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(A) Let the capacity of each capacitor be $C$.When three capacitors of capacity $C$ are connected in parallel,their equivalent capacity is $C_p = C +

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$5$

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(A) Let the capacity of each capacitor be $C$.When three capacitors of capacity $C$ are connected in parallel,their equivalent capacity is $C_p = C + C + C = 3C$.Now,this combination is connected in series with another capacitor of capacity $C$.The resultant equivalent capacity $C_{eq}$ is given by the series formula:$\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C} = \frac{1}{3C} + \frac{1}{C} = \frac{1 + 3}{3C} = \frac{4}{3C}$.Therefore,$C_{eq} = \frac{3C}{4}$.Given $C_{eq} = 3.75 \mu F$,we have:$3.75 = \frac{3C}{4}$$C = \frac{3.75 	imes 4}{3} = 1.25 	imes 4 = 5.00 \mu F$.Thus,the capacity of each capacitor is $5 \mu F$.

When three capacitors of equal capacities are connected in parallel and one of the same capacitor is connected in series with this combination,the resultant capacity is $3.75 \mu F$. The capacity of each capacitor is: (in $\mu F$)

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