Fig. 16

The total impedance of the circuit is \(R_{eq}=20+R+j\omega L-\frac{1}{\omega C}\)

The maximum current is following through the circuit at the time of resonance. For the resonance \(X_L\) and \(X_C\) is same.

Therefore \(\omega L=\frac{1}{\omega C}\),

At the resonance the total current passing through the circuit \(I=\frac{V}{20+R}\)

Or \(0.5=\frac{25}{20+R}\) , \(R=30\ \mathrm{\Omega}\)

The voltage drop across the capacitor \(V_C=I\times X_C=I/\omega C\)

Or \(C=\frac{0.5}{150\times2\times\pi\times400}=1.33\ \mu F\)

\(\omega^2=LC\)

\(L=\frac{\omega^2}{C}=\frac{1\times1.33}{\left(2\pi\times400\right)^2\times{10}^{-6}}=210\ mH\)