In the circuit shown, the AC source has voltage \(\mathrm{V}=20 \cos (\omega \mathrm{t})\) volt with \(\omega=2000 \mathrm{rad} / \mathrm{s}\). The magnitude of amplitude current will be nearly

Current through the circuit is given by
\(i=\frac{V_0}{\sqrt{R^2+\left(X_C-X_L\right)^2}}\) where \(X_L=\omega L\) and \(X C=\frac{1}{\omega C}\).
Given: \(\mathrm{V}_0=20 \mathrm{~V},=6 \Omega, \mathrm{L}=5 \times 10^{-3} \mathrm{H}, \omega=2000 \frac{\mathrm{rad}}{\mathrm{s}}\) and \(\mathrm{C}=50 \times 10^{-6} \mathrm{~F}\)
\(\therefore \mathrm{X}_{\mathrm{C}}=\frac{1}{50 \times 10^{-6} \times 2000} \Omega 10 \Omega \text { and } \mathrm{X}_{\mathrm{L}}=\) \(2000 \times 5 \times 10^{-3} \Omega=10 \Omega\)
\(\because \mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}\)
The current is simply: \(\mathrm{I}=\frac{\mathrm{v}_0}{\mathrm{R}}\)
\(\therefore \mathrm{i}=\frac{20}{6} \mathrm{~A}=3.3 \mathrm{~A}\)