Physics - EMI/AC Question with Solution | TestHub

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

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

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

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