Question

For the displacement current through the plates of a parallel plate capacitor of capacitance 30 µF to be 150 µA, the potential difference across the plates of the capacitor has to vary at the rate of:

• A. 10 V/s
• B. 5 V/s
• C. 15 V/s
• D. 20 V/s

Hint:

The displacement current is directly related to the rate of change of the potential difference across a capacitor. Using the formula \( I_D = C \frac{dV}{dt} \), you can easily calculate the required rate of change of voltage.

Answer 1

The Correct Option is B

Step 1: Formula for Displacement Current The displacement current \( I_D \) in a parallel plate capacitor is given by the equation: \[ I_D = C \frac{dV}{dt} \] Where: - \( I_D \) is the displacement current, - \( C \) is the capacitance of the capacitor, - \( \frac{dV}{dt} \) is the rate of change of the potential difference across the capacitor. 

Step 2: Substitute Given Values We are given the following values: - \( C = 30 \, \mu\text{F} = 30 \times 10^{-6} \, \text{F} \), - \( I_D = 150 \, \mu\text{A} = 150 \times 10^{-6} \, \text{A} \). Substitute these values into the formula: \[ 150 \times 10^{-6} = 30 \times 10^{-6} \times \frac{dV}{dt} \] 

Step 3: Solve for \( \frac{dV}{dt} \) Now, solve for the rate of change of the potential difference: \[ \frac{dV}{dt} = \frac{150 \times 10^{-6}}{30 \times 10^{-6}} = 5 \, \text{V/s} \] Thus, the rate of change of the potential difference across the plates of the capacitor is: \[ \mathbf{5 \, \text{V/s}} \]