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GATE 2017 EE – Question 51

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 · Multiple choice

The switch in the figure below was closed for a long time. It is opened at $t = 0$. The current in the inductor of 2 H for $t \geq 0$, is

A 50 V source feeds a $6\ \Omega$ resistor and the switch. After the switch there is a node with a vertical $8\ \Omega$ resistor to ground, a $32\ \Omega$ resistor connecting to ground diagonally, an $8\ \Omega$ resistor along the top to the right-hand node, a $32\ \Omega$ resistor from the right-hand node to ground, and the 2 H inductor from the right-hand node to ground.
  1. $2.5e^{-4t}$
  2. $5e^{-4t}$
  3. $2.5e^{-0.25t}$
  4. $5e^{-0.25t}$

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Correct answer: (A) $2.5e^{-4t}$

Explanation

With the switch closed for a long time the inductor is a short circuit. The source sees $6 + (8 \parallel 8) = 10\ \Omega$, so the current is 5 A and the node voltage is 20 V. The top $8\ \Omega$ resistor carries $\frac{20}{8} = 2.5$ A into the inductor, so $i(0) = 2.5$ A. After the switch opens, the inductor sees $32 \parallel 32 \parallel (8 + 8) = 8\ \Omega$. The time constant is $\frac{L}{R} = \frac{2}{8} = 0.25$ s, so $i(t) = 2.5e^{-4t}$ A.