Question

A \(100\,\mu\text{F}\) capacitor is connected to a \(100\,\text{V}\), \(50\,\text{Hz}\) AC supply. The rms value of the current is:

• A. \(3.14\,\text{A}\)
• B. \(4.75\,\text{A}\)
• C. \(2.33\,\text{A}\)
• D. \(5.5\,\text{A}\)

Hint:

For a pure capacitor in an AC circuit, \[ X_C=\frac{1}{2\pi fC} \] and \[ I_{\text{rms}}=\frac{V_{\text{rms}}}{X_C}. \] Also, current leads voltage by \(90^\circ\) in a purely capacitive circuit.

Answer 1

The Correct Option is A

Step 1: Calculate capacitive reactance.
\[ X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi\times50\times100\times10^{-6}} = \frac{1}{0.01\pi} = \frac{100}{\pi}\,\Omega \]

Step 2: Find rms current.
\[ I_{rms} = \frac{V_{rms}}{X_C} = \frac{100}{100/\pi} = \pi \approx 3.14\,\text{A} \] \[ \boxed{3.14\,\text{A}} \]