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A series $\mathrm{L}-\mathrm{C}-\mathrm{R}$ circuit containing a resistance of $120 \Omega$ has angular frequency $4 \times 10^5 \mathrm{rad} \mathrm{s}^{-1}$. At resonance the voltage across resistance and inductor are $60 \mathrm{~V}$ and $40 \mathrm{~V}$ respectively, then the value of inductance will be
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Verified Answer
The correct answer is:
$0.2 \mathrm{mH}$
At resonance, the impedance of the circuit is equal to resistance
$$
\therefore \mathrm{I}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}}=\frac{60}{120}=0.5 \mathrm{~A}
$$
Inductive reactance, $X_L=\frac{V_L}{I}=\frac{40}{0.5}=80 \Omega$
$$
\begin{aligned}
& \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L} \\
& \therefore \mathrm{L}=\frac{\mathrm{X}_{\mathrm{L}}}{\omega}=\frac{80}{4 \times 10^5}=20 \times 10^{-5} \mathrm{H} \\
& =0.2 \mathrm{mH}
\end{aligned}
$$
$$
\therefore \mathrm{I}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}}=\frac{60}{120}=0.5 \mathrm{~A}
$$
Inductive reactance, $X_L=\frac{V_L}{I}=\frac{40}{0.5}=80 \Omega$
$$
\begin{aligned}
& \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L} \\
& \therefore \mathrm{L}=\frac{\mathrm{X}_{\mathrm{L}}}{\omega}=\frac{80}{4 \times 10^5}=20 \times 10^{-5} \mathrm{H} \\
& =0.2 \mathrm{mH}
\end{aligned}
$$
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