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  4. An alternating voltage E=200 √2 sin (100 t) V is connected to a 1 μ F capacitor through an AC ammeter. The reading of ammeter is

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Q. An alternating voltage $E=200 \sqrt{2} \sin (100\, t) V$ is connected to a $1\, \mu F$ capacitor through an $AC$ ammeter. The reading of ammeter is

ManipalManipal 2011Alternating Current

Solution:

Given $E=200 \sqrt{2} \sin (100\, t)$ ...(i)
Comparing Eq. (i) with
$E=E_{0} \sin \omega t$
Peak voltage $E_{0}=200 \sqrt{2}, \omega=100$
$C=1\, \mu F =1 \times 10^{-6} F$
Reading of ammeter, $i=\frac{E_{ rms }}{X_{C}}$
$=\frac{E_{0} \omega_{C}}{\sqrt{2}}=\frac{E_{ rms }}{1 / \omega C}=\frac{E_{0} \omega C}{\sqrt{2}}$
$=\frac{200 \sqrt{2} \times 100 \times 1 \times 10^{-6}}{\sqrt{2}}$
$=20 \times 10^{-3} A$
$=20\, mA$