A parallel plate capacitor having plates of r | vedclass

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(A) The r.m.s. current $I_{rms}$ in a capacitive circuit is given by $I_{rms} = \frac{V_{rms}}{X_C}$,where $X_C = \frac{1}{\omega C}$ is the capacitiv

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$6.9 	imes 10^{-6} 	ext{ A}$

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(A) The r.m.s. current $I_{rms}$ in a capacitive circuit is given by $I_{rms} = \frac{V_{rms}}{X_C}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.Substituting $X_C$,we get $I_{rms} = V_{rms} \cdot \omega \cdot C$.Given values are $V_{rms} = 230 	ext{ V}$,$\omega = 300 	ext{ rad/s}$,and $C = 100 	ext{ pF} = 100 	imes 10^{-12} 	ext{ F}$.$I_{rms} = 230 	imes 300 	imes 100 	imes 10^{-12} 	ext{ A}$.$I_{rms} = 230 	imes 3 	imes 10^4 	imes 10^{-12} 	ext{ A}$.$I_{rms} = 690 	imes 10^{-8} 	ext{ A} = 6.9 	imes 10^{-6} 	ext{ A}$.

$A$ parallel plate capacitor having plates of radius $6 \text{ cm}$ has capacitance $100 \text{ pF}$. It is connected to a $230 \text{ V}$ a.c. supply with angular frequency $300 \text{ rad/s}$. The r.m.s. value of current is:

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