Question

An AC voltage \( V = 50\sqrt{2} \sin(100t) \) is applied across a capacitor of capacitance \( C = 1 \mu F \). What is the rms value of the current through the capacitor?

• A. \( 0.0025 \, \text{A} \)
• B. \( 0.01 \, \text{A} \)
• C. \( 0.005 \, \text{A} \)
• D. \( 0.007 \, \text{A} \)

Hint:

The current through a capacitor in an AC circuit is given by \( I = C \frac{dV}{dt} \). To find the rms value, divide the maximum current by \( \sqrt{2} \).

Answer 1

The Correct Option is C

AC Voltage Applied to a Capacitor

Given the AC voltage: \[ V(t) = 50\sqrt{2} \sin(100t) \]

Capacitance: \[ C = 1\ \mu F = 1 \times 10^{-6}\ F \]

Step 1: RMS Voltage

The given voltage expression is \( V(t) = V_0 \sin(\omega t) \), with \( V_0 = 50\sqrt{2} \). The RMS voltage is calculated as: \[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{50\sqrt{2}}{\sqrt{2}} = 50\ V \]

Step 2: Capacitive Reactance

Capacitive reactance is determined by: \[ X_C = \frac{1}{\omega C} \] Using \( \omega = 100\ \text{rad/s} \) and \( C = 1 \times 10^{-6} \ \text{F} \): \[ X_C = \frac{1}{100 \times 1 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 = 10000\ \Omega \]

Step 3: RMS Current

The RMS current through the capacitor in an AC circuit is: \[ I_{rms} = \frac{V_{rms}}{X_C} = \frac{50}{10000} = 0.005\ A \]

✅ Final Answer:

0.005 A (Option 3)