In the figure the equivalent resistance of the following combination of resistors $r_1,\, r_2,\, r_3$ and $r_4$, if $r_1\, =\, r_2\, =\,r_3\, =\, r_4\, =\, 2.0\, \Omega$, between the points A and B is

In this case, the first two resistors of 2 ohms each are connected in series and the third and fourth resistors of 2 ohms each are connected in parallel. The first set is connected in series with the second set of resistors.

The resistances R3 and R4 that are connected in parallel have a resistance given as $\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =\frac { 1 }{ 1 } =1\quad ohm$. That is, these two resistances is reduced to a single resistance of 1 ohm.

The resistances R1 and R2 that are connected in series have a resistance given as 2+2 = 4 ohms.

This combined resistance when connected in series with reduced resistance of 1 ohm gives a total resistance of 4+1 = 5 ohms.

Hence, the total resistance in the circuit is 5 ohms.