A 30,mu F capacitor is charged by a constant | vedclass

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(D) The charge $Q$ on a capacitor is given by the formula $Q = CV$,where $C$ is the capacitance and $V$ is the potential difference.Given $C = 30\,\mu

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$0.4$

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(D) The charge $Q$ on a capacitor is given by the formula $Q = CV$,where $C$ is the capacitance and $V$ is the potential difference.Given $C = 30\,\mu F = 30 	imes 10^{-6}\,F$ and $V = 400\,V$,the total charge required is:$Q = 30 	imes 10^{-6} 	imes 400 = 12 	imes 10^{-3}\,C$.Since the capacitor is charged by a constant current $I = 30\,mA = 30 	imes 10^{-3}\,A$,the time $t$ taken is given by $Q = I 	imes t$,or $t = \frac{Q}{I}$.Substituting the values:$t = \frac{12 	imes 10^{-3}}{30 	imes 10^{-3}} = \frac{12}{30} = 0.4\,s$.

$A$ $30\,\mu F$ capacitor is charged by a constant current of $30\, mA$. If the capacitor is initially uncharged,how long does it take for the potential difference to reach $400\, V$?

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