ISRO EC 2025 Q.80: solution and answer

ISRO 2025 · ECQ.80 · 1 mark · MCQ

A unit step voltage is applied at t = 0 to a series RL circuit with R = 2Ω, L = 2 H. What is the voltage across the inductor at t = 0+?

ISRO EC 2025 question 80: A step-voltage source (sketch of a unit step waveform beside the source symbol) in series with a resistor R = 2 Ω and an… — Figure from the ISRO EC 2025 paper. It is part of the question.

(A)
0 V
(B)
1 V
(C)
2 V
(D)
Infinity

Answer and solution

ANSWER

The voltage across the inductor at $t = 0^{+}$ is $1 V$ , option (B).

Official answer: BSource: ISRO's final answer key of 15 May 2026

The source is a unit step: $v (t) = 1 V$ for $t \geq 0$ , and before the step the inductor carries no current, $i (0^{-}) = 0$ .
The current through an inductor cannot change instantaneously, so $i (0^{+}) = i (0^{-}) = 0$ .
The voltage across the resistor at $0^{+}$ is $v_{R} (0^{+}) = R i (0^{+}) = 2 \times 0 = 0 V$ .
Apply KVL around the series loop at $t = 0^{+}$ : $v_{L} (0^{+}) = 1 - v_{R} (0^{+}) = 1 - 0 = 1 V$ .
Check with the full solution: $i (t) = \frac{1}{R} (1 - e^{- R t / L}) = 0.5 (1 - e^{- t})$ , so $v_{L} = L \frac{d i}{d t} = 2 \times 0.5 e^{- t} = e^{- t} V$ . At $t = 0^{+}$ this is $1 V$ , and it decays with time constant $L / R = 1 s$ .
So 0 V would need the inductor to act as a short circuit at $0^{+}$ , and 2 V would need the resistor to drop the full source value (not the case, since there is no current yet). Infinity would need an impulse source, but a step is finite.

KEY IDEA

At $t = 0^{+}$ an inductor with zero initial current behaves like an open circuit: its current stays at zero, so the resistor drops nothing and the whole source voltage appears across the inductor.

WATCH OUT FOR

Treating the inductor as a short circuit at $0^{+}$ and answering 0 V, which is how it behaves in steady state, not at the instant of switching.

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 2 ISRO Transients questions, with notes