GATE 2024 EE Q.14: solution and answer

GATE 2024 · EEQ.14 · 1 mark · MCQ

The circuit shown in the figure with the switch $S$ open, is in steady state. After the switch $S$ is closed, the time constant of the circuit in seconds is

GATE 2024 EE question 14: Two horizontal rails: a top wire and a bottom (reference) wire. — Figure from the GATE 2024 EE paper. It is part of the question.

(A)
1.25
(B)
0
(C)
1
(D)
1.5

Answer and solution

ANSWER

The time constant after S closes is 1.25 s, which is option (A).

Official answer: ASource: Official answer key, GATE 2024 EE (IISc Bengaluru)

Time constant of a source-free network: set the independent source to zero. A $1 A$ current source becomes an open circuit, so it drops out of the circuit.
With $S$ closed and the source open, only one loop is left: node A $\to$ left branch ( $1 Ω$ + $1 H$ ) $\to$ bottom rail $\to$ the parallel inductors $\to$ node B $\to$ $1 H$ $\to$ $1 Ω$ $\to$ $S$ $\to$ node A.
Parallel inductors: $1 H ∥ 1 H = \frac{1 \times 1}{1 + 1} = 0.5 H$ . Before switching, the right-hand part was isolated from the source, so both carried zero current; being identical and sharing the same voltage, they carry equal currents after closing and act as a single $0.5 H$ element.
Everything in the one loop is in series: $R_{e q} = 1 + 1 = 2 Ω$ , $L_{e q} = 1 + 1 + 0.5 = 2.5 H$ .
$τ = \frac{L _{e q}}{R _{e q}} = \frac{2.5}{2} = 1.25 s$
Check with KCL/KVL: let $i_{2}$ be the current through $S$ . By KCL at node A, the left-branch current is $i_{1} = 1 - i_{2}$ . Equating the voltage of node A (above the bottom rail) along both paths: $(1 - i_{2}) - \frac{d i _{2}}{d t} = i_{2} + \frac{d i _{2}}{d t} + 0.5 \frac{d i _{2}}{d t}$ . This rearranges to $2.5 \frac{d i _{2}}{d t} + 2 i_{2} = 1$ , so $τ = 2.5/2 = 1.25 s$ . Option (A).

KEY IDEA

To find a time constant, set the independent sources to zero. A current source becomes an open circuit, so the left R–L branch ends up in series with the right-hand path. That leaves one R–L loop with R_eq = 2 Ω and L_eq = 1 + 1 + (1 ∥ 1) = 2.5 H, so τ = L_eq/R_eq = 1.25 s.

WATCH OUT FOR

Replacing the current source with a short circuit, as you would a voltage source. The short then bypasses the left 1 Ω + 1 H branch and leaves R = 1 Ω and L = 1 + 0.5 = 1.5 H, giving τ = 1.5 s, which is the distractor (D). The same wrong answer comes from ignoring the left branch because it 'belongs to the source'. A second mistake is adding the two right-hand 1 H inductors in series (as 2 H) when they are actually in parallel (0.5 H).

From the GATE 2024 EE paper. Every question of this paper · Download the full paper · All 20 Transients questions, with notes

GATE 2024 EE Q.14: solution and answer

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