Three capacitors of capacitance 3,mu F are co | vedclass

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(A) To obtain the maximum capacitance,the capacitors must be connected in parallel.In parallel combination,the equivalent capacitance is given by $C_{

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$9\,\mu F$,$1\,\mu F$

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(A) To obtain the maximum capacitance,the capacitors must be connected in parallel.In parallel combination,the equivalent capacitance is given by $C_{eq} = C_1 + C_2 + C_3$.Since all capacitors have the same capacitance $C = 3\,\mu F$,we have $C_{max} = 3C = 3 	imes 3\,\mu F = 9\,\mu F$.To obtain the minimum capacitance,the capacitors must be connected in series.In series combination,the equivalent capacitance is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.Thus,$C_{min} = \frac{C}{3} = \frac{3\,\mu F}{3} = 1\,\mu F$.Therefore,the maximum and minimum capacitances are $9\,\mu F$ and $1\,\mu F$ respectively.

Three capacitors of capacitance $3\,\mu F$ are connected in a circuit. Then their maximum and minimum capacitances will be

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