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Physics
Find the peak current and resonant frequency of the following circuit (as shown in figure). <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/3233934ef7c8294d36357f09523bda18-.png />
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Q. Find the peak current and resonant frequency of the following circuit (as shown in figure).
JEE Main
JEE Main 2021
Alternating Current
A
$0.2 A$ and $50 Hz$
48%
B
$0.2 A$ and $100 Hz$
41%
C
$2 A$ and $100 Hz$
5%
D
$2 A$ and $50 Hz$
6%
Solution:
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 $\sqrt{10} \approx \pi$