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

Answer 1

The correct option is A $50mA$

Given : $C=10\mu F={10}^{-5}F$

The source voltage is given as : $V=50\sqrt{2}\sin 100t$

On comparing it with $V=V_o\sin wt$,

We get: $w=100$ and $V_o=50\sqrt{2}$ volts

$\therefore$ Rms voltage, $V_{rms}=\frac{V_o}{\sqrt{2}}=50$ volts

Thus reading of ammeter, $I_{rms}=\frac{V_{rms}}{X_c}=wC{V}_{rms}$ $(\because X_c=\frac{1}{wC})$

$\therefore$ $I=100\times {10}^{-5}\times 50=5\times {10}^{-2}A$ $=50$ $mA$