Question

An alternating voltage of amplitude 40 V and frequency 4 kHz is applied directly across the capacitor of 12 μF. The maximum displacement current between the plates of the capacitor is nearly:

• A. 13 A
• B. 8 A
• C. 10 A
• D. 12 A

Answer 1

The Correct Option is D

The displacement current in a capacitor is the same as the conduction current. The capacitive reactance \(X_c\) is given by:

\[ X_c = \frac{1}{\omega C} = \frac{1}{2 \pi f C} \]

Substituting the given values:

\[ X_c = \frac{1}{2 \pi \times 4 \times 10^3 \, \text{Hz} \times 12 \times 10^{-6} \, \text{F}} \approx 3.317 \, \Omega \]

The current \(I\) is given by:

\[ I = \frac{V}{X_c} = \frac{40 \, \text{V}}{3.317 \, \Omega} \approx 12 \, \text{A} \]

Answer 2

Step 1: Given data
Peak voltage (amplitude), \( V_0 = 40\,\text{V} \)
Frequency, \( f = 4\,\text{kHz} = 4\times10^3\,\text{Hz} \)
Capacitance, \( C = 12\,\mu\text{F} = 12\times10^{-6}\,\text{F} \).

Step 2: Relation for displacement current
For a capacitor, the maximum displacement current is given by: \[ I_0 = \omega C V_0 \] where \( \omega = 2\pi f \).

Step 3: Substitute the values
\[ \omega = 2\pi(4\times10^3) = 8\pi\times10^3 \] \[ I_0 = (8\pi\times10^3)(12\times10^{-6})(40) \] \[ I_0 = 8\pi \times 10^3 \times 480\times10^{-6} = 8\pi \times 0.48 \times 10^0 \] \[ I_0 \approx 12.06\,\text{A}. \] Thus, the maximum displacement current is approximately 12 A.

Final answer

12 A