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A coil having an inductance of $\frac{1}{\pi} H$ is connected in series with a resistance of $300 \Omega$. If A.C. source $(20 \mathrm{~V}-200 \mathrm{~Hz})$ is connected across the combination, the phase angle between voltage and current is
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The correct answer is:
$\tan ^{-1}\left(\frac{4}{3}\right)$
For LR circuit:
Phase angle $\tan \phi=\frac{\omega L}{R}=\frac{2 \pi \times 200}{300} \times \frac{1}{\pi}=\frac{4}{3}$
$\therefore \phi=\tan ^{-1} \frac{4}{3}$
Phase angle $\tan \phi=\frac{\omega L}{R}=\frac{2 \pi \times 200}{300} \times \frac{1}{\pi}=\frac{4}{3}$
$\therefore \phi=\tan ^{-1} \frac{4}{3}$
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