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University circuits, how do circuits like this get uab? (Assuming the bridge is unbalanced)
Solution: The external power supply structure of electric furnace is: R =(R 1+R3)∩(R2+R4).

Therefore, the voltage across R3 of R 1 series is U, and the voltage across R3 is U3 = U× R3/(R1+R3);

In R2 series, the voltage across R4 is u, and the voltage across R4 is U4=U×R4/(R2+R4).

UAB = U3-U4 = UR3/(r 1+R3)-UR4/(R2+R4).

It can also be understood that if the lowest node is m, then U3=Uam and U4=-Umb.

Uab=Uam+Umb=U3-U4 .