Two capacitors of capacities 1 mu F and C mu | vedclass

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(B) When capacitors are connected in series,the charge on each capacitor is the same as the total charge on the combination.Given,$q = 80 \mu C$ for b

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$1600$

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(B) When capacitors are connected in series,the charge on each capacitor is the same as the total charge on the combination.Given,$q = 80 \mu C$ for both capacitors.The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.First,we find the value of $C$ using the equivalent capacitance formula: $C_{	ext{eq}} = \frac{q}{V} = \frac{80 \mu C}{120 \ V} = \frac{2}{3} \mu F$.For series connection,$\frac{1}{C_{	ext{eq}}} = \frac{1}{1} + \frac{1}{C}$.$\frac{3}{2} = 1 + \frac{1}{C} \implies \frac{1}{C} = \frac{1}{2} \implies C = 2 \mu F$.Now,the energy stored in the capacitor $C$ is $U = \frac{q^2}{2C} = \frac{(80 \mu C)^2}{2 	imes 2 \mu F} = \frac{6400 	imes 10^{-12}}{4 	imes 10^{-6}} \ J = 1600 	imes 10^{-6} \ J = 1600 \ \mu J$.

Two capacitors of capacities $1 \mu F$ and $C \mu F$ are connected in series and the combination is charged to a potential difference of $120 \ V$. If the charge on the combination is $80 \mu C$,the energy stored in the capacitor of capacity $C$ in $\mu J$ is

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