Help needed on wheatstone bridge problem
R1 and R2 in series total 3K, while R3 and R4 in series total 6K, so the current will split 2/3rds to R1+R2, and 1/3rd to R3+R4. This is a fairly straightforward application of resistors in series and parallel.

You can choose a ground arbitrarily. It's just a reference point. I'm going to put it at the bottom of the schematic.

With 2 amps through R1+R2, V(A) will be:

$$V_A = R_2 \cdot 2 \ \text{A} = 4000 \ \text{V}$$

And with 1 amp through R3+R4, V(B) will be:

$$V_B = R_4 \cdot 1 \ \text{A} = 3000 \ \text{V}$$

The voltage difference will be:

$$v_o = V_A - V_B = 1000 \ \text{V}$$

This is easily confirmed by a quick simulation:







simulate this circuit &ndash; Schematic created using CircuitLab

Click to open the schematic, then run the DC simulation. You'll see the expression "V(A) - V(B)" computed as 1.000 kV.