Question

The resultant capacitance (in $\mu F$) between the points A and B in the figure is

Answer 1

If we redraw this circuit by looking at the potentials at points A,B,C,D,E and F then the circuit can be represented somewhat like this
Now, we can clearly say, which all circuits are parallel and which all are in series.
For simplication let us rearrange it a bit further
Thus, the total effective capacitance in between points A and B is $\frac{1}{C_{eq}}=\frac{1}{30}\mu F+\frac{1}{30}\mu F$
$⟹C_{eq}=15\mu F$