Question

An alternating voltage of 220 V, 50 Hz frequency is applied across a capacitor of capacitance 2 μF. The impedance of the circuit is

• A. \( \frac{\pi}{5000} \)
• B. \( \frac{1000}{5000 \pi} \)
• C. \( 500 \pi \)
• D. \( 500 \pi \)

Hint:

The impedance of a capacitor in an AC circuit is \( Z = \frac{1}{\omega C} \), where \( \omega \) is the angular frequency and \( C \) is the capacitance.

Answer 1

The Correct Option is C

Step 1: Use the formula for the impedance of a capacitor.
The impedance of a capacitor in an AC circuit is given by: \[ Z = \frac{1}{\omega C} \] where \( \omega = 2\pi f \), \( f = 50 \, \text{Hz} \), and \( C = 2 \, \mu\text{F} = 2 \times 10^{-6} \, \text{F} \).
Step 2: Calculate the impedance.
First, calculate the angular frequency: \[ \omega = 2\pi \times 50 = 314.16 \, \text{rad/s} \] Now, substitute the values to get the impedance: \[ Z = \frac{1}{314.16 \times 2 \times 10^{-6}} = 500 \pi \]
Final Answer: \[ \boxed{500 \pi} \]