Question

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

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

Answer 1

Answer: (C)

Given,
Alternating voltage
$V=200\sqrt{2}\sin \left(100t\right)$
$\therefore \omega =100$
$C=1\mu F=1\times 10^{-6}F$
Capacitive reactance,
$X_c=\frac{1}{\omega c}=\frac{1}{100\times 10^{-6}}=10^4\Omega$
$V_0=200\sqrt{2}$
$\therefore V_{rms}=\frac{200\sqrt{2}}{\sqrt{2}}$
$V_{rms}=200V$
Reading of ammeter,
$I=\frac{V_{rms}}{X_C}=\frac{200}{10^4}$
$=0.02A$
$=20mA$