A 500 ,mu F capacitor is charged at a steady | vedclass

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(D) Given,capacitance $C = 500 \,\mu F = 500 	imes 10^{-6} \,F$.Charging rate $I = \frac{dQ}{dt} = 100 \,\mu C/s = 100 	imes 10^{-6} \,C/s$.Target p

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) Given,capacitance $C = 500 \,\mu F = 500 	imes 10^{-6} \,F$.Charging rate $I = \frac{dQ}{dt} = 100 \,\mu C/s = 100 	imes 10^{-6} \,C/s$.Target potential difference $V = 10 \,V$.The charge $Q$ required to reach this potential is given by $Q = CV$.$Q = (500 	imes 10^{-6} \,F) 	imes (10 \,V) = 5000 	imes 10^{-6} \,C = 5 	imes 10^{-3} \,C$.Since the capacitor is charged at a steady rate,$Q = I 	imes t$.Therefore,$t = \frac{Q}{I} = \frac{5 	imes 10^{-3} \,C}{100 	imes 10^{-6} \,C/s} = \frac{5 	imes 10^{-3}}{10^{-4}} \,s = 50 \,s$.

$A$ $500 \,\mu F$ capacitor is charged at a steady rate of $100\, \mu C/s$. The potential difference across the capacitor will be $10\, V$ after an interval of.....$s$.

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