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:
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:
Updated On: Feb 2, 2026
- 13 A
- 8 A
- 10 A
- 12 A
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The Correct Option is D
Approach Solution - 1
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:
\[ I = \frac{V}{X_c} = \frac{40 \, \text{V}}{3.317 \, \Omega} \approx 12 \, \text{A} \]
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Approach Solution -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
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
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