An alternating voltage of amplitude $40 \mathrm{~V}$ and frequency $4 \mathrm{~kHz}$ is applied directly across the capacitor of $12 \mu \mathrm{F}$. The maximum displacement current between the plates of the capacitor is nearly :

<p>Let's calculate the maximum displacement current between the plates of the capacitor when an alternating voltage is applied directly across it.</p> <p>The formula for the capacitive reactance $X_C$ of a capacitor is given by:</p> <p>$X_C = \frac{1}{2\pi fC}$</p> <p>where</p> <ul> <li>$f$ is the frequency of the alternating voltage, and</li> <li>$C$ is the capacitance of the capacitor.</li> </ul> <p>Given:</p> <ul> <li>The amplitude of the voltage ($V_{max}$) = $40$ V</li> <li>Frequency ($f$) = $4000$ Hz (or $4 \mathrm{~kHz}$)</li> <li>Capacitance ($C$) = $12 \mu \mathrm{F} = 12 \times 10^{-6} \mathrm{~F}$</li> </ul> <p>First, we calculate $X_C$:</p> <p>$X_C = \frac{1}{2\pi(4000)(12 \times 10^{-6})} \approx \frac{1}{2\pi \cdot 4 \cdot 12 \cdot 10^{-3}} \approx \frac{1}{96\pi \cdot 10^{-3}}$</p> <p>$X_C \approx \frac{1}{3.14 \cdot 96 \cdot 10^{-3}} \approx \frac{1}{0.30144} \approx 3.316 \Omega$</p> <p>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:</p> <p>$I_{max} = \frac{V_{max}}{X_C}$</p> <p>$I_{max} = \frac{40}{3.316} \approx 12.06 \mathrm{~A}$</p> <p>Hence, the maximum displacement current between the plates of the capacitor is nearly:</p> <p>Option D: 12 A</p>