Question

The rms value of the conduction current in a parallel plate capacitor is 6.9 μA. 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 solve this problem, we need to find the capacitance (C) of a parallel plate capacitor when it is connected to a 230 V AC supply with an angular frequency of 600 rad/s and a given RMS conduction current of 6.9 μA. Let's break down the solution step-by-step:

1. First, we know that the conduction current in a capacitor is given by the formula: I_c = V \cdot \omega \cdot C, where I_c is the RMS value of the conduction current, V is the RMS voltage across the capacitor, \omega is the angular frequency, and C is the capacitance.
2. We can rearrange the formula to solve for the capacitance C: C = \frac{I_c}{V \cdot \omega}.
3. Substitute the given values into the equation:
   • RMS conduction current, I_c = 6.9\, \mu A = 6.9 \times 10^{-6}\, A
   • RMS voltage, V = 230\, V
   • Angular frequency, \omega = 600\, rad/s
4. Calculate the capacitance: C = \frac{6.9 \times 10^{-6}}{230 \times 600}
5. Continue with the calculation: C = \frac{6.9 \times 10^{-6}}{138000} = 50 \times 10^{-12}\, F = 50\, pF.

Thus, the capacitance of the capacitor is 50 pF, which is the correct answer.

Explanation: In an AC circuit, the conduction current through a capacitor is determined by its capacitive reactance, which in turn depends on the angular frequency and the capacitance. Using the given conduction current and the AC voltage, we calculated the capacitance by rearranging and substituting the known values into the conduction current formula.