JEE Main 2022 — Alternating Current Question with Solution

From the diagram given,

Angular frequency $\omega =100 \mathrm{rad} {\mathrm{s}}^{-1}$ and $V_{rms}=\frac{V_0}{\sqrt{2}}=\frac{20\sqrt{2}}{\sqrt{2}}=20 \mathrm{V}$.

Now, $X_L=\omega L=100\times 100\times {10}^{-3}=10 \mathrm{\Omega }$ and $X_C=\frac{1}{\omega C}=\frac{1}{100\times 100\times {10}^{-6}}=100 \mathrm{\Omega }$

$\left|\left(X_L-X_C\right)\right|=\left|10-100\right|=90 \mathrm{\Omega }$

The impedance of the circuit,

$Z=\sqrt{{\left(X_L-X_C\right)}^2+R^2}=\sqrt{{\left(90\right)}^2+{\left(120\right)}^2}=150 \mathrm{\Omega }$

The current in the circuit,

$i_{\mathrm{mrs}}=\frac{V_{\mathrm{rms}}}{Z}=\left(\frac{20}{150}\right) \mathrm{A}$ 

Power factor $\cos \varphi =\frac{R}{Z}$

Now

$$\begin{gathered} \left(V_{rms}{i}_{rms}\cos \varphi \right)\Delta t=ms\left(\Delta T\right) \\ \Rightarrow \Delta t=\frac{ms\left(\Delta T\right)}{V_{rms}\times {\displaystyle \frac{V_{rms}}{Z}}\times {\displaystyle \frac{R}{Z}}}=\frac{ms\left(\Delta T\right)Z^2}{{V_{rms}}^2\times R}=\frac{2\times 16\times 22500}{400\times 120}=15 \mathrm{s} \end{gathered}$$

$\Rightarrow \Delta t=15 \mathrm{s}$