A parallel plate capacitor has a capacitance | vedclass

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(D) Given: Capacitance $C = 200 \ pF = 200 	imes 10^{-12} \ F$,Voltage $V_{rms} = 230 \ V$,Angular frequency $\omega = 300 \ rad/s$.The capacitive re

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$13.8 \ \mu A$ and $13.8 \ \mu A$

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(D) Given: Capacitance $C = 200 \ pF = 200 	imes 10^{-12} \ F$,Voltage $V_{rms} = 230 \ V$,Angular frequency $\omega = 300 \ rad/s$.The capacitive reactance is given by $X_C = \frac{1}{\omega C} = \frac{1}{300 	imes 200 	imes 10^{-12}} = \frac{1}{6 	imes 10^{-8}} = \frac{10^8}{6} \ \Omega$.The $rms$ conduction current $I_c$ is $I_c = \frac{V_{rms}}{X_C} = 230 	imes 300 	imes 200 	imes 10^{-12} \ A$.$I_c = 230 	imes 6 	imes 10^{-8} \ A = 1380 	imes 10^{-8} \ A = 13.8 	imes 10^{-6} \ A = 13.8 \ \mu A$.According to the Maxwell-Ampere law,the displacement current $I_d$ in a capacitor is equal to the conduction current $I_c$ in the connecting wires. Therefore,$I_d = I_c = 13.8 \ \mu A$.

List-$I$	List-$II$
$(A)$ Gauss's law	$(I)$ Surface charge density
$(B)$ Faraday's law	$(II)$ Electric charge and energy conservation
$(C)$ Ampere's law	$(III)$ Change in magnetic flux
$(D)$ Kirchhoff's law	$(IV)$ Change in electric flux
	$(V)$ Total electric flux

$A$ parallel plate capacitor has a capacitance $C = 200 \ pF$. It is connected to a $230 \ V$ $AC$ supply with an angular frequency $\omega = 300 \ rad/s$. The $rms$ value of the conduction current in the circuit and the displacement current in the capacitor,respectively,are:

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