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A constant current source is supplying 10A to a circuit shown in Figure. The switch S, which is initially closed for a sufficiently long time, is suddenly opened. Obtain obtain the complete time response of the inductor current. What is the energy stored in L, a long time after the switch is opened?
E=250J
E=250J
E=250J
E=350J
When the switch is closed for a long time the entire current flows through the short circuit and hence initial inductor current,
When the switch is opened current gets divided into
resistance and inductance,
---------(1)
Since, both R and L are connected in parallel, 
From (1)
Solving the differential as a linear DE, we get
The steady state current in Inductor is,
Hence, energy stored after a long time or in Steady State,
Note: This circuit can also be simplified by using source transformation and converting the current source to a voltage source and then standard equation of charging RL circuit can be applied.