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GATE 2016 EE – Question 49

Electric Circuits · Transient response of DC and AC networks, sinusoidal steady-state analysis, resonance, two port networks, balanced three phase circuits, star-delta transformation, complex power and power factor in AC circuits · 2 marks · Numerical answer

In the circuit shown, switch $S_2$ has been closed for a long time. At time $t = 0$ switch $S_1$ is closed. At $t = 0^+$, the rate of change of current through the inductor, in amperes per second, is ________.

A 3 V source feeds switch $S_1$ and a 1 Ω resistor to a node. From that node, a 1 H inductor goes to ground. Switch $S_2$ connects the node to a 2 Ω resistor in series with a 3 V source, also to ground.

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Show answer and explanation

Correct answer: 1.9 to 2.1

Explanation

Before $S_1$ closes the inductor is steady, acting as a short, so it carries $\frac{3}{2} = 1.5$ A from the right-hand 3 V source through the 2 Ω resistor. The inductor current cannot change suddenly, so at $t = 0^+$ it is still 1.5 A. With the node voltage $v$, KCL gives $\frac{3 - v}{1} + \frac{3 - v}{2} = 1.5$, so $(3 - v)(1.5) = 1.5$ and $v = 2$ V. The inductor voltage is 2 V, so $\frac{di}{dt} = \frac{v}{L} = \frac{2}{1} = 2$ A/s.