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In a series resonant R-L-C circuit, the voltage across $R$ is $100 \mathrm{~V}$ and the value of $\mathrm{R}=1000 \Omega$. The capacitance of the capacitor is $2 \times 10^{-6} \mathrm{~F}$; angular frequency of AC is $200 \mathrm{rad} \mathrm{s}^{-1}$. Then the potential difference across the inductance coil is
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Verified Answer
The correct answer is:
$250 \mathrm{~V}$
The current in the circuit
$$
\begin{aligned}
\mathrm{i} &=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}} \\
&=\frac{100}{1000}=0.1 \mathrm{~A}
\end{aligned}
$$
At resonance,
$$
\begin{aligned}
\mathrm{V}_{\mathrm{L}} &=\mathrm{V}_{\mathrm{C}}=\mathrm{iX}_{\mathrm{C}}=\frac{\mathrm{i}}{\omega \mathrm{C}} \\
&=\frac{0.1}{200 \times 2 \times 10^{-6}} \\
&=250 \mathrm{~V}
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{i} &=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}} \\
&=\frac{100}{1000}=0.1 \mathrm{~A}
\end{aligned}
$$
At resonance,
$$
\begin{aligned}
\mathrm{V}_{\mathrm{L}} &=\mathrm{V}_{\mathrm{C}}=\mathrm{iX}_{\mathrm{C}}=\frac{\mathrm{i}}{\omega \mathrm{C}} \\
&=\frac{0.1}{200 \times 2 \times 10^{-6}} \\
&=250 \mathrm{~V}
\end{aligned}
$$
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