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A coil has resistance 30 ohm and inductive reactance 20 ohm at $50 \mathrm{~Hz}$ frequency. If an ac source, of 200 volt, $100 \mathrm{~Hz},$ is connected across the coil, the current in the coil will be
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The correct answer is:
$4.0 \mathrm{~A}$
If $\omega=50 \times 2 \pi$ then $\omega \mathrm{L}=20 \Omega$ If $\omega^{\prime}=100 \times 2 \pi$ then $\omega^{\prime} \mathrm{L}=40 \Omega$
Current flowing in the coil is
$$
\mathrm{I}=\frac{200}{\mathrm{Z}}=\frac{200}{\sqrt{\mathrm{R}^{2}+\left(\omega^{\prime} \mathrm{L}\right)^{2}}}=4 \mathrm{~A}
$$
Current flowing in the coil is
$$
\mathrm{I}=\frac{200}{\mathrm{Z}}=\frac{200}{\sqrt{\mathrm{R}^{2}+\left(\omega^{\prime} \mathrm{L}\right)^{2}}}=4 \mathrm{~A}
$$
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