Three capacitors each of capacitance C and br | vedclass

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(D) For capacitors connected in series,the equivalent capacitance $C_{eq}$ is given by the formula: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

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$\frac{C}{3}, 3 V$

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(D) For capacitors connected in series,the equivalent capacitance $C_{eq}$ is given by the formula: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.Since $C_1 = C_2 = C_3 = C$,we have $\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$.Thus,$C_{eq} = \frac{C}{3}$.When capacitors are connected in series,the total breakdown voltage is the sum of the individual breakdown voltages of each capacitor,provided they are identical.Therefore,the total breakdown voltage $V_{total} = V + V + V = 3 V$.Hence,the equivalent capacitance is $\frac{C}{3}$ and the breakdown voltage is $3 V$.

Three capacitors each of capacitance $C$ and breakdown voltage $V$ are connected in series. The capacitance and breakdown voltage of the series combination will be respectively:

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