JEE Main 2024 — Alternating Current Question with Solution

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