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

1. 3C, V/3
2. C/3, 3V
3. 3C, 3V
4. C/3, V/3

Correct answer: C/3, 3V

Solution

When capacitors are connected in series, the equivalent capacitance is given by 1/C_eq = 1/C + 1/C + 1/C, which simplifies to C_eq = C/3. The breakdown voltage in series adds up, so the total breakdown voltage is 3V.

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