Question

The rms value of conduction current in a parallel plate capacitor is $69\, \mu A$ The capacity of this capacitor, if it is connected to $230 \, V$ ac supply with an angular frequency of $600\, rad / s$, will be :

• A. $5 \, pF$
• B. $50 \, pF$
• C. $100 \, pF$
• D. $200\, pF$

Answer 1

The Correct Option is B

To determine the capacity of a parallel plate capacitor connected to an AC supply, we can use the relationship between the root mean square (RMS) current, voltage, angular frequency, and capacitance.

The current through a capacitor in an AC circuit is given by the formula:

\(I_{\text{rms}} = V_{\text{rms}} \cdot \omega \cdot C\)

where:

• \(I_{\text{rms}}\) is the RMS value of the current.
• \(V_{\text{rms}}\) is the RMS voltage across the capacitor.
• \(\omega\) is the angular frequency.
• \(C\) is the capacitance of the capacitor.

From the given data:

• \(I_{\text{rms}} = 69\, \mu A = 69 \times 10^{-6}\, A\)
• \(V_{\text{rms}} = 230\, V\)
• \(\omega = 600\, \text{rad/s}\)

Substituting these values into the formula:

\(69 \times 10^{-6} = 230 \times 600 \times C\)

Solving for \(C\):

\(C = \frac{69 \times 10^{-6}}{230 \times 600}\)

Calculating the above expression:

\(C = \frac{69 \times 10^{-6}}{138000}\)

\(C = 5 \times 10^{-11} \, F = 50 \, pF\)

Thus, the capacitance of the capacitor is 50 pF. This solution verifies that the correct option is

$50 \, pF$