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

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Question

(A) From the given diagram,the current $i$ leads the voltage $e$ by a phase angle $\phi = \pi/4$. This indicates that the circuit is an $RC$ series ci

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) From the given diagram,the current $i$ leads the voltage $e$ by a phase angle $\phi = \pi/4$. This indicates that the circuit is an $RC$ series circuit.For an $RC$ circuit,the phase angle $\phi$ is given by $	an \phi = \frac{X_C}{R} = \frac{1}{\omega CR}$.Given $\phi = \pi/4$,we have $	an(\pi/4) = 1$,so $1 = \frac{1}{\omega CR}$,which implies $\omega CR = 1$.Given $\omega = 100 	ext{ rad/s}$,we have $100 	imes C 	imes R = 1$,or $CR = 1/100 = 0.01 	ext{ s}$.Checking option $(A)$: $R = 1000 \, \Omega$ and $C = 10 	imes 10^{-6} 	ext{ F} = 10^{-5} 	ext{ F}$.Then $CR = 1000 	imes 10^{-5} = 10^{-2} = 0.01 	ext{ s}$.This matches the condition,so option $(A)$ is correct.

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