GATE 2023 IN Q.65: solution and answer

Official answer: 99.0 to 101.0Source: Official answer key, GATE 2023 IN (IIT Kanpur)

1. Read the source. $V_s=10\sqrt{2}\sin (20000\pi t)\text{V}$ has peak value $10\sqrt{2}\text{V}$, so $V_{rms}=\frac{10\sqrt{2}}{\sqrt{2}}=10\text{V}$ and $\omega =20000\pi \text{rad/s}$. Taking the source as the reference, the rms phasor is $V_s=10\angle 0^{\circ }\text{V}$.

2. Inductive reactance, with $L=\frac{1}{20\pi }\text{mH}=\frac{1}{20\pi }\times 10^{-3}\text{H}$: $$X_L=\omega L=20000\pi \times \frac{1}{20\pi }\times 10^{-3}=1\Omega$$

3. Capacitive reactance, with $C=\frac{1}{20\pi }\text{mF}=\frac{1}{20\pi }\times 10^{-3}\text{F}$: $\omega C=20000\pi \times \frac{1}{20\pi }\times 10^{-3}=1\text{S}$, so $$X_C=\frac{1}{\omega C}=1\Omega$$

4. Because $X_L=X_C$, the series loop is at resonance. The $+j1\Omega$ of the inductor and the $-j1\Omega$ of the capacitor cancel, so the loop impedance is purely resistive.

5. The circuit is a single series loop, so the source's 50 Ω internal resistance is in series with the drawn 25 Ω resistor and $R=25\Omega$: $$Z=50+25+25+j(1-1)=100\Omega$$

6. The same current flows through every element of the loop, including R: $$I=\frac{V_s}{Z}=\frac{10\angle 0^{\circ }}{100}=0.1\angle 0^{\circ }\text{A (rms)}$$ So the rms current through R is $0.1\text{A}=100.0\text{mA}$.