Circuit and potential difference?

When AB is connected to 12 v, CD is 4 v. Since there is no current through the 2-ohm resistor, no voltage is set up across it. The potential difference across R is thus equal to 4 v.
Hence, current through R, I = 4/R

In the circuit, current flows from A to the upper 4-ohm resistor, then to R, and back to B through the lower 4-ohm resistor. Hence, we have, using Ohm's Law for resistors in series,
12 = (4/R).(4+4+R)
solve for R gives R = 4 ohms

Now, a voltage of 12 v is applied across CD, there is no current flowing through the upper and lower 4-ohm resistors (because AB is open). The circuit only consists of the 2-ohm resistor and R(= 4 ohms) in series.

Hence, voltage across R = 12 x[R/(R + 2)]
But we have already found R = 4 ohms, voltage across R thus
= 12 x [4/(4+2)] v = 12 x (4/6) v = 8 v

Because voltage across AB equals to voltage across R,
hence, voltage across AB = 8 v