Question

An alternating voltage given as, \(E=100\sqrt{2}\sin (100t)\) volt is applied to a capacitor of \(1\ \mu \text{F}\). The current reading of the ammeter will be equal to

• A. \(20\ \text{mA}\)
• B. \(10\ \text{mA}\)
• C. \(40\ \text{mA}\)
• D. \(80\ \text{mA}\)

Hint:

Find the capacitive reactance, divide the rms voltage by it, and remember an AC ammeter shows the rms current.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In a purely capacitive circuit, $I_{rms}=E_{rms}\,\omega C$.

Step 2: Calculation
$I_{rms}=100\times 100\times 10^{-6}=10^{-2}\ \text{A}=10\ \text{mA}$.

Step 3: Why the other options fail
$20\ \text{mA}$ comes from using the peak voltage $100\sqrt{2}$ with a wrong factor, and $40\ \text{mA}$ and $80\ \text{mA}$ from unit errors in $\omega C$.

Final Answer:
The ammeter reading is $10\ \text{mA}$. \[ \boxed{\text{(B) }10\ \text{mA}} \]