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The switch in the following circuit, shown in fig 3, has been connected to the 12 V source for a long time. At t=0, the switch is thrown to 24V. The value of L=2H, 
(a) Determine 
and 
(b) Write the differential equation governing 
for t>0
(c) Compute the steady state value of 
(a) 1A, 2V
(b)
(c) =4V
(a) 1A, 1V
(b)
(c) =4V
(a) 1A, 2V
(b) 
(c) =4V
(a) 1A, 2V
(b)
(c) =8V
The given system is shown below,
L = 2H
(a) at t<0:- Since switch is connected to 12V for long time. Hence, the circuits enters steady state. So ‘L’ behaves as short circuit while capacitor behaves as open circuit.
This condition is shown below,
(b) For t>0, the switch is moved to other position
Using KVL
Putting 
(c) At steady state
Inductors becomes short circuited and capacitor becomes open circuited.
