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Solved Problems-5

Problems-5
The circuit shown in the figure is initially in the steady state with the switch S open. At t = 0, the switch S is closed. Obtain the current through the inductor for t > 0. Take R1 = R2 = R4 =1Ω and R3 = 2Ω and L = 1H.
 
Description: Description: Description: 1192.png
Solution
When the switch S is open and steady state exists, the current through the inductor is,
 
Description: Description: Description: 1187.png
 
Description: Description: 7365.png
 
After S is closed, for t > 0, by KVL,
2i1 – i2 – i3 = 1
 i1 + 2i2 + Description: Description: 7371.png – i3 = 0
 i1 – i2 + 4i3 = 0
 
Taking Laplace transform,
2I1(s) – I2(s) – I3(s) = Description: Description: 7377.png
 I1(s) + I2(s)[s + 2] – I3(s) = i2(0 –) = 1
 I1(s) – I2(s) + 4I3(s) = 0
 
By Cramer’s Rule,
I2(s) = Description: Description: 7383.png
 
Taking inverse Laplace transform,
i2(t) = Description: Description: 7389.png 
 




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