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GATE 2016 EC (EC3) – Question 42

Networks, Signals and Systems · Time and Frequency Domain Analysis of Linear Circuits · 2 marks · Numerical answer

The circuit has reached steady state before t=0, when the 3 Ω resistor burns out and becomes open. The current i(t), in amperes, at $t=0^+$ is ________.

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Correct answer: 0.98 to 1.02 OR -1.02 to -0.98 OR 1.96 to 2.04 OR -2.04 to -1.96

Explanation

At the initial DC steady state the capacitors are open. The 1 Ω, upper 2 Ω and lower 3 Ω resistors carry $12/(1+2+3)=2$ A. With the bottom node at zero, the top node is 10 V and the bridge midpoint is 6 V.

Capacitor-voltage continuity holds those top and midpoint voltages at 0+. The right node now connects through equal 2 Ω resistors to 10 V and 6 V, so it is at 8 V. Current from the left midpoint to the right is $(6-8)/2=-1$ A, or +1 A in the opposite direction.

**Official grading:** the source does not show a clear current arrow and accepts intervals around ±1 and ±2 A. All four original accepted intervals are retained; this derivation gives the ±1 pair rather than asserting the extra pair follows from the drawn network.

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