GATE 2023 EC Q.35: solution and answer

GATE 2023 · ECQ.35 · 1 mark · NAT

In the circuit shown below, switch S was closed for a long time. If the switch is opened at $t = 0$ , the maximum magnitude of the voltage $V_{R}$ , in volts, is ________ (rounded off to the nearest integer).

GATE 2023 EC question 35: A single loop. — Figure from the GATE 2023 EC paper. It is part of the question.

Answer and solution

ANSWER

The maximum magnitude of $V_{R}$ is $4 V$ . It occurs just after the switch opens ( $V_{R} (0^{+}) = - 4 V$ ) and then decays as $- 4 e^{- 2 t} V$ .

Official answer: 04Source: Official answer key, GATE 2023 EC (IIT Kanpur)

Before the switch opens ( $t < 0$ , DC steady state): the inductor behaves as a short circuit. It is in parallel with the $2 Ω$ resistor, so the resistor has zero voltage across it and carries no current: $V_{R} (0^{-}) = 0$ .
Inductor current in steady state (the short bypasses the $2 Ω$ , leaving only the $1 Ω$ in the loop): $i_{L} (0^{-}) = \frac{V _{s}}{1 Ω} = \frac{2}{1} = 2 A$ , flowing through the inductor from the left node to the right node.
An inductor current cannot change instantaneously: $i_{L} (0^{+}) = i_{L} (0^{-}) = 2 A$ , still flowing left to right through the inductor.
For $t > 0$ the open switch disconnects the source and the $1 Ω$ resistor. The inductor and the $2 Ω$ resistor now form a closed loop on their own, so the inductor current must return through the resistor from the right node to the left node.
Voltage across the resistor at $t = 0^{+}$ ( $+$ at the left node): current flows through the resistor from right to left, so the right node is at the higher potential and $V_{R} (0^{+}) = - i_{L} (0^{+}) \times 2 Ω = - 2 \times 2 = - 4 V$ .
Source-free RL decay with $τ = \frac{L}{R} = \frac{1}{2} s$ : $V_{R} (t) = - 4 e^{- 2 t} V, t > 0$ The magnitude falls monotonically, so it is largest at $t = 0^{+}$ (and $V_{R} = 0$ for all $t < 0$ ).
Maximum magnitude: $∣ V_{R} ∣_{m a x} = 4 V$ .

KEY IDEA

The inductor current stays at its pre-switching value (2 A, set only by the 1 Ω resistor because the inductor shorts the 2 Ω). Once the switch opens, that current has nowhere to go except back through the 2 Ω resistor, so the resistor voltage jumps from 0 to 2 A × 2 Ω = 4 V in magnitude and then decays with τ = L/R = 0.5 s.

WATCH OUT FOR

Answering 0 V because the 2 Ω resistor carried no current before switching. Resistor voltage can jump; only the inductor current is continuous. A second mistake is using the full 3 Ω series path to find i_L(0⁻) (giving 2/3 A and |V_R| ≈ 1.33 V). In DC steady state the inductor shorts out the 2 Ω, so i_L(0⁻) = 2 A. The sign of V_R(0⁺) is negative, but the question asks for the magnitude.

From the GATE 2023 EC paper. Every question of this paper · Download the full paper · All 20 Transients questions, with notes

GATE 2023 EC Q.35: solution and answer

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