The capacity between the points A and B in the given circuit will be:

\frac{2 C_{1} C_{2} + C_{3} (C_{1} + C_{2})}{C_{1} + C_{2} + 2 C_{3}}

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\frac{C_{1} C_{2} + C_{2} C_{3} + C_{3} C_{1}}{C_{1} + C_{2} + C_{3}}

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\frac{C_{1} (C_{2} + C_{3}) + C_{2} (C_{1} + C_{3})}{C_{1} + C_{2} + 3 C_{3}}

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\frac{C_{1} C_{2} C_{3}}{C_{1} C_{2} + C_{2} C_{3} + C_{3} C_{1}}

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Solution

The correct option is A $\frac{2 C_{1} C_{2} + C_{3} (C_{1} + C_{2})}{C_{1} + C_{2} + 2 C_{3}}$
The charge distribution is as shown in the figure.
Using Kirchhoff's law,
$V - q 1 / C_{1} + (q 2 - q 1) / C_{3} - q 1 / C_{1} = 0 \Rightarrow V = 2 q 1 / C_{1} - (q 2 - q 1) / C_{3}$
$V - q 2 / C_{2} + (q 1 - q 2) / C_{3} - q 2 / C_{2} = 0 \Rightarrow V = 2 q 2 / C_{2} - (q 1 - q 2) / C_{3}$
Therefore,
$C_{1} C_{3} V = 2 C_{3} q 1 - C_{2} q 2 + C_{2} q 1 = (2 C_{3} + C_{1}) q 1 - C_{1} q 2$
$C_{2} C_{3} V = 2 C_{3} q 2 - C_{2} q 1 + C_{2} q 2 = (2 C_{3} + C_{2}) q 2 - C_{2} q 1$
Solving these two equations we get,
$q 1 = \frac{C_{1} (C_{2} + C_{3}) V}{C_{1} + C_{2} + 2 C_{3}}$
$q 2 = \frac{C_{2} (C_{1} + C_{3}) V}{C_{1} + C_{2} + 2 C_{3}}$
Equivalent capacitance $= Q_{n e t} / V = (q 1 + q 2) / V = \frac{2 C_{1} C_{2} + C_{1} C_{3} + C_{2} C_{3}}{C_{1} + C_{2} + 2 C_{3}}$

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