For the RLC circuit shown, v i(t)=2(200t)+4(500t) V. The output voltage is

GATE 2016 EC (EC3) – Question 19

Networks, Signals and Systems · Sinusoidal Steady State Analysis · 1 mark · Multiple choice

For the RLC circuit shown, $v_i(t)=2\cos(200t)+4\sin(500t)$ V. The output voltage is

Diagram for GATE 2016 EC (EC3) question 19

$\cos(200t)+2\sin(500t)$
$2\cos(200t)+4\sin(500t)$
$\sin(200t)+2\cos(500t)$
$2\sin(200t)+4\cos(500t)$

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Correct answer: (B) $2\cos(200t)+4\sin(500t)$

Explanation

At ω=200 rad/s, the 0.25 H and 100 μF series LC branch resonates: its impedance is zero. In parallel with 2 Ω it shorts the series path, so the 200 rad/s output equals the input component.

At ω=500 rad/s, the 0.4 H and 10 μF parallel LC branch resonates and becomes open. No current flows through the preceding series impedance, so the 500 rad/s output also equals its input component.

By superposition, the output is **$2\cos200t+4\sin500t$, B**.

GATE 2016 EC (EC3) – Question 19

Explanation

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