The given data are \(R=6\ \mathrm{\Omega}\), L=0.4H, capacitance is variable.

Supply voltage is 100V

Supply frequency is 50 Hz

Angular frequency \(\omega=2\pi f=2\pi\times50=100\pi\ rad/sec\)

At resonance

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

The capacitance at resonance \(C=\frac{1}{\left(100\pi\right)^2\times0.4}=\frac{100\times{10}^{-6}}{\pi^2\times0.4}=25.33\ \mu F\)

The current at resonance \(I=\frac{100}{6}=16.67\ Amp\).

The voltage drop across the capacitor during the resonance is

\(X_c\times I=\frac{1}{\omega C}\times I=\frac{{10}^6}{100\pi\times25.33}\times16.67=2094.8\ V\)

The Q factor of coil is \(\frac{\omega L}{R}=\frac{100\pi\times0.4}{6}=20.9\)