The equivalent capacitance of the combination shown in the figure is:

The correct answer is Option (3) → 8 µF

All three capacitors (2 µF, 4 µF and 2 µF) are connected between the same two junctions, which are joined to A and B. They have the same potential difference across them, so they are in parallel, and capacitances in parallel add.

    \( C_{\mathrm{eq}} = C_1 + C_2 + C_3 = 2 + 4 + 2 = 8\,\mu\mathrm{F} \)