QJEEM24Ph8a64: JEE Main 2024 Physics - Class 12 JEEM24Ph8a64 Solution: An alternating voltage of amplitude 40   V and frequency 4 …

Let's calculate the maximum displacement current between the plates of the capacitor when an alternating voltage is applied directly across it.

The formula for the capacitive reactance $$X_C$$ of a capacitor is given by:

$X_C = \frac{1}{2\pi fC}$

where

• $$f$$ is the frequency of the alternating voltage, and
• $$C$$ is the capacitance of the capacitor.

Given:

• The amplitude of the voltage ($$V_{max}$$) = $$40$$ V
• Frequency ($$f$$) = $$4000$$ Hz (or $$4\mathrm{kHz}$$)
• Capacitance ($$C$$) = $$12\mu \mathrm{F}=12\times {10}^{-6}\mathrm{F}$$

First, we calculate $$X_C$$:

$X_C = \frac{1}{2\pi(4000)(12 \times 10^{-6})} \approx \frac{1}{2\pi \cdot 4 \cdot 12 \cdot 10^{-3}} \approx \frac{1}{96\pi \cdot 10^{-3}}$

$X_C \approx \frac{1}{3.14 \cdot 96 \cdot 10^{-3}} \approx \frac{1}{0.30144} \approx 3.316 \Omega$

Now, the maximum current ($$I_{max}$$) in the circuit can be calculated using Ohm's law, considering the maximum voltage across the capacitor and its capacitive reactance:

$I_{max} = \frac{V_{max}}{X_C}$

$I_{max} = \frac{40}{3.316} \approx 12.06 \mathrm{~A}$

Hence, the maximum displacement current between the plates of the capacitor is nearly:

Option D: 12 A