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I have searched high and low for a diagram similar to this in order to help me understand what little I do know (in an intro college electronics course) but I cannot find anyway to understand this. So am hoping I can find some help.

Everywhere I search for an answer or guide I get nice, beautiful complete circuits (A nice obvious loop so I know how to follow along with Ohm's Law, KCL, KVL, Voltage Divider equation, etc. (Like this, this, this, etc)

The actual assignment question is different than this - but if I can understand what is happening here than I will be able to proceed. I just can't follow because I don't know what is implied here?

What I've been told is that Va and Vb are not ground, and for my purposes I need to know what is happening at Vo. 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 thats best way I can describe it)

enter image description here

So my intuition tells me that since Va and Vb are in opposite directions, that theres something like a Va - Vb going on here. I'd love to use KVL but this doesn't look like a loop to me - so no idea what can do. Because it also don't look like a loop, I don't know what the current would be. 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 "meet" - but that doesn't make sense (I still see current using Ohms law as \$\frac{V}{R1+R2}\$, but there are two sources here so no idea whats really going on.

Basically am lost on what this diagram means, and I'm sure it must be valid (it's not a trick question - idea is to find Vo) - but I just don't know what to make of it since every other example has a nice complete circuit. This thing must be like a complete circuit in some way? Like closed in some way? There must be something happening between Va and Vb implied?

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  • \$\begingroup\$ Why don't you redraw the circuit and include the sources for Va and Vb. I think that will help you see how KVL can be used. \$\endgroup\$ Feb 4 at 23:27
  • \$\begingroup\$ @ElliotAlderson I am actually not sure how to do that really though. Like I draw the battery symbol instead on each but there is still a gap. Now maybe there is something implied here - like maybe the sources Va and Vb are connected by an ideal wire with voltage 0 (since I know the positive side of each source is facing ‘in’ towards their respective resistors) but that’s kinda why I mentioned convention - since I’m not sure what this represents in a more traditional diagram. \$\endgroup\$
    – backslash
    Feb 4 at 23:43
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This simple graphic solution with two examples may help.

schematic

simulate this circuit – 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 &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 "meet" ...

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.)

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  • \$\begingroup\$ Hi I am going through your answer and it is helping a lot. One thing to clarify as I work through it: you’ve added grounds ‘before’ (negative terminal?) of the v1 and v2 voltage sources. Is this something that is just necessary for any such voltage source arranged in this way? A lot of my issue with this is the little implied things in this schematics. Seeing it here def helps me articulate what is going on and wish it was in the diagram I was given. \$\endgroup\$
    – backslash
    Feb 5 at 3:54
  • \$\begingroup\$ @backslash When the voltage values Va and Vb are defined, they're with respect to a zero of voltage. That zero is 'ground'. Drawing the voltage sources and connecting them to ground is the best way to show that. It also shows on the diagram what has been left out of the schematic. Another way to do that would be connected a voltage source of Va-Vb between those two points. It too restores the current loop, but then you can only compute V0 with respect to Va and Vb, and not to 'ground'. \$\endgroup\$
    – Neil_UK
    Feb 5 at 6:07
  • \$\begingroup\$ @neil_uk thank you that helps. I think I am following that part now finally. \$\endgroup\$
    – backslash
    Feb 5 at 6:58
  • \$\begingroup\$ @backslash: Think of the voltages being referenced to ground in the same way that altitudes are referenced to sea-level. See another answer I wrote here. \$\endgroup\$
    – Transistor
    Feb 5 at 8:21
  • \$\begingroup\$ ... You can then have Ha and Hb (H = height) referenced to sea level or you can use Hb - Ha which will tell you how much higher b is relative to a. If there's a steady incline going between the two points you can calculate the height at any point along that road if you know the fraction of the distance between the two. \$\endgroup\$
    – Transistor
    Feb 5 at 15:24
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\$V_O\$ will be a weighted average of \$V_A\$ and \$V_B\$ if there is no significant loading at \$V_O\$.

The weighting depends upon the resistors.

\$V_O = \frac{R_2V_A + R_1V_B}{R_1+R_2}\$

You can derive this equation from the assumption that the current is the same through each resistor. That is the same as the assumption that there is no significant loading at \$V_O\$

If the current is the same through each resistor, then

\$ I = \frac{V_A - V_O}{R1} = \frac{V_O - V_B}{R2}\$

which can be rearranged to give the formula above.

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    \$\begingroup\$ To whomever voted against this answer, could you please offer an explanation why? \$\endgroup\$ Feb 5 at 2:30
  • \$\begingroup\$ i can't vote otherwise i'd vote it up. I didn't know how to use these equations (or how you were arriving at them) until i figured out a few other things - but it is correct and useful (i arrived at both equations when I was trying to solve various values in terms of the other values). thank you. \$\endgroup\$
    – backslash
    Feb 14 at 19:26
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A short way of doing it is to connect the two open nodes with ... your intuition is spot on ... a (Va - Vb) source.

Now you have a loop and you can do the analysis within the loop as you have learned. This gives Vo relative to Vb (which you can regard as a local GND) or to Va if you prefer your schematics upside down.

The choice of local GND is arbitrary, whichever point you choose, you should get the same answer as long as you use it consistently; Vb seems the most logical choice since we are used to ground being down...

When you need absolute voltages, if you took Vb as your local ground, simply add Vb to your other local nodes ... thus Va is then Vb + (Va - Vb) exactly as it should be.

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