An inductor of inductance 1000 mH is connected in series with a resistance, a variable capacitance and an AC source of frequency 2.0 kHz. What value of the capacitance result in drawing the maximum current from the circuit?

$63nF$

The correct answer is option (2) : $63nF$

At resonance current is maximum

$⇒omega L =\frac{1}{\omega C}⇒C=\frac{1}{\omega^2L}⇒C=\frac{1}{(2\pi f)^2\times L}$

$⇒C=\frac{1}{(2\times \pi \times 2 \times 10^3}=63nF$