A circuit connected to an ac source of emf e | vedclass

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Question

(A) The phase difference $\phi$ in an $RL$ or $RC$ circuit is given by $	an \phi = \frac{X}{R}$.Given $\phi = \frac{\pi}{4}$,we have $	an(\frac{\pi}

Vedclass · Accepted Answer

$RC$ circuit with $R = 1 	ext{ k}\Omega$ and $C = 1 \mu	ext{F}$

Vedclass · Answer

(A) The phase difference $\phi$ in an $RL$ or $RC$ circuit is given by $	an \phi = \frac{X}{R}$.Given $\phi = \frac{\pi}{4}$,we have $	an(\frac{\pi}{4}) = 1$,which implies $X = R$.Here,$\omega = 1000 	ext{ rad/s}$ and $R = 1000 \Omega$.For an $RC$ circuit,$X_C = \frac{1}{\omega C} = R \implies C = \frac{1}{\omega R} = \frac{1}{1000 	imes 1000} = 10^{-6} 	ext{ F} = 1 \mu	ext{F}$.For an $RL$ circuit,$X_L = \omega L = R \implies L = \frac{R}{\omega} = \frac{1000}{1000} = 1 	ext{ H}$.Checking the options:Option $(A)$: $R = 1000 \Omega$,$C = 1 \mu	ext{F}$. $X_C = \frac{1}{1000 	imes 10^{-6}} = 1000 \Omega$. Since $X_C = R$,the phase difference is $\frac{\pi}{4}$.Thus,option $(A)$ is correct.

$A$ circuit connected to an $ac$ source of $emf$ $e = e_0 \sin(1000t)$,where $t$ is in seconds,exhibits a phase difference of $\frac{\pi}{4}$ between the $emf$ $e$ and current $i$. Which of the following circuits will exhibit this?

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