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(B) The potential difference across the battery is $V = 100 \, V$.The equivalent capacitance $C$ of the two capacitors $C_1 = 4 \, \mu F$ and $C_2 = 8

Vedclass · Accepted Answer

$1.33 	imes 10^{-2} \, J$

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(B) The potential difference across the battery is $V = 100 \, V$.The equivalent capacitance $C$ of the two capacitors $C_1 = 4 \, \mu F$ and $C_2 = 8 \, \mu F$ connected in series is given by:$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{4} + \frac{1}{8} = \frac{2 + 1}{8} = \frac{3}{8} \, \mu F^{-1}$Therefore,$C = \frac{8}{3} \, \mu F = \frac{8}{3} 	imes 10^{-6} \, F$.The energy $E$ stored in the series combination is given by the formula:$E = \frac{1}{2} C V^2$Substituting the values:$E = \frac{1}{2} 	imes \left( \frac{8}{3} 	imes 10^{-6} ight) 	imes (100)^2$$E = \frac{1}{2} 	imes \frac{8}{3} 	imes 10^{-6} 	imes 10^4$$E = \frac{4}{3} 	imes 10^{-2} \, J$$E \approx 1.33 	imes 10^{-2} \, J$

$A$ $100 \, V$ battery is connected across the series combination of two capacitors of $4 \, \mu F$ and $8 \, \mu F$. The energy stored in the series combination is

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