Question
An alternating voltage $\text{E}=50\sqrt{2}\sin(100\text{t})$ V is connected to a $1\mu\text{F}$ capacitor through an ac ammeter. What will be the reading of the ammeter?
✓
Answer
- $5\text{mA}$
$\text{X}_\text{C}=\frac{1}{\text{C}\omega}=10000\Omega$
ammeater reading $=\text{I}_\text{rms}=\frac{\text{V}_\text{rms}}{\mid\text{jX}_\text{c}\mid}=\frac{50}{10000}=5\text{mA}$
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