GATE EE 2017 Set 1 — Question 41
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Electric Circuits → Transient & AC Steady-State → First-Order RL & RC
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Question
The switch in the figure below was closed for a long time. It is opened at . The current in the inductor of for , is
Correct answer
(A) 2.5e^(-4t)
Solution
1. Steady state before the switch opens ():
In steady state the inductor behaves as a short circuit, so the right-hand node is tied to the bottom rail and both resistors are short-circuited.
Seen from the switch, the circuit is the shunt resistor in parallel with the series resistor (whose far end is now at rail potential):Source current , so the node voltage after the switch is .
The current in the series branch flows entirely through the inductor:2. After the switch opens ():
The source and the resistor are disconnected. Looking out from the inductor terminals, the two resistors are in parallel with the series pair :3. Current for :The initial value rules out options (B) and (D), and the time constant (not ) rules out option (C).Answer: (A)
In steady state the inductor behaves as a short circuit, so the right-hand node is tied to the bottom rail and both resistors are short-circuited.
Seen from the switch, the circuit is the shunt resistor in parallel with the series resistor (whose far end is now at rail potential):Source current , so the node voltage after the switch is .
The current in the series branch flows entirely through the inductor:2. After the switch opens ():
The source and the resistor are disconnected. Looking out from the inductor terminals, the two resistors are in parallel with the series pair :3. Current for :The initial value rules out options (B) and (D), and the time constant (not ) rules out option (C).Answer: (A)
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