Physics - EMI/AC Question with Solution | TestHub

Given here, $f=2 \mathrm{kHz}=2000 \mathrm{Hz}, C=62.5 \mathrm{nF}=62.5\times {10}^{-9} \mathrm{F} \text{and} R=50 \mathrm{\Omega }$

Resonance frequency (amplitude of the current will be maximum in this case) is given by $f=\frac{1}{2\pi \sqrt{LC}}$, here, $L$ is the inductance.

Putting the values, 

$2000 \mathrm{Hz}=\frac{1}{2\pi \sqrt{L\times 62.5\times {10}^{-9}}}$

$L=\frac{1}{4{\pi }^2\times {2000}^2\times 62.5\times {10}^{-9}}=0.1 \mathrm{H}=100 \mathrm{mH}$