When an AC source of e.m.f. e = E_0 sin(100t | vedclass

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Question

(A) The diagram shows that the current $i$ leads the $e.m.f. \ e$ by a phase angle $\phi = \frac{\pi}{4}$.In an $AC$ circuit,the current leads the vol

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The diagram shows that the current $i$ leads the $e.m.f. \ e$ by a phase angle $\phi = \frac{\pi}{4}$.In an $AC$ circuit,the current leads the voltage only in an $RC$ circuit.For an $RC$ series circuit,the phase angle $\phi$ is given by $	an \phi = \frac{X_C}{R} = \frac{1}{\omega C R}$.Given $\phi = \frac{\pi}{4}$,we have $	an(\frac{\pi}{4}) = 1$,so $\frac{1}{\omega C R} = 1$,which implies $CR = \frac{1}{\omega}$.From the given equation $e = E_0 \sin(100t)$,the angular frequency is $\omega = 100 	ext{ rad/s}$.Thus,$CR = \frac{1}{100} = 0.01 	ext{ s}$.Given $R = 1 	ext{ k}\Omega = 10^3 \ \Omega$,we find $C = \frac{1}{\omega R} = \frac{1}{100 	imes 10^3} = \frac{1}{10^5} = 10^{-5} 	ext{ F} = 10 \mu	ext{F}$.Therefore,the correct values are $R = 1 	ext{ k}\Omega$ and $C = 10 \mu	ext{F}$.

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