An inductor of 500 mH is in series with a resistance and a variable capacitor connected to a source of frequency 0.4 kHz. The value of capacitance of the capacitor to get a maximum current will be

0.32 μF

The correct answer is Option (2) → 0.32 μF

$\text{Given: } L = 500\,\text{mH} = 0.5\,\text{H}, \, f = 0.4\,\text{kHz} = 400\,\text{Hz}$

$\text{At resonance, } \omega L = \frac{1}{\omega C}$

$\Rightarrow C = \frac{1}{\omega^2 L}$

$\omega = 2\pi f = 2\pi \times 400 = 2513.27\,\text{rad/s}$

$C = \frac{1}{(2513.27)^2 \times 0.5}$

$C = \frac{1}{3.16 \times 10^6}$

$C = 3.17 \times 10^{-7}\,\text{F} = 0.317\,\mu\text{F}$

$\text{Answer: } C = 0.317\,\mu\text{F}$