GATE 2020 EC Q.16: solution and answer

Official answer: 2.80 to 2.85Source: Official answer key, GATE 2020 EC (IIT Delhi)

1. Write the source and the current as phasors at $\omega =5\text{rad/s}$ (rms values, amplitude $÷\sqrt{2}$): $V=\frac{200}{\sqrt{2}}\angle 0^{\circ }=141.42\angle 0^{\circ }\text{V}$ and $I=\frac{10}{\sqrt{2}}\angle -45^{\circ }=7.071\angle -45^{\circ }\text{A}$, because $\pi /4=45^{\circ }$.

2. $R$ and $L$ are in series with the source, and $i(t)$ leaves the source's + terminal. So the source voltage equals $I$ times the total impedance: $$Z=\frac{V}{I}=\frac{141.42\angle 0^{\circ }}{7.071\angle -45^{\circ }}=20\angle 45^{\circ }\Omega$$ (Peak phasors give the same ratio, $200/10=20$.) The current lags the voltage by $45^{\circ }$, as it should with an inductive load.

3. Convert $Z$ to rectangular form and match it to the series RL impedance $R+j\omega L$: $$R+j\omega L=20\cos 45^{\circ }+j20\sin 45^{\circ }=14.142+j14.142\Omega$$

4. Equate the imaginary parts: $\omega L=14.142\Omega$. The real parts also give $R=14.142\Omega$, but $R$ is not needed.

5. Solve for the inductance: $$L=\frac{14.142}{\omega }=\frac{14.142}{5}=2\sqrt{2}=2.828\text{H}\approx 2.83\text{H}$$