GATE 2023 IN Q.47: solution and answer

GATE 2023 · INQ.47 · 2 marks · NAT

The R-L circuit with $R = 10 k Ω$ and $L = 1 mH$ is excited by a step current $I_{0} u (t)$ . At $t = 0^{-}$ , there is a current $I_{L} = I_{0} /5$ flowing through the inductor. The minimum time taken for the current through the inductor to reach 99% of its final value is ______ $μ s$ (rounded off to two decimal places).

GATE 2023 IN question 47: Three elements connected in parallel between a top rail and a bottom rail. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

ANSWER

The inductor current reaches 99% of its final value $I_{0}$ after $t = τ ln 80 = 0.1 μ s \times 4.382 \approx 0.44 μ s$ .

Official answer: 0.43 to 0.45Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Time constant. With the current source open-circuited, the inductor sees only $R$ , so $τ = \frac{L}{R} = \frac{1 \times 1 0 ^{- 3}}{10 \times 1 0 ^{3}} = 1 0^{- 7} s = 0.1 μ s$
Initial condition. Inductor current cannot change instantaneously, so $i_{L} (0^{+}) = i_{L} (0^{-}) = \frac{I _{0}}{5}$ .
Final value. For $t > 0$ the source delivers $I_{0}$ . In steady state the inductor behaves as a short circuit, so it carries all of the source current and none flows in $R$ : $i_{L} (\infty) = I_{0}$ .
First-order response. $i_{L} (t) = i_{L} (\infty) + [i_{L} (0^{+}) - i_{L} (\infty)] e^{- t / τ} = I_{0} - \frac{4 I _{0}}{5} e^{- t / τ}$ for $t \geq 0$ .
Condition for 99% of the final value. $I_{0} - \frac{4 I _{0}}{5} e^{- t / τ} = 0.99 I_{0} \Rightarrow \frac{4}{5} e^{- t / τ} = 0.01 \Rightarrow e^{- t / τ} = \frac{1}{80}$ .
Solve for t. $t = τ ln 80 = 0.1 μ s \times 4.382 = 0.4382 μ s \approx 0.44 μ s$ The current rises monotonically from $I_{0} /5$ towards $I_{0}$ , so this is the first (minimum) time it reaches 99% of its final value.

KEY IDEA

This is a first-order RL circuit, so $i_{L} (t) = i_{L} (\infty) + [i_{L} (0^{+}) - i_{L} (\infty)] e^{- t / τ}$ with $τ = L / R$ , where $R$ is the resistance the inductor sees with the current source open-circuited. The non-zero initial current matters: the gap still to close is only $\frac{4}{5} I_{0}$ . The 99% point is where that gap has shrunk to $0.01 I_{0}$ , so $e^{- t / τ} = \frac{0.01}{0.8} = \frac{1}{80}$ .

WATCH OUT FOR

Ignoring the initial current and using the zero-state rule $t = τ ln 100 = 0.46 μ s$ , which falls outside the key. A second mistake is shorting the current source when finding $τ$ (that would give $τ = 0$ ). An ideal current source is replaced by an open circuit, so the inductor sees $R$ alone.

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

GATE 2023 IN Q.47: solution and answer

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