In the given circuit $L = \frac{20}{\pi} H, R = 2kΩ$ and $C =\frac{5}{\pi}μF$ are in series across a 220 V, 50 Hz ac source.

The peak voltage drop between A and B is:

440 V

The correct answer is Option (1) → 440 V

Given: $L=\frac{20}{\pi}\,\mathrm{H}$, $R=2\times10^3\,\Omega$, $C=\frac{5}{\pi}\,\mu\mathrm{F}$, $f=50\,\mathrm{Hz}$, $V_{\text{rms}}=220\,\mathrm{V}$.

$\omega=2\pi f=100\pi\ \mathrm{rad/s}$.

$X_L=\omega L=100\pi\cdot\frac{20}{\pi}=2000\ \Omega,\quad X_C=\frac{1}{\omega C}=\frac{1}{100\pi\cdot\frac{5}{\pi}\times10^{-6}}=2000\ \Omega$.

Impedance magnitude: $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{(2000)^2+0^2}=2000\ \Omega$.

$I_{\text{rms}}=\frac{V_{\text{rms}}}{Z}=\frac{220}{2000}=0.11\ \mathrm{A}$.

$V_{R,\text{rms}}=I_{\text{rms}}R=0.11\times2000=220\ \mathrm{V}$.

$V_{C,\text{rms}}=I_{\text{rms}}X_C=220\ \mathrm{V}$ (lags by $90^\circ$).

$V_{AB,\text{rms}}=\sqrt{V_{R,\text{rms}}^2+V_{C,\text{rms}}^2}=220\sqrt{2}\ \mathrm{V}$.

Peak value: $V_{AB,\text{peak}}=\sqrt{2}\,V_{AB,\text{rms}}=\sqrt{2}\cdot220\sqrt{2}=440\ \mathrm{V}$.

Answer: $440\ \mathrm{V}$