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A coil has a resistance of $30 \Omega$ and an inductive reactance of $20 \Omega$ at $50 \mathrm{~Hz}$ frequency. If an AC source of $200 \mathrm{~V}$, $100 \mathrm{~Hz}$ is connected across the coil, the current in the coil is
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$4 \mathrm{~A}$
Inductive reactance, $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L} \therefore 20=2 \pi$ (50) L...(1)
When AC source of $200 \mathrm{~V}, 100 \mathrm{~Hz}$ is connected across the coil then
$\begin{aligned} & \mathrm{X}_{\mathrm{L}}=2 \pi(\mathrm{f}) \mathrm{L}=2 \pi(100) \mathrm{L}=2 \pi(50) \mathrm{L} \times 2 \\ & \therefore \mathrm{X}_{\mathrm{L}}^{\prime}=20 \times 2=40 \Omega\end{aligned}$
$\therefore$ Impedance,
$\mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}\right)^2}=\sqrt{(30)^2+(40)^2}=50 \Omega$
$\therefore$ Current in the coil, $I=\frac{V}{Z}=\frac{200}{50}=4 A$
When AC source of $200 \mathrm{~V}, 100 \mathrm{~Hz}$ is connected across the coil then
$\begin{aligned} & \mathrm{X}_{\mathrm{L}}=2 \pi(\mathrm{f}) \mathrm{L}=2 \pi(100) \mathrm{L}=2 \pi(50) \mathrm{L} \times 2 \\ & \therefore \mathrm{X}_{\mathrm{L}}^{\prime}=20 \times 2=40 \Omega\end{aligned}$
$\therefore$ Impedance,
$\mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}\right)^2}=\sqrt{(30)^2+(40)^2}=50 \Omega$
$\therefore$ Current in the coil, $I=\frac{V}{Z}=\frac{200}{50}=4 A$
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