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A coil has resistance $30 \Omega$ and inductive reactance $20 \Omega$ at $50 \mathrm{~Hz}$ frequency. If an $\mathrm{AC}$ source of $200 \mathrm{~V}, 100 \mathrm{~Hz}$, is connected across the coil, the current in the coil will be
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Verified Answer
The correct answer is:
$4.0 \mathrm{~A}$
If $\omega=50 \times 2 \pi$ then $\omega L=20 \Omega$
Similarly $\omega^{\prime}=100 \times 2 \pi$ then $\omega^{\prime} L=40 \Omega$
$$
\begin{aligned}
i & =\frac{200}{Z}=\frac{200}{\sqrt{R^2+\left(\omega^{\prime} L\right)^2}} \\
& =\frac{200}{\sqrt{(30)^2+(40)^2}}=4 \mathrm{~A}
\end{aligned}
$$
Similarly $\omega^{\prime}=100 \times 2 \pi$ then $\omega^{\prime} L=40 \Omega$
$$
\begin{aligned}
i & =\frac{200}{Z}=\frac{200}{\sqrt{R^2+\left(\omega^{\prime} L\right)^2}} \\
& =\frac{200}{\sqrt{(30)^2+(40)^2}}=4 \mathrm{~A}
\end{aligned}
$$
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