GATE 2023 EE Q.29: solution and answer

GATE 2023 · EEQ.29 · 1 mark · NAT

The value of parameters of the circuit shown in the figure are
$R_{1} = 2 Ω$ , $R_{2} = 2 Ω$ , $R_{3} = 3 Ω$ , $L = 10 mH$ , $C = 100 μ F$
For time $t < 0$ , the circuit is at steady state with the switch 'K' in closed condition. If the switch is opened at $t = 0$ , the value of the voltage across the inductor ( $V_{L}$ ) at $t = 0^{+}$ in Volts is ________ (Round off to 1 decimal place).

GATE 2023 EE question 29: A 10 A DC current source (arrow pointing up) is connected between a bottom wire (node B) and a top wire (node A). — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

ANSWER

$V_{L} (0^{+}) = 8.0 V$

Official answer: 7.9 to 8.1Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Topology: the source's top node $A$ and bottom node $B$ carry three parallel branches, $(K + R_{3})$ , $(R_{1} + L)$ and $(R_{2} + C)$ . The right ends of the $L$ and $C$ branches are wired straight back to $B$ .
Steady state for $t < 0$ (DC): $L$ acts as a short and $C$ as an open, so the $10 A$ splits between $R_{3}$ and $R_{1}$ only. By current division, $i_{L} (0^{-}) = 10 \times \frac{3}{3 + 2} = 6 A$ , flowing from $A$ through $R_{1}$ and $L$ .
No current flows in $R_{2}$ , so the capacitor holds the full node voltage: $v_{C} (0^{-}) = V_{A B} = R_{1} i_{L} = 2 \times 6 = 12 V$ . Check: $R_{3}$ carries $12/3 = 4 A$ , and $6 + 4 = 10 A$ .
Continuity at switching: $i_{L} (0^{+}) = i_{L} (0^{-}) = 6 A$ and $v_{C} (0^{+}) = v_{C} (0^{-}) = 12 V$ .
At $t = 0^{+}$ , $K$ is open, so the $R_{3}$ branch carries nothing. KCL at node $A$ : $i_{C} (0^{+}) = 10 - i_{L} (0^{+}) = 10 - 6 = 4 A$ through $R_{2}$ and $C$ .
KVL on the $R_{2}$ – $C$ branch gives the node voltage: $V_{A B} (0^{+}) = R_{2} i_{C} + v_{C} = 2 \times 4 + 12 = 20 V$ . The node voltage jumps from $12 V$ to $20 V$ . Only $v_{C}$ and $i_{L}$ have to be continuous.
KVL on the $R_{1}$ – $L$ branch: $V_{A B} = R_{1} i_{L} + V_{L} \Rightarrow V_{L} (0^{+}) = 20 - 2 \times 6 = 8.0 V$ , positive at the marked $+$ terminal (the $R_{1}$ side).
Sanity check: $\frac{d i _{L}}{d t} (0^{+}) = \frac{V _{L}}{L} = \frac{8}{10 \times 1 0 ^{- 3}} = 800 A/s > 0$ . So $i_{L}$ is rising, as it must: in the new steady state $C$ is open and the whole $10 A$ flows through $R_{1}$ and $L$ .

KEY IDEA

At $t = 0^{+}$ , treat the inductor as a $6 A$ current source and the capacitor as a $12 V$ voltage source, both values taken from the $t < 0$ DC steady state. Once $K$ opens, the $10 A$ source current has only two paths. The inductor fixes its branch at $6 A$ , so the $R_{2}$ – $C$ branch must take $4 A$ . That sets the node voltage at $2 \times 4 + 12 = 20 V$ , and $V_{L}$ is whatever $R_{1}$ leaves over: $20 - 12 = 8 V$ .

WATCH OUT FOR

Treating the node voltage $V_{A B}$ as continuous (keeping $12 V$ at $0^{+}$ ) gives $V_{L} = 12 - 12 = 0$ . Only $v_{C}$ and $i_{L}$ are continuous; $V_{A B}$ jumps to $20 V$ because the capacitor branch now takes $4 A$ through $R_{2}$ . Other slips: forgetting the $R_{2}$ drop, sending current through the $R_{2}$ – $C$ branch in the $t < 0$ DC analysis (the capacitor is open), or letting $i_{L}$ jump straight to $10 A$ .

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

GATE 2023 EE Q.29: solution and answer

Answer and solution

More Transients questions