Page 91 - LAPLACE TRANSFORM

2
2 { − s+5 }
I(s) = 7 + 7
2
(s − 2) (s + 10)

2
2
I(s) = 7 + { (− s+5 ) }
2
2
(s − 2) 7 (s + (√10) )
2 1 2 − s 5
I(s) = { } + { + }
2
2
2
2
7 (s − 2) 7 (s + (√10) ) (s + (√10) )
2 1 2 s 5 √10
I(s) = { } + { (−1) + ( )( ) }
2
2
2
2
7 (s − 2) 7 (s + (√10) ) √10 (s + (√10) )

By applying the inverse Laplace transform method;

2 2t 2 5
i(t) = e + [ (−1) cos (√10) t + ( )sin(√10) ]
7 7 √10

2t
( ) = [ e − cos (√ ) t + ( ) (√ ) ]
√

Example 6 (LC Series circuit)

50mF

-3t
12e V 4H

Solve the current i(t) by using Laplace transform when S1 switch is closed. Calculate

the value of VL (t) and VC (t) when t = 0.1sec (Assume that the initial condition is

zero).

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