GATE 2020 IN Q.33: solution and answer

Official answer: 1Source: Official answer key, GATE 2020 IN (IIT Delhi)

1. Source phasor: $v(t)=\sqrt{2}100\cos (1000t)\text{V}$ has a peak of $100\sqrt{2}\text{V}$, so its rms phasor is $\mathbf{V}=100\angle 0^{\circ }\text{V}$ (rms), with $\omega =1000\text{rad/s}$.

2. Inductive reactance: $X_L=\omega L=1000\times 100\times 10^{-3}=100\Omega$.

3. Capacitive reactance: $X_C=\frac{1}{\omega C}=\frac{1}{1000\times 10\times 10^{-6}}=100\Omega$.

4. Series impedance: $Z=R+j(X_L-X_C)=100+j(100-100)=100\Omega$. Because $X_L=X_C$, the circuit is at series resonance: ${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.1\times 10^{-5}}}=1000\text{rad/s}$, which is the source frequency.

5. Ohm's law in phasor form: $\mathbf{I}=\frac{\mathbf{V}}{Z}=\frac{100\angle 0^{\circ }}{100}=1\angle 0^{\circ }\text{A}$ (rms). The arrow's direction can only flip the sign, which shifts the phase by $180^{\circ }$. It cannot change the magnitude.

6. rms value of $\mathbf{I}$: $|\mathbf{I}|=1\text{A}$.