If a capacitor of capacitance 100 mu F is cha | vedclass

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(D) The capacitance of the capacitor is $C = 100 \mu F = 100 	imes 10^{-6} \ F = 10^{-4} \ F$. The potential difference required is $V = 100 \ V$. Th

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$100$

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(D) The capacitance of the capacitor is $C = 100 \mu F = 100 	imes 10^{-6} \ F = 10^{-4} \ F$. The potential difference required is $V = 100 \ V$. The total charge $Q$ required to be stored on the capacitor is given by $Q = CV$. Substituting the values,$Q = (10^{-4} \ F) 	imes (100 \ V) = 10^{-2} \ C$. The rate of charging is given as $I = 100 \mu C s^{-1} = 100 	imes 10^{-6} \ C s^{-1} = 10^{-4} \ C s^{-1}$. Since the charge is supplied at a steady rate,$Q = I 	imes t$,where $t$ is the time taken. Therefore,$t = \frac{Q}{I} = \frac{10^{-2} \ C}{10^{-4} \ C s^{-1}} = 10^{2} \ s = 100 \ s$.

If a capacitor of capacitance $100 \mu F$ is charged at a steady rate of $100 \mu C s^{-1}$,then the time taken to produce a potential difference of $100 \ V$ between the capacitor plates is (in $s$)

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