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Q. Find the peak current and resonant frequency of the following circuit (as shown in figure).
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JEE MainJEE Main 2021Alternating Current

Solution:

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as given $z=\sqrt{\left(x_{L}-x_{C}\right)^{2}+R^{2}}$
$x _{ L }=\omega_{ L }=100 \times 100 \times 10^{-3}=10 \Omega$
$x _{ C }=\frac{1}{\omega_{ C }}=\frac{1}{100 \times 100 \times 10^{-6}}=10 \Omega$
$z =\sqrt{(10-100)^{2}+ R ^{2}}=\sqrt{90^{2}+120^{2}}$
$=30 \times 5=150 \Omega$
$i _{ peak }=\frac{\Delta v }{ z }=\frac{30}{150}=\frac{1}{5} amp =0.2 amp$
& For resonant frequency
$\Rightarrow \omega L =\frac{1}{\omega C } \Rightarrow \omega^{2}=\frac{1}{ LC } \Rightarrow \omega=\frac{1}{\sqrt{ LC }}$
$\& f =\frac{1}{2 \pi \sqrt{ LC }} \Rightarrow \frac{1}{2 \pi \sqrt{100 \times 10^{-3} \times 100 \times 10^{-6}}}$
$=\frac{100 \sqrt{10}}{2 \pi}=\frac{100 \pi}{2 \pi}=50 Hz$
as $\sqrt{10} \approx \pi$