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THE LAPLACE TRANSFORM METHOD
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Applying the method of partial fractions, 104 A B s s 2500 s s 2500 104 104 4 A and B 4 s 2500 s 0 s s 2500 I s 4 4 2 4 6 s s 2500 s 2500 s s 2500 13
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Taking the inverse Laplace transform of (14), we obtain i 4 6e 2500t (A).
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In the series RL circuit of Fig. 16-7, an exponential voltage v 50e 100t (V) is applied by closing the switch at t 0. Find the resulting current.
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The time-domain equation for the given circuit is Ri L In the s-domain, (15) has the form RI s sLI s Li 0 V s Substituting the circuit constants and the transform of the source, V s 50= s 100 , in (16), 10I s s 0:2 I s By the Heaviside expansion formula, l 1 I s l 1   X P an a t P s en Q s Q 0 an n 1:2 Then, 5 s 100 or I s 250 s 100 s 50 17 16 di v dt 15
Here, P s 250, Q s s2 150s 5000, Q 0 s 2s 150, a1 100, and a2 50. i l 1 I s 250 100t 250 50t e e 5e 100t 5e 50t 50 50 A
The series RC circuit of Fig. 16-8 has a sinusoidal voltage source v 180 sin 2000t  (V) and an initial charge on the capacitor Q0 1:25 mC with polarity as shown. Determine the current if the switch is closed at a time corresponding to  908.
Fig. 16-7
Fig. 16-8
Fig. 16-9
The time-domain equation of the circuit is 40i t   t 1 1:25 10 3 i  d 180 cos 2000t 25 10 6 0 18
THE LAPLACE TRANSFORM METHOD
[CHAP. 16
The Laplace transform of (18) gives the s-domain equation 40I s 1:25 10 3 4 104 180s I s 2 s 25 10 6 s s 4 106 4:5s2 1:25 4 106 s 103 s 103 19
I s
(20)
Applying the Heaviside expansion formula to the rst term on the right in (20), we have P s 4:5s2 , Q s s3 103 s2 4 106 s 4 109 , Q 0 s 3s2 2 103 s 4 106 , a1 j2 103 , a2 j2 103 , and a3 103 . Then, i
3 P j2 103 j2 103 t P j2 103 j2 103 t P 103 103 t 0 0 1:25e 10 t e e e 0 3 3 Q j 10 Q j2 10 Q 103 3 3 3
1:8 j0:9 e j2 10 t 1:8 j0:9 e j2 10 t 0:35e 10 1:8 sin 2000t 3:6 cos 2000t 0:35e 4:02 sin 2000t 116:68 0:35e 10
21
103 t
At t 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 10 3 i0 180 sin 908 40 3:25 A 25 10 6 The same result is obtained if we set t 0 in (21).
16.10 In the series RL circuit of Fig. 16-9, the source is v 100 sin 500t  (V). resulting current if the switch is closed at a time corresponding to  0.
The s-domain equation of a series RL circuit is RI s sLI s Li 0 V s The transform of the source with  0 is V s 100 500 s2 500 2
Determine the
22
Since there is no initial current in the inductance, Li 0 0. Substituting the circuit constants into (22), 5I s 0:01sI s 5 104 s2 25 104 or I s 5 106 s2 25 104 s 500 23
Expanding (23) by partial fractions,     1 j 1 j 10 I s 5 5 s j500 s j500 s 500 The inverse Laplace transform of (24) is i 10 sin 500t 10 cos 500t 10e 500t 10e 500t 14:14 sin 500t 458 A 24
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