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When a d.c voltage of $200 \mathrm{~V}$ is applied to a coil of self-inductance $\left(\frac{2 \sqrt{3}}{\pi}\right) \mathrm{H}$, a current of 1A flows through it. But by replacing d.c. source with a.c. source of $200 \mathrm{~V}$, the current in the coil is reduced to $0.5 \mathrm{~A}$. Then the frequency of a.c. supply is
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Verified Answer
The correct answer is:
$50 \mathrm{~Hz}$
When a d.c. voltage is applied
$$
\mathrm{R}=\frac{\mathrm{V}}{1}=\frac{200}{1}=200 \Omega
$$
When a.c. voltage is applied
$$
\begin{aligned}
& \mathrm{Z}=\frac{\mathrm{V}}{1}=\frac{200}{0.5}=400 \Omega \\
& \mathrm{Z}^2=\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2 \\
& (400)^2=(200)^2+\mathrm{X}_{\mathrm{L}}^2 \\
& \mathrm{X}_{\mathrm{L}}=200 \sqrt{3} \Omega \\
& 2 \pi \mathrm{fL}=2 \pi \mathrm{f} \times \frac{2 \sqrt{3}}{\pi}=200 \sqrt{3} \\
& \mathrm{f}=50 \mathrm{~Hz}
\end{aligned}
$$
$$
\mathrm{R}=\frac{\mathrm{V}}{1}=\frac{200}{1}=200 \Omega
$$
When a.c. voltage is applied
$$
\begin{aligned}
& \mathrm{Z}=\frac{\mathrm{V}}{1}=\frac{200}{0.5}=400 \Omega \\
& \mathrm{Z}^2=\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2 \\
& (400)^2=(200)^2+\mathrm{X}_{\mathrm{L}}^2 \\
& \mathrm{X}_{\mathrm{L}}=200 \sqrt{3} \Omega \\
& 2 \pi \mathrm{fL}=2 \pi \mathrm{f} \times \frac{2 \sqrt{3}}{\pi}=200 \sqrt{3} \\
& \mathrm{f}=50 \mathrm{~Hz}
\end{aligned}
$$
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