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

3 C, 3 V

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

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3 C, \frac{V}{3}

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\frac{C}{3}, 3 V

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Solution

The correct option is D $\frac{C}{3}, 3 V$
They are connected in series,
$\frac{1}{C_{e q}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$
$\Rightarrow C_{e q} = \frac{C}{3}$
Voltage will be divided equally between them.
So, we can connect battery of emf $3 V$ .

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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

The correct option is D C3,3VThey are connected in series, 1Ceq=1C+1C+1C=3C ⇒Ceq=C3 Voltage will be divided equally between them. So, we can connect battery of emf 3V.

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

The correct option is D $\frac{C}{3}, 3 V$
They are connected in series,
$\frac{1}{C_{e q}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$
$\Rightarrow C_{e q} = \frac{C}{3}$
Voltage will be divided equally between them.
So, we can connect battery of emf $3 V$ .