An alternating e.m.f. of \(0 \cdot 2 \mathrm{~V}\) is applied across an LCR series circuit having \(\mathrm{R}=4 \Omega, \mathrm{C}=\) \(80 \mu \mathrm{F}\) and \(\mathrm{L}=200 \mathrm{mH}\). At resonance the voltage drop across the inductor is

At resonance \(\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\)
\(\therefore\) Current \(=\frac{0.2}{R}=\frac{0.2}{4}=\frac{0.1}{2}=0.05 \mathrm{~A}\)
Now, \(\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \mathrm{f} \mathrm{L}\)
\(
\begin{aligned}
\mathrm{f} &=\frac{1}{2 \pi \sqrt{\mathrm{LC}}} \\
\therefore \quad \mathrm{X}_{\mathrm{L}} &=\frac{2 \pi \mathrm{L}}{2 \pi \sqrt{\mathrm{LC}}}=\sqrt{\frac{\mathrm{L}}{\mathrm{C}}} \\
\mathrm{X}_{\mathrm{L}} &=\sqrt{\frac{200 \times 10^{-3}}{80 \times 10^{-6}}}=\frac{10^{2}}{2}=50 \Omega \\
\therefore \quad \mathrm{V}_{\mathrm{L}} &=\mathrm{i} \times \mathrm{L}=0.05 \times 50=2.5 \mathrm{~V}
\end{aligned}
\)