Question
In the given $AC$ circuit, when switch $S$ is at $position \,1$, the source $emf$ leads current by $\pi /6$. Now, if the switch is at $position \,2$, then
$\tan \frac{\pi}{6}=\frac{\omega \mathrm{L}}{\mathrm{R}}=\frac{(1000)\left(\sqrt{3} \times 10^{-3}\right)}{\mathrm{R}}$ or $\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{\mathrm{R}}$
$\therefore $ $\mathrm{R}=3 \Omega$
In Position - $2$
$\tan \phi =\frac{X_{\mathrm{C}}}{\mathrm{R}}=\frac{(1 / \omega \mathrm{C})}{\mathrm{R}}$
$=\frac{1 /\left(1000 \times \frac{1000}{3} \times 10^{-6}\right)}{3}=1 $
$\therefore $ $\phi =\frac{\pi}{4}$
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