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

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(D) Given: Capacitance $C = 500\,\mu F = 500 	imes 10^{-6}\,F$.Rate of charging $\frac{dq}{dt} = 100\,\mu C/s = 100 	imes 10^{-6}\,C/s$.Final potent

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) Given: Capacitance $C = 500\,\mu F = 500 	imes 10^{-6}\,F$.Rate of charging $\frac{dq}{dt} = 100\,\mu C/s = 100 	imes 10^{-6}\,C/s$.Final potential difference $V = 10\,V$.We know that the charge $q$ on a capacitor is given by $q = CV$.Differentiating with respect to time $t$,we get $\frac{dq}{dt} = C \frac{dV}{dt}$.However,since the charging rate is constant,we can use the relation $q = I 	imes t$,where $I = \frac{dq}{dt}$.So,$CV = I 	imes t$.Substituting the values: $(500 	imes 10^{-6}\,F) 	imes (10\,V) = (100 	imes 10^{-6}\,C/s) 	imes t$.$5000 	imes 10^{-6} = 100 	imes 10^{-6} 	imes t$.$t = \frac{5000}{100} = 50\,s$.Therefore,the time interval is $50\,s$.

$A$ $500\,\mu F$ 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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