When an AC source of emf E with angular frequ | vedclass

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(D) From the figure, the current $I$ leads the emf $E$ by a phase angle $\phi = \frac{\pi}{4}$. This indicates that the circuit is a capacitive circui

Vedclass · Accepted Answer

$R=1 	ext{ k}\Omega, C=10 \mu	ext{F}$

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(D) From the figure, the current $I$ leads the emf $E$ by a phase angle $\phi = \frac{\pi}{4}$. This indicates that the circuit is a capacitive circuit ($RC$ series circuit).In an $RC$ series circuit, the phase angle $\phi$ is given by $	an \phi = \frac{X_C}{R}$.Given $\phi = \frac{\pi}{4}$, we have $	an(\frac{\pi}{4}) = 1$, so $X_C = R$.We know $X_C = \frac{1}{\omega C}$.Given $\omega = 100 	ext{ rad/s}$ and $R = 1 	ext{ k}\Omega = 1000 	ext{ }\Omega$.Substituting these values: $1000 = \frac{1}{100 	imes C}$.$C = \frac{1}{100 	imes 1000} = \frac{1}{10^5} = 10 	imes 10^{-6} 	ext{ F} = 10 \mu	ext{F}$.Thus, $R = 1 	ext{ k}\Omega$ and $C = 10 \mu	ext{F}$.

When an $AC$ source of emf $E$ with angular frequency $\omega = 100 \text{ rad/s}$ 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 consists of only $RC$ or $RL$ in series, then:

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