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A sinusoidal voltage of peak value $300 \mathrm{~V}$ and an angular frequency $\omega=400 \mathrm{rads}^{-1}$ is applied to series $L-C-R$ circuit, in which $R=3 \Omega$, $L=20 \mathrm{mH}$ and $C=625 \mu \mathrm{F}$. The peak current in the circuit is
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Verified Answer
The correct answer is:
$60 \mathrm{~A}$
Given, $V_{m}=300 \mathrm{~V}, \omega=400 \mathrm{rads}^{-1}$
$$
R=3 \Omega, L=20 \mathrm{mH}=20 \times 10^{-3} \mathrm{H}
$$
and $C=625 \mu \mathrm{F}=625 \times 10^{-6} \mathrm{~F}$
Impedance of the circuit,
$$
Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}
$$
Here,
$$
X_{L}=\omega L=400 \times 20 \times 10^{-3}=8 \Omega
$$
and
$$
X_{C}=\frac{1}{\omega C}=\frac{1}{400 \times 625 \times 10^{-6}}=4 \Omega
$$
$$
Z=\sqrt{3^{2}+(8-4)^{2}}=\sqrt{25}=5 \Omega
$$
$\therefore$ Peak current, $I_{m}=\frac{V_{m}}{Z}=\frac{300}{5}=60 \mathrm{~A}$
$$
R=3 \Omega, L=20 \mathrm{mH}=20 \times 10^{-3} \mathrm{H}
$$
and $C=625 \mu \mathrm{F}=625 \times 10^{-6} \mathrm{~F}$
Impedance of the circuit,
$$
Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}
$$
Here,
$$
X_{L}=\omega L=400 \times 20 \times 10^{-3}=8 \Omega
$$
and
$$
X_{C}=\frac{1}{\omega C}=\frac{1}{400 \times 625 \times 10^{-6}}=4 \Omega
$$
$$
Z=\sqrt{3^{2}+(8-4)^{2}}=\sqrt{25}=5 \Omega
$$
$\therefore$ Peak current, $I_{m}=\frac{V_{m}}{Z}=\frac{300}{5}=60 \mathrm{~A}$
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