GATE 2026 EC Q.31: solution and answer

Official answer: 71.50 to 72.50Source: Official answer key, GATE 2026 EC (IIT Guwahati)

1. Topology check. The ideal source, the $1\mu \text{F}$ capacitor, the $2.2\text{mH}$ inductor and the $1\text{k}\Omega$ resistor all connect between the same two nodes, so all four are in parallel. The voltage across the resistor is therefore the source voltage, $v_R(t)=v_i(t)=12\sin (\omega t)\text{V}$, at every frequency.

2. Resonant frequency, for reference. ${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{2.2\times 10^{-3}\times 10^{-6}}}\approx 2.132\times 10^4\text{rad/s}$, about $3.39\text{kHz}$. At ${\omega }_0$ the currents $I_L$ and $I_C$ are equal and opposite, so the parallel LC pair draws no net current. That changes the current the source supplies. It does not change the voltage across R, because an ideal voltage source fixes that voltage.

3. Peak and rms. In $12\sin (\omega t)$ the amplitude is a peak value: $V_m=12\text{V}$, so $V_{\text{rms}}=\frac{12}{\sqrt{2}}\text{V}$.

4. Average power in the resistor. $$P=\frac{V_{\text{rms}}^2}{R}=\frac{V_m^2}{2R}=\frac{12^2}{2\times 1000}=\frac{144}{2000}\text{W}$$

5. $P=0.072\text{W}=72.00\text{mW}$.