# Voltage Divider-like circuit with multiple sources

I'm extremely new to circuits and am having some trouble with a variation on the voltage divider problems I've been doing.

I'm trying to determine $V_{out}$ in the following circuit relative to some $V_{GND}$ that I need to determine the location of .

I know that in a standard voltage divider, the line to $V_{out}$ carries negligible current. In this case, I can't quite determine the current flow because the lines from $V_2$ and $V_1$ carry current that is equal to $V_2/R_2$ and $V_1/R_1$ at the point in between the two resistors. How does this return to ground, and how would I begin to apply Kirchoff's Law for the loop? simulate this circuit – Schematic created using CircuitLab

I would think that ground node positions would be like this, but I'm nevertheless confused about the current flow. Any advice would be appreciated.

• Why is your vout connected to the ground? – Alexander Sabiona Apr 6 '15 at 1:27
• Right, that wouldn't make sense. So V_out goes to some voltmeter that is connected to ground, but not to ground itself? – Henry H. Apr 6 '15 at 1:39
• yes! It means you're taking the voltage at that node with respect to the ground. – Alexander Sabiona Apr 6 '15 at 2:04

## 2 Answers

I can't quite determine the current flow because the lines from $V_2$ and $V_1$ carry current that is equal to $V_2/R_2$ and $V_1/R_1$ at the point in between the two resistors. How does this return to ground, and how would I begin to apply Kirchoff's Law for the loop?

Hmm. I don't agree with this. I think you came up with these equations with the old schematic, which had an extra ground connection.

The actual current through the resistors is $$I_1 = \frac{V_1 - V_{out}}{R_1}$$ and $$I_2 = \frac{V_{out} - V_2}{R_2}$$ Since there is no path for current to take another loop, these currents must be the same! That means that $$\frac{V_1 - V_{out}}{R_1} = \frac{V_{out} - V_2}{R_2}$$ and some math gives you $$V_{out} = \frac{V_1 R_2 + V_2 R_1}{R_1 + R_2} = \frac{V_1 R_2 + V_1 R_1 - V_1 R_1 + V_2 R_1}{R_1 + R_2} = V_1 + (V_2 - V_1)\frac{R_1}{R_1 + R_2}$$ Notice that if $V_1 = 0$, you get back the equation that you're used to: $$V_{out}|_{V_1 = 0} = V_2\frac{R_1}{R_1 + R_2}$$

• That makes perfect sense! I'll try to do the calculation again myself, keeping this in mind. Thank you! – Henry H. Apr 6 '15 at 2:29
• Awesome. Hit those upvote/accept buttons if it works out :) – Greg d'Eon Apr 6 '15 at 2:31

It's still the same voltage divider. The only difference is that now your Vin is (V2-V1) and your Vout is with respect to V1. So Vout = (V2-V1)*R1/(R1+R2)+V1.