Question

An AC source \( V = 282 \sin(100t) \) volt is connected across a \( 1 \mu F \) capacitor. The RMS value of current in the circuit will be (Take \( \sqrt{2} = 1.41 \)):

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

Hint:

For a capacitor in AC circuit, current leads voltage by \( 90^\circ \), and the capacitive reactance is \( X_C = \frac{1}{\omega C} \).

Answer 1

The Correct Option is B

RMS Current Calculation

1: Given Data
- Peak voltage: \( V_0 = 282V \)
- Angular frequency: \( \omega = 100 \)
- Capacitance: \( C = 1 \mu F = 10^{-6} F \)

2: Capacitor Reactance
\[ X_C = \frac{1}{\omega C} = \frac{1}{(100) (10^{-6})} = 10^4 \Omega \]

3: Peak Current Calculation
Using:
\[ I_0 = \frac{V_0}{X_C} \]
\[ I_0 = \frac{282}{10^4} = 0.0282 A = 28.2 mA \]

4: RMS Current
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{28.2}{1.41} = 20 mA \]

Therefore, the result is (B) 20 mA.