GATE 2023 EC Q.44: solution and answer

GATE 2023 · ECQ.44 · 2 marks · MCQ

The switch $S_{1}$ was closed and $S_{2}$ was open for a long time. At $t = 0$ , switch $S_{1}$ is opened and $S_{2}$ is closed, simultaneously. The value of $i_{c} (0^{+})$ , in amperes, is

GATE 2023 EC question 44: Parallel circuit with a common bottom rail and a top rail. — Figure from the GATE 2023 EC paper. It is part of the question.

(A)
1
(B)
$- 1$
(C)
0.2
(D)
0.8

Answer and solution

ANSWER

$i_{c} (0^{+}) = - 1 A$ , option (B). The capacitor, still at $20 V$ , supplies $0.8 A$ to the $25 Ω$ resistor and $0.2 A$ to the inductor, so its current flows opposite to the marked arrow.

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

Steady state for $t < 0$ ( $S_{1}$ closed, $S_{2}$ open): the capacitor acts as an open circuit and the inductor as a short circuit. The 1 A source then feeds two parallel paths: the $100 Ω$ resistor (in series with the shorted inductor) and the $25 Ω$ resistor.
Top-rail voltage, which is also the capacitor voltage: $v_{C} (0^{-}) = 1 \times \frac{100 \times 25}{100 + 25} = 20 V$
Inductor current, flowing downward through the $100 Ω$ branch: $i_{L} (0^{-}) = \frac{20}{100} = 0.2 A$ . Check: the $25 Ω$ resistor carries $\frac{20}{25} = 0.8 A$ , and $0.2 + 0.8 = 1 A$ , which is the source current.
Continuity rules: $v_{C} (0^{+}) = v_{C} (0^{-}) = 20 V$ and $i_{L} (0^{+}) = i_{L} (0^{-}) = 0.2 A$ . At $t = 0^{+}$ the capacitor therefore behaves as a $20 V$ source and the inductor as a $0.2 A$ current source.
Circuit at $t = 0^{+}$ : $S_{1}$ is in series with the current source, so opening it disconnects the source from the network. Closing $S_{2}$ shorts the $100 Ω$ resistor, so that resistor carries no current and the inductor is connected directly between the rails. The capacitor, the inductor and the $25 Ω$ resistor are now in parallel, all at $20 V$ . The inductor voltage jumps from $0$ to $20 V$ , which is allowed: only the inductor current has to be continuous.
Current in the $25 Ω$ resistor: $i_{25} (0^{+}) = \frac{v _{C} ( 0 ^{+} )}{25} = \frac{20}{25} = 0.8 A$ , flowing downward.
KCL at the top node, taking every branch current downward (the direction of the marked $i_{c}$ ). No source current enters, because $S_{1}$ is open: $i_{c} (0^{+}) + i_{L} (0^{+}) + i_{25} (0^{+}) = 0$
$i_{c} (0^{+}) = - (0.2 + 0.8) = - 1 A$ The minus sign means the capacitor is discharging, so current flows upward through it, against the marked arrow. This is option (B). The values $0.01 F$ and $1 H$ are not needed to find the $t = 0^{+}$ value.

KEY IDEA

Work out $v_{C} (0^{-})$ and $i_{L} (0^{-})$ from the DC steady state, with the capacitor open and the inductor shorted. At $t = 0^{+}$ , replace the capacitor with a voltage source of $v_{C} (0^{-})$ and the inductor with a current source of $i_{L} (0^{-})$ , redraw the circuit for the new switch positions, and apply KCL. The sign of $i_{c}$ is set by the marked arrow direction.

WATCH OUT FOR

Answering $+ 1$ by dropping the sign: the capacitor current actually flows upward, against the marked $i_{c}$ . Other traps: choosing $0.8$ because you forgot the inductor keeps drawing its $0.2 A$ after the switch; choosing $0.2$ because you ignored the $25 Ω$ resistor; still treating the capacitor as open at $t = 0^{+}$ and getting $i_{c} = 0$ ; and thinking the open $S_{1}$ leaves the $1 A$ source in the circuit.

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.44: solution and answer

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