GATE 2024 IN Q.33: solution and answer

GATE 2024 · INQ.33 · 1 mark · NAT

The switch in the following figure has been closed for a long time ( $t < 0$ ). It is opened at $t = 0$ seconds. The value of $d v_{c} / d t$ at $t = 0^{+}$ is ________ V/s (rounded off to nearest integer).

GATE 2024 IN question 33: A single-loop-plus-branch circuit. — Figure from the GATE 2024 IN paper. It is part of the question.

Answer and solution

ANSWER

$\frac{d v _{c}}{d t}_{0^{+}} = 15 V/s$

Official answer: 15Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

Steady state for $t < 0$ (switch closed). In DC steady state the inductor acts as a short circuit and the capacitor as an open circuit. The source therefore drives one current through the series 4 Ω resistor, the inductor and the lower 4 Ω resistor: $i_{L} (0^{-}) = \frac{12}{4 + 4} = 1.5 A$
Capacitor voltage before switching. The capacitor is in parallel with the lower 4 Ω resistor, so $v_{c} (0^{-}) = 1.5 \times 4 = 6 V$ . The slope does not need this value, but it fixes the state of the circuit.
Continuity at the switching instant. Neither the inductor current nor the capacitor voltage can jump, so $i_{L} (0^{+}) = i_{L} (0^{-}) = 1.5 A$ and $v_{c} (0^{+}) = v_{c} (0^{-}) = 6 V$ .
KCL at the top node at $t = 0^{+}$ . With the switch open, the 4 Ω branch carries no current. All of the inductor current therefore flows into the capacitor: $i_{C} (0^{+}) = i_{L} (0^{+}) = 1.5 A$ , entering the + terminal of $v_{c}$ .
Capacitor law $i_{C} = C \frac{d v _{c}}{d t}$ . $\frac{d v _{c}}{d t}_{0^{+}} = \frac{i _{C} ( 0 ^{+} )}{C} = \frac{1.5}{0.1} = 15 V/s$

KEY IDEA

The inductor current cannot change instantly. Once the switch opens, the capacitor is the only path left for that current, so $i_{C} (0^{+}) = i_{L} (0^{-})$ and $d v_{c} / d t ∣_{0^{+}} = i_{L} (0^{-}) / C$ .

WATCH OUT FOR

Some students treat the capacitor current as zero at $t = 0^{+}$ because it was zero in steady state, and so answer 0. Only the capacitor voltage is continuous, not its current. Others keep the lower 4 Ω resistor in the circuit after the switch opens, splitting the current and getting a smaller slope. A third error is to take $i_{L} (0^{-}) = 12/4 = 3 A$ by forgetting the second resistor, which gives 30.

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

GATE 2024 IN Q.33: solution and answer

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