A network of four resistances is connected to 9   V battery, as shown in figure. The magnitude of…

The resistance in each branch is $6 \mathrm{\Omega }$. They are in parallel. So, net resistance is $R\text{'}=\frac{6}{2}=3 \mathrm{\Omega }$.

The total current flowing through the loop,

$I=\frac{V}{R\text{'}}=\frac{9 \mathrm{V}}{3 \mathrm{\Omega }}=3 \mathrm{A}$

Thus, through each branch the current is $1.5 \mathrm{A}$.

Hence, the voltage difference across the points is

$V_A-V_B=\left(4\times 1.5\right)-\left(2\times 1.5\right)=3 \mathrm{V}$