$\dfrac{35}{6}\mu F$
None of these
$15\mu F$
$\dfrac{25}{6}\mu F$

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The given circuit is the same as Circuit 1 in the above figure.

Adding up the two parallel capacitors between $A$ and $C$ , and $B$ and $D$ , we have $10+15=25 \mu F$ and $1+1 = 2 \mu F$ respectively and we get equivalent circuit as Circuit 2.

We note that $V_{CB} = \dfrac{25}{30} V_{AB} = \dfrac{5}{6} V_{AB}$ and that $V_{DB} = \dfrac{10}{12} V_{AB} = \dfrac{5}{6} V_{AB}$ . This means that point $C$ and point $D$ are at the same potential, and $\color{#3D99F6}{\text{we can consider the 13 } \mu F \text{ capacitor as a short circuit.}}$ Therefore, the resultant equivalent circuit is as Circuit 3.

Therefore, the equivalent capacitance across $AB$ is given by:

$\begin{aligned} C_{AB} & = (25||10) \color{#3D99F6}{\$} (5||2) \quad \quad \small \color{#3D99F6}{\$ \text{ means "in series", as we can't use + here.}} \\ & = 35 \$ 7 \\ & = \frac{35 \times 7}{35+7} \\ & = \boxed{\dfrac{35}{6}} \end{aligned}$