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When an ac source of $e.m.f.\ e = E_0 sin(100t)$ is connected across a circuit, the phase difference between the $e.m.f.\ e$ and the current $i$ in the circuit is observed to be $\frac{\pi }{4}$ as shown in the diagram. If the circuit consists possibly only of $ RC$ or $ RL $ or $ LC$ in series, find the relationship between the two elements
$R = 1K\Omega ,$ $C = 10\mu F$
$R = 1K\Omega ,$ $C = 1\mu F$
$R = 1K\Omega ,$ $L = 10H$
$R = 1K\Omega ,$ $L = 1H$
Solution
Fig. shows that current $i$ leads the emf $e$ by a phase angle $\pi / 4 .$ Therefore, the circuit can be $R-C$ circuit alone.
$\tan \phi=\frac{X_{C}}{R}=\frac{1}{\omega C R}=\tan \frac{\pi}{4}=1$ or $C R=\frac{1}{\omega}$
From $e=E_{0} \sin 100 t, \omega=100 \mathrm{rad} / \mathrm{s}$
$\therefore C R=\frac{1}{\omega}=\frac{1}{100}$
when $R=1 K \Omega=10^{3} \Omega$
$C=\frac{1}{\omega R}=\frac{1}{10^{5}}=10^{-5} F=10 \mu F$
