CHAP. 16]

THE LAPLACE TRANSFORM METHOD

411

Using partial fractions, IðsÞ ¼

10  j10 10 þ j10 þ s  j500 s þ 500

and inverting, i ¼ ð10  j10Þe j500t þ ð10 þ j10Þe500t ¼ 14:14e jð500t=4Þ þ ð10 þ j10Þe500t

ðAÞ

ð26Þ

The actual voltage is the imaginary part of (25); hence the actual current is the imaginary part of (26). i ¼ 14:14 sin ð500t  =4Þ þ 10e500t

ðAÞ

16.12 In the series RLC circuit shown in Fig. 16-10, there is no initial charge on the capacitor. If the switch is closed at t ¼ 0, determine the resulting current. The time-domain equation of the given circuit is di 1 Ri þ L þ dt C

ðt iðÞ d ¼ V

ð27Þ

1 V ¼ IðsÞ sC s

ð28Þ

0

Because ið0þ Þ ¼ 0, the Laplace transform of (27) is RIðsÞ þ sLIðsÞ þ

or

2IðsÞ þ 1sIðsÞ þ

Hence,

IðsÞ ¼

1 50 IðsÞ ¼ 0:5s s

(29)

50 50 ¼ s2 þ 2s þ 2 ðs þ 1 þ jÞðs þ 1  jÞ

(30)

Expanding (30) by partial fractions, IðsÞ ¼

j25 j25  ðs þ 1 þ jÞ ðs þ 1  jÞ

ð31Þ

and the inverse Laplace transform of (31) gives i ¼ j25feð1jÞt  eð1þjÞt g ¼ 50et sin t

Fig. 16-10

ðAÞ

Fig. 16-11

16.13 In the two-mesh network of Fig. 16-11, the two loop currents are selected as shown. Write the sdomain equations in matrix form and construct the corresponding circuit. Writing the set of equations in the time domain,   ðt 1 di 5i1 þ Q0 þ i1 ðÞd þ 5i2 ¼  and 10i2 þ 2 2 þ 5i1 ¼  2 dt 0 Taking the Laplace transform of (32) to obtain the corresponding s-domain equations,

ð32Þ

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An