Question

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

• A. 10 mA
• B. 20 mA
• C. 40 mA
• D. 80 mA

Answer 1

Answer: (B)

Given $E=200\sqrt{2}\sin (100t)$ ...(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, $i=\frac{E_{rms}}{X_C}$
$=\frac{E_0{\omega }_C}{\sqrt{2}}=\frac{E_{rms}}{1/\omega C}=\frac{E_0\omega C}{\sqrt{2}}$
$=\frac{200\sqrt{2}\times 100\times 1\times 10^{-6}}{\sqrt{2}}$
$=20\times 10^{-3}A$
$=20mA$