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

The correct answer is 3C.

Based on the provided circuit diagram, we observe a combination of three capacitors, each labeled with a capacitance of C. They are positioned with one on the top horizontal wire, one on the middle vertical wire, and one on the bottom horizontal wire. To determine the overall equivalent capacitance, we must systematically trace the electrical nodes (points of equal potential) from the left input terminal to the right output terminal.

Let's designate the left terminal as Node A and the right terminal as Node B. By carefully tracing the solid conducting wires, which represent paths of zero resistance, we can conceptually simplify the circuit's topology:

1. The left terminal (Node A) is directly connected to the left plate of the top capacitor. Following the wire geometry of this classic circuit topology, the equipotential line of Node A also connects to the bottom plate of the middle vertical capacitor and the left plate of the bottom capacitor.

2. The right terminal (Node B) is directly connected to the right plate of the top capacitor, the top plate of the middle vertical capacitor, and the right plate of the bottom capacitor.

When we map out these connections, we find that the potential difference across each of the three capacitors is exactly the same, spanning directly from Node A to Node B. Because all three capacitors share the exact same two common nodes across their plates, they are connected entirely in a parallel configuration.

The formula for the equivalent capacitance ( $C_{eq}$ ) of capacitors connected in parallel is simply the sum of their individual capacitances:

$C_{eq}=C_1+C_2+C_3$

Since each capacitor in the figure has an identical value of C, we substitute this into our formula:

$C_{eq}=C+C+C$

$C_{eq}=3C$

Thus, the total equivalent capacitance of the combination is exactly 3C.