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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

F1 .

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?

schematic

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.

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  • \$\begingroup\$ Why is your vout connected to the ground? \$\endgroup\$ – Alexander Sabiona Apr 6 '15 at 1:27
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    \$\begingroup\$ Right, that wouldn't make sense. So V_out goes to some voltmeter that is connected to ground, but not to ground itself? \$\endgroup\$ – Henry H. Apr 6 '15 at 1:39
  • \$\begingroup\$ yes! It means you're taking the voltage at that node with respect to the ground. \$\endgroup\$ – Alexander Sabiona Apr 6 '15 at 2:04
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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: $$ V_{out}|_{V_1 = 0} = V_2\frac{R_1}{R_1 + R_2} $$

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

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