Circuit Analysis Schematics Conventions in Voltage Divider
This simple graphic solution with two examples may help.



simulate this circuit &ndash; Schematic created using CircuitLab
Figure 1. Consider the voltages as heights referenced to 0 V. Consider the resistors as distances from the common junction.

In (a) VO will be 1/5 of the way between V1 and V2. Since there is a rise of 5 V between V1 and V2 then VO will be V1 + 1 = 2 + 1 = 3 V.
In (b) VO will be 3/5 of the way between V1 and V2. Since there is a fall of 10 V between V1 and V2 then VO will be V1 - 10 &amp;times (3 / 5) = 4 - 6 = -2 V.


What I've been told is that Va and Vb are not ground,

They are not ground but they share a common ground and this is the reference from which the voltages are measured.

... and for my purposes I need to know what is happening at Vo.

Hopefully the diagrams help.

I know in this case that Va and Vb are 'sources' - so they are 'pushing' 'towards' each other and 'meeting in the middle' (that is my description of it, since I'm new to this that's best way I can describe it).

Almost. Since one is higher than the other the current will flow downhill from that to the other. Think of connecting a single AA cell to three AA cells in series by a pair of resistors. The three will discharge into the one which will charge.

So my intuition tells me that since Va and Vb are in opposite directions, ...

They can be in the same direction (a) or opposite (b).

... that there's something like a Va - Vb going on here.

Correct.

I'd love to use KVL but this doesn't look like a loop to me ...

When you draw in the ground reference you complete the loop. (All points with GND symbols are connected together.)

Because it also don't look like a loop, I don't know what the current would be.

Hopefully that's clear now too. The current will be
(VA - VB) / (RA + RB).

It seems intuitively wrong to have two currents (in my head it seems like there would be a current from Va across R1 and a current from Vb across R2 - and in the middle they'd &quot;meet&quot; ...

With nothing else connected at VO the current out of A goes into B.

From the comments:

Thank you. If i may ask for clarification: in situation (a) above I get the following:


Current = (7 - 2) / (1000 + 4000) = 5 / 5000 or = 1/1000 A (or 1 mA);
Voltage Drop at R1 = 1/1000 * 1000 = 1 V;

Correct and voltage drop is from right to left in your approach. (We have to be consistent elsewhere now.)

Voltage Drop at R2 (similarly) = 4 V, therefore voltage in between them is 4 - 1 V = 3 V this results in same but I arrive at it the wrong way I think ...

No, you've happened to get the same answer because of the easy numbers I chose. Your 4 V drop on R2 is correct but VO is then calculated by V0 = 7 - 4 = 3 V. And for R1 it's V0 = 2 + 1 = 3 V. (Both have to give the same answer.)