Question

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

• A. $ 40 \, mA $
• B. $ 20 \sqrt{2}\,mA $
• C. $ 20\, mA $
• D. $ 10\sqrt{2}\,mA $

Answer 1

Answer: (C)

Given,
Alternating voltage
$V=200\sqrt{2}\sin\left(100 t\right)$
$\therefore \omega=100$
$C=1\,\mu F=1\times10^{-6}F$
Capacitive reactance,
$X_{c}=\frac{1}{\omega c}=\frac{1}{100\times10^{-6}}=10^{4} \Omega$
$V_{0}=200\sqrt{2}$
$\therefore V_{rms}=\frac{200\sqrt{2}}{\sqrt{2}}$
$V_{rms}=200\,V $
Reading of ammeter,
$I=\frac{V_{rms}}{X_{C}}=\frac{200}{10^{4}}$
$=0.02\,A $
$=20\,mA$