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16.6

THE LAPLACE TRANSFORM METHOD

[CHAP. 16

In the series RC circuit of Fig. 16-5, the capacitor has an initial charge 2.5 mC. At t ¼ 0, the switch is closed and a constant-voltage source V ¼ 100 V is applied. Use the Laplace transform method to find the current. The time-domain equation for the given circuit after the switch is closed is   ðt 1 RiðtÞ þ Q0 þ iðÞ d ¼ V C 0 or

10iðtÞ þ

  ðt 1 3 ð2:5  10 Þ þ iðÞ d ¼V 50  106 0

(6)

Q0 is opposite in polarity to the charge which the source will deposit on the capacitor. Taking the Laplace transform of the terms in (6), we obtain the s-domain equation 10IðsÞ 

or

2:5  103 IðsÞ 100 þ ¼ 6 6 s 50  10 s 50  10 s IðsÞ ¼

15 s þ ð2  103 Þ

(7)

The time function is now obtained by taking the inverse Laplace transform of (7):   3 15 ¼ 15e210 t ðAÞ iðtÞ ¼ l1 s þ ð2  103 Þ

Fig. 16-5

16.7

ð8Þ

Fig. 16-6

In the RL circuit shown in Fig. 16-6, the switch is in position 1 long enough to establish steadystate conditions, and at t ¼ 0 it is switched to position 2. Find the resulting current. Assume the direction of the current as shown in the diagram. i0 ¼ 50=25 ¼ 2 A. The time-domain equation is 25i þ 0:01

The initial current is then

di ¼ 100 dt

ð9Þ

Taking the Laplace transform of (9), 25IðsÞ þ 0:01sIðsÞ  0:01ið0þ Þ ¼ 100=s

ð10Þ

25IðsÞ þ 0:01sIðsÞ þ 0:01ð2Þ ¼ 100=s

ð11Þ

Substituting for ið0þ Þ,

and

IðsÞ ¼

100 0:02 104 2  ¼  sð0:01s þ 25Þ 0:01s þ 25 sðs þ 2500Þ s þ 2500

(12)

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