Three capacitors each of capacity $$4\mu F$$  are to be connected in such a way that the effective capacitance is $$6\mu F.$$  This can be done by

Solution :
Given, $${C_1} = {C_2} = {C_3} = 4\mu F$$
(A) The network of three capacitors can be shown as

Here, $${C_1}$$ and $${C_2}$$ are in series and the combination of two is in parallel with $${C_3}.$$
$$\eqalign{ & {C_{{\text{net}}}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} + {C_3} = \left( {\frac{{4 \times 4}}{{4 + 4}}} \right) + 4 \cr & = 2 + 4 = 6\mu F \cr} $$
(B) The corresponding network is shown in figure below

Here, $${C_1}$$ and $${C_2}$$ are in parallel and this combination is in series with $${C_3}.$$
So, $${C_{{\text{net}}}} = \frac{{\left( {{C_1} + {C_2}} \right) \times {C_3}}}{{\left( {{C_1} + {C_2}} \right) + {C_3}}} = \frac{{\left( {4 + 4} \right) \times 4}}{{\left( {4 + 4} \right) + 4}}$$
$$ = \frac{{32}}{{12}} = \frac{8}{3}\mu F$$
(C) The corresponding network is shown below. All of three capacitors are in series.
So, $$\frac{1}{{{C_{{\text{net}}}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$$
$$\therefore C = \frac{4}{3}\mu F$$
(D) The corresponding network is shown below.

All of them are in parallel.
So, $${C_{{\text{net}}}} = {C_1} + {C_2} + {C_3}$$
$$ = 4 + 4 + 4 = 12\,\mu F$$
Thus, options (A) is correct.