When an ac source of e.m.f. e=E_0 (100 t) is connected across a c… - Vidyadip

MCQ

When an ac source of e.m.f. $e=E_0 \sin (100 t)$ 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 $\pi / 4$, as shown in the diagram. If the circuit consists possibly only of $R C$ or $L C$ in series, find the relationship between the two elements

✓
(a) $R=1 k \Omega, C=10 \mu F$
B
(b) $R=1 k \Omega, C=1 \mu F$
C
(c) $R=1 k \Omega, L=10 H$
D
(d) $\quad R=1 k \Omega, L=1 H$

✓

Answer

Correct option: A.

(a) $R=1 k \Omega, C=10 \mu F$

(a) As the current $i$ leads the voltage by $\frac{\pi}{4}$, it is an $R C$ circuit, hence$\begin{aligned}& \tan \phi=\frac{X_C}{R} \Rightarrow \tan \frac{\pi}{4}=\frac{1}{\omega C R} \\& \Rightarrow \omega C R=1 \text { as } \omega=100 \mathrm{rad} / \mathrm{sec}\end{aligned}$$\Rightarrow C R=\frac{1}{100} \sec ^{-1} \text {. }$From all the given options only option (a) is correct.

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