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THE LAPLACE TRANSFORM METHOD

[CHAP. 16

The Laplace transform of (18) gives the s-domain equation 40IðsÞ þ

or

1:25  103 4  104 180s IðsÞ ¼ 2 þ s 25  106 s s þ 4  106

ð19Þ

4:5s2 1:25  þ 4  106 Þðs þ 103 Þ s þ 103

(20)

IðsÞ ¼

ðs2

Applying the Heaviside expansion formula to the first term on the right in (20), we have PðsÞ ¼ 4:5s2 , QðsÞ ¼ s3 þ 103 s2 þ 4  106 s þ 4  109 , Q 0 ðsÞ ¼ 3s2 þ 2  103 s þ 4  106 , a1 ¼ j2  103 , a2 ¼ j2  103 , and a3 ¼ 103 . Then, i¼

3 Pðj2  103 Þ j2103 t Pð j2  103 Þ j2103 t Pð103 Þ 103 t þ 0 þ 0  1:25e10 t e e e 0 3 3 Q ðj  10 Þ Q ð j2  10 Þ Q ð103 Þ 3

3

¼ ð1:8  j0:9Þej210 t þ ð1:8 þ j0:9Þe j210 t  0:35e10

3

t

ð21Þ

103 t

¼ 1:8 sin 2000t þ 3:6 cos 2000t  0:35e ¼ 4:02 sin ð2000t þ 116:68Þ  0:35e10

3

t

ðAÞ

At t ¼ 0, the current is given by the instantaneous voltage, consisting of the source voltage and the charged capacitor voltage, divided by the resistance. Thus, !, 1:25  103 i0 ¼ 180 sin 908  40 ¼ 3:25 A 25  106 The same result is obtained if we set t ¼ 0 in (21).

16.10 In the series RL circuit of Fig. 16-9, the source is v ¼ 100 sin ð500t þ Þ (V). resulting current if the switch is closed at a time corresponding to  ¼ 0.

Determine the

The s-domain equation of a series RL circuit is RIðsÞ þ sLIðsÞ  Lið0þ Þ ¼ VðsÞ

ð22Þ

The transform of the source with  ¼ 0 is VðsÞ ¼

ð100Þð500Þ s2 þ ð500Þ2

Since there is no initial current in the inductance, Lið0þ Þ ¼ 0. Substituting the circuit constants into (22), 5IðsÞ þ 0:01sIðsÞ ¼

5  104 s2 þ 25  104

or

IðsÞ ¼

5  106 ðs2 þ 25  104 Þðs þ 500Þ

ð23Þ

Expanding (23) by partial fractions,     1 þ j 1  j 10 IðsÞ ¼ 5 þ5 þ s þ j500 s  j500 s þ 500

ð24Þ

The inverse Laplace transform of (24) is i ¼ 10 sin 500t  10 cos 500t þ 10e500t ¼ 10e500t þ 14:14 sin ð500t  458Þ

ðAÞ

16.11 Rework Problem 16.10 by writing the voltage function as v ¼ 100e j500t

ðVÞ

ð25Þ

Now VðsÞ ¼ 100=ðs  j500Þ, and the s-domain equation is 5IðsÞ þ 0:01sIðsÞ ¼

100 s  j500

or

IðsÞ ¼

104 ðs  j500Þðs þ 500Þ

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