Five equal resistances are connected in a network as shown in figure. The net resistance between the points A and B is:

The network consists of five equal resistances $$R$$ arranged in a Wheatstone bridge configuration between points A and B. Four resistors form the four arms of the bridge, and the fifth resistor is the galvanometer arm connecting the two middle junctions.

In a Wheatstone bridge with all equal resistances, the bridge is balanced because the ratio of resistances in the two arms on each side is the same: $$\frac{R}{R} = \frac{R}{R} = 1$$.

When the bridge is balanced, no current flows through the middle (fifth) resistor. Therefore, it can be removed without affecting the circuit.

The remaining circuit has two paths from A to B: each path consists of two resistors $$R$$ in series, giving a resistance of $$2R$$ per path. These two paths are in parallel, so the net resistance is $$\frac{2R \times 2R}{2R + 2R} = \frac{4R^2}{4R} = R$$.

Hence the correct answer is Option 1: $$R$$.