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

The current \( I \) through a capacitor in an AC circuit is given by: \[ I = C \frac{dV}{dt} \] Where: - \( V = 50\sqrt{2} \sin(100t) \) is the applied AC voltage, - \( C = 1 \mu F = 1 \times 10^{-6} \, F \). First, we differentiate \( V \) with respect to time: \[ \frac{dV}{dt} = 50\sqrt{2} \times 100 \cos(100t) = 5000\sqrt{2} \cos(100t) \] Now, the current is: \[ I = C \times 5000\sqrt{2} \cos(100t) \] Substituting the values of \( C \): \[ I = (1 \times 10^{-6}) \times 5000\sqrt{2} \cos(100t) \] The RMS value of current is given by: \[ I_{\text{rms}} = \frac{I_{\text{max}}}{\sqrt{2}} = \frac{5000\sqrt{2}}{\sqrt{2}} \times 10^{-6} = 0.005 \, \text{A} \] Thus, the rms value of the current is \( 0.005 \, \text{A} \).