100 capacitors,each having a capacity of 10,m | vedclass

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(C) The total capacitance of $100$ capacitors connected in parallel is $C_{eq} = 100 	imes 10\,\mu F = 1000\,\mu F = 10^{-3}\,F$.The potential differ

Vedclass · Accepted Answer

$5 	imes 10^6\;J$ and $150\;paise$

Vedclass · Answer

(C) The total capacitance of $100$ capacitors connected in parallel is $C_{eq} = 100 	imes 10\,\mu F = 1000\,\mu F = 10^{-3}\,F$.The potential difference is $V = 100\,kV = 10^5\,V$.The energy stored in the capacitors is $U = \frac{1}{2} C_{eq} V^2$.$U = \frac{1}{2} 	imes 10^{-3} 	imes (10^5)^2 = \frac{1}{2} 	imes 10^{-3} 	imes 10^{10} = 0.5 	imes 10^7 = 5 	imes 10^6\,J$.Given that $1\,kWh = 3.6 	imes 10^6\,J$ and the cost is $108\;paise\;per\;kWh$.The cost of charging is $	ext{Cost} = \frac{U}{3.6 	imes 10^6} 	imes 108$.$	ext{Cost} = \frac{5 	imes 10^6}{3.6 	imes 10^6} 	imes 108 = \frac{5}{3.6} 	imes 108 = 5 	imes 30 = 150\;paise$.

$100$ capacitors,each having a capacity of $10\,\mu F$,are connected in parallel and are charged by a potential difference of $100\,kV$. The energy stored in the capacitors and the cost of charging them,if electrical energy costs $108\;paise\;per\;kWh$,will be

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