At this time, the circuit structure is two independent loops:

1. Left loop:12v-3Ω-6Ω-6v, current i2, counterclockwise. Find i2=2A.

Therefore, the "+"of u2 corresponds to the voltage of the lower common node C, that is, the voltage of the 6 Ω series 6V voltage source branch is UAC = 6I2-6 = 6× 2-6 = 6V;

Of course, it can also be obtained by using the branch voltage of 3ω series 12V voltage source: UC =-3i2+12 =-3x2+12 = 6 (v).

2. Right loop: 2A current source is connected in series with1Ω resistor, so the voltage across1Ω resistor is UCB = 2×1Ω = 2v, and the voltage polarity is positive at the top and negative at the bottom.

So U2=Uab=Uac+Ucb=6+2=8V.