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

The given values are the amplitude $$V_0 = 40$$ V, the frequency $$f = 4$$ kHz = $$4000$$ Hz, and the capacitance $$C = 12 \mu$$F $$= 12 \times 10^{-6}$$ F.

The displacement current through a capacitor equals the conduction current, so

$$ i = C\frac{dV}{dt}$$

For $$V = V_0 \sin(\omega t)$$, the maximum current is

$$ i_{max} = C \omega V_0 = C \times 2\pi f \times V_0$$

$$ i_{max} = 12 \times 10^{-6} \times 2\pi \times 4000 \times 40$$

$$ i_{max} = 12 \times 10^{-6} \times 8\pi \times 10^3 \times 40$$

$$ = 12 \times 8\pi \times 40 \times 10^{-3}$$

$$ = 3840\pi \times 10^{-3}$$

$$ = 3.84\pi \approx 3.84 \times 3.14 \approx 12.06 \text{ A}$$

Thus the maximum displacement current is nearly $$12$$ A.

The correct answer is Option (2): 12 A.