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When an $A C$ source of emf $E$ with frequency $\omega=100 \mathrm{~Hz}$ is connected across a circuit, the phase difference between
$E$ and current I in the circuit is observed to be $\frac{\pi}{4}$ as shown in the figure. If the circuit consist of only $\mathrm{RC}$ or $\mathrm{RL}$ in series, then
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$E$ and current I in the circuit is observed to be $\frac{\pi}{4}$ as shown in the figure. If the circuit consist of only $\mathrm{RC}$ or $\mathrm{RL}$ in series, then
Solution:
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Verified Answer
The correct answer is:
$\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=10 \mu \mathrm{F}$
.. Since supply voltage lags the current $\operatorname{By}(\phi=\pi / 4)$
$$
\begin{aligned}
&\tan \phi=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}=1, \mathrm{X}_{\mathrm{C}}=\mathrm{R} \ldots \ldots(1) \\
&\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 10 \times 10^{-6}}=1 \mathrm{k} \Omega=\mathrm{R}
\end{aligned}
$$
$$
\begin{aligned}
&\tan \phi=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}=1, \mathrm{X}_{\mathrm{C}}=\mathrm{R} \ldots \ldots(1) \\
&\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 10 \times 10^{-6}}=1 \mathrm{k} \Omega=\mathrm{R}
\end{aligned}
$$
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