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An alternating voltage $\varepsilon=30 \sin 200 t$ (in volts) is applied to the circuit below. The amplitude of the current through the circuit is
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Verified Answer
The correct answer is:
$3 \mathrm{~A}$
Given, applied emf is
$E=30 \sin 200 t$
In given circuit,
$L=0.05 \mathrm{H}, C=500 \mu \mathrm{F}$
$R=10 \Omega, \omega=200 \mathrm{rad} \mathrm{s}^{-1}$
Now, inductive reactance, $X_L=L \omega=0.05 \times 200$ $=10 \Omega$
Also, capacitive reactance,
$X_C=\frac{1}{C \omega}=\frac{1}{500 \times 10^{-6} \times 200}=10 \Omega$
As $X_L=X_C$, circuit is in resonance, so amplitude of current is
$I_{\max }=\frac{E_{\max }}{R}=\frac{30}{10}=3 \mathrm{~A}$
$E=30 \sin 200 t$
In given circuit,
$L=0.05 \mathrm{H}, C=500 \mu \mathrm{F}$
$R=10 \Omega, \omega=200 \mathrm{rad} \mathrm{s}^{-1}$
Now, inductive reactance, $X_L=L \omega=0.05 \times 200$ $=10 \Omega$
Also, capacitive reactance,
$X_C=\frac{1}{C \omega}=\frac{1}{500 \times 10^{-6} \times 200}=10 \Omega$
As $X_L=X_C$, circuit is in resonance, so amplitude of current is
$I_{\max }=\frac{E_{\max }}{R}=\frac{30}{10}=3 \mathrm{~A}$
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