Three capacitors,each of capacitance C and br | vedclass

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

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$C/3, 3V$

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(C) When capacitors are 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 $\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$.Therefore,$C_{eq} = \frac{C}{3}$.For capacitors in series,the total breakdown voltage $V_{total}$ is the sum of the individual breakdown voltages of each capacitor: $V_{total} = V_1 + V_2 + V_3$.Given $V_1 = V_2 = V_3 = V$,we get $V_{total} = V + V + V = 3V$.Thus,the equivalent capacitance is $C/3$ and the equivalent breakdown voltage is $3V$.

Three capacitors,each of capacitance $C$ and breakdown voltage $V$,are connected in series. Find the equivalent capacitance and the equivalent breakdown voltage.

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