 Apply the initial- and final-value theorems in Problem 16.14. The initial value of i1 is given by    s þ 250 i1 ð0þ Þ ¼ lim ½sI1 ðsÞ ¼ lim 6:667 ¼ 6:67 A s!1 s!1 s þ 166:7

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
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