Question

An alternating voltage \(E = 100\sqrt{2} \sin(50t)\) is connected to a \(2\mu\text{F}\) capacitor through an a.c. ammeter. The ammeter reading will be

• A. \(10\text{ mA}\)
• B. \(5\text{ mA}\)
• C. \(20\text{ mA}\)
• D. \(30\text{ mA}\)

Hint:

Always convert peak voltage to RMS before using AC formulas.

Answer 1

The Correct Option is A

Concept:
For capacitor in AC: \[ I_{rms} = \frac{V_{rms}}{X_C}, \quad X_C = \frac{1}{\omega C} \] Step 1: Find \(V_{rms}\) and \(\omega\). \[ V = 100\sqrt{2} \Rightarrow V_{rms} = 100 \text{ V}, \quad \omega = 50 \]
Step 2: Calculate capacitive reactance. \[ X_C = \frac{1}{\omega C} = \frac{1}{50 \times 2 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \Omega \]
Step 3: Find current. \[ I = \frac{100}{10^4} = 10^{-2} \text{ A} = 10 \text{ mA} \]
Step 4: Conclusion. Ammeter reading = \(10\text{ mA}\)