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

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(B) When capacitors are connected in series,the equivalent capacitance $C_{	ext{eq}}$ is given by $\frac{1}{C_{	ext{eq}}} = \frac{1}{C_1} + \frac{1}

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$1600$

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(B) When capacitors are connected in series,the equivalent capacitance $C_{	ext{eq}}$ is given by $\frac{1}{C_{	ext{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$.For $C_1 = 1 \mu F$ and $C_2 = C \mu F$,we have $C_{	ext{eq}} = \frac{1 	imes C}{1 + C} = \frac{C}{C+1} \mu F$.Given the total charge $q = 80 \mu C$ and potential difference $V = 120 \ V$,we use the relation $q = C_{	ext{eq}} V$.Substituting the values: $80 = \left( \frac{C}{C+1} ight) 	imes 120$.Dividing both sides by $40$: $2 = \left( \frac{C}{C+1} ight) 	imes 3$.$2(C+1) = 3C \implies 2C + 2 = 3C \implies C = 2 \mu F$.In a series combination,the charge on each capacitor is the same as the total charge $q = 80 \mu C$.The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.For the capacitor $C = 2 \mu F$: $U = \frac{(80 \mu C)^2}{2 	imes 2 \mu F} = \frac{6400}{4} \mu J = 1600 \mu J$.

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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