An alternating voltage of amplitude 40 , V an | vedclass

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(D) The displacement current $(I_d)$ in a capacitor is equal to the conduction current $(I_c)$ flowing through the circuit.The capacitive reactance $(

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$12$

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(D) The displacement current $(I_d)$ in a capacitor is equal to the conduction current $(I_c)$ flowing through the circuit.The capacitive reactance $(X_C)$ is given by the formula:$X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$Given:$V_{max} = 40 \, V$$f = 4 \, kHz = 4 	imes 10^3 \, Hz$$C = 12 \, \mu F = 12 	imes 10^{-6} \, F$Calculating $X_C$:$X_C = \frac{1}{2 	imes 3.1416 	imes 4 	imes 10^3 	imes 12 	imes 10^{-6}}$$X_C = \frac{1}{8 \pi 	imes 12 	imes 10^{-3}} = \frac{1}{0.3016} \approx 3.317 \, \Omega$The maximum current $(I_{max})$ is:$I_{max} = \frac{V_{max}}{X_C} = \frac{40}{3.317} \approx 12.06 \, A$Thus, the maximum displacement current is nearly $12 \, A$.

An alternating voltage of amplitude $40 \, V$ and frequency $4 \, kHz$ is applied directly across a capacitor of $12 \, \mu F$. The maximum displacement current between the plates of the capacitor is nearly: (in $ \, A$)

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