Welcome To The House Of The Resistor

The three resistors that form the top of the house are in series, so the sum of those resistors (1 + 2 + 2 = 5) gives the equivalent resistance for that section of the circuit. Now our circuit looks something like this:

Fortunately, the top two resistances are in the same ratio (5/3 = 1.667) as the bottom two resistances (2.5/1.5 = 1.667), so no current flows through a bridge in this configuration. In fact, we can treat it as though it wasn't here.

The resistances in the top are in series and they sum to 8 while the resistances in the bottom sum to 4. These two resistances are in parallel, so we use the relationship $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}$ to find the equivalent resistance between A and B:

$\frac{1}{R_{total}} = \frac{1}{8} + \frac{1}{4} \longrightarrow \frac{1}{R_{total}} = \frac{3}{8} \longrightarrow R_{total} = \boxed{\frac{8}{3}}$

The important thing to solving this problem is realizing the potential difference across the 4Ω resistor is 0 V. Once you know that, you can simplify the diagram.

This is a good problem to help you practice the principles of adding resistors in series and parallel. You will need to use both in this case, since there are two strings of series resistors, finally connected in parallel.

The final answer is found by looking at: $(\frac{1}{1+2+2+3}+\frac{1}{2.5+1.5})^{-1}=\frac{8}{3}$