Question

A capacitor of capacitance $10\mu F$ is connected to an $AC$ source and an $AC$ Ammeter. If the source voltage varies as $V=50\sqrt{2}\sin 100t$, the reading of the ammeter is

• A. 50 mA
• B. 70.7 mA
• C. 5.0 mA
• D. 7.07 mA

Answer 1

Answer: (A)

$C=10\mu F=10\times 10^{-6}F$
$V=50\sqrt{2}\sin 100t$
Current, $I=\frac{V}{R},\omega =100$
Current $=\frac{V_{rms}}{X_C}$
$X_C=$ Impedance of capacitor
$X_C=\frac{1}{\omega C}$
$I=\frac{V_{rms}}{1/\omega C}$
$V_{rma}=\frac{V}{\sqrt{2}}=\frac{50\sqrt{2}}{\sqrt{2}}=50V$
Reading of ammeter
$=50\times 100\times 10\times 10^{-6}A$
$=50\times 10^{-3}A\Rightarrow 50mA$