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  • An AC source is connected with a resistance (R) and an uncharged capacitance C, in series. The potential difference across the resistor is in phase with the initial potential difference across the

An AC source is connected with a resistance (R) and an uncharged capacitance C, in series. The potential difference across the resistor is in phase with the initial potential difference across the capacitor for the first time at the instant (assume that at t = 0, emf is zero).

Option: 1

\frac{\pi}{4 \omega}


Option: 2

\frac{2 \pi}{\omega}


Option: 3

\frac{\pi}{2 \omega}


Option: 4

\frac{3 \pi}{2 \omega}


Answers (1)

best_answer

Let \quad V=V_0 \sin \omega t
[as \mathrm{V}=0 at \mathrm{t}=0 ]
Then \quad V_R=V_0 \sin \omega t and \quad \mathrm{V}_{\mathrm{C}}=\mathrm{V}_0 \sin (\omega \mathrm{t}-\pi / 2) \mathrm{V} and \mathrm{V}_{\mathrm{R}} are in same phase. While \mathrm{V}_{\mathrm{C}}\, \, lags \mathrm{V}\, (or \mathrm{V}_{\mathrm{R}} ) by \, 90^{\circ}.
Now V_R is in same phase with initial potential difference across the capacitor for the first time when,
\begin{aligned} & \quad \omega t=-\frac{\pi}{2}+2 \pi=\frac{3 \pi}{2} \\ & \therefore \quad t=\frac{3 \pi}{2 \omega} \end{aligned}

Posted by

seema garhwal

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