# Help needed on wheatstone bridge problem

My circuits 1 professor gave us some practice problems for an upcoming exam, and I'm not sure that I know how to do this one. My approach was to apply the circuit divider formula for the left and right sides, and that gave me the current across both parallel branches of the bridge, but I'm unsure of how to proceed. I was also wondering if I can place ground at some point on this circuit? Or would I need a voltage source to be able to do so? Thanks in advance.

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 – 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.

Your approach is correct. After finding the left-hand and right-hand currents $I_{left}$ and $I_{right}$. You apply KVL and write a loop around Vo, R2, and R4 as,

$$+V_o + I_{right}R_4 - I_{left}R_2 = 0$$

Rearrange for $V_o$