How do I find the voltage drop in a circuit with two batteries and three resistors?

[MonkeySage]

V1= 9V and is in series with R1=10Ω
V2= 3V and is in series with R2=15Ω
They converge at node N3 and the current continues through R3=24Ω

I need to find the Voltage through N3 and total current through R3.


____R1___N3____
|................|.........|
|................R2.......|
+................+.......R3
9v...............3v......|
-..................-.......|
|_________|_____|

Somehow, when I try to calculate V at R3, by using Ohms law to get I2 and I1, magically I get a higher voltage than I started with. I know that is impossible.

[Erloas]

Different numbers but same exact circuit as this example

It has been a long time since I've done nodal analysis (about 14-15 years) but I think I did it correctly.

I came up with 5.28V at N3.

And to double check I tried the loop analysis method
Which gave the current through R3 as 0.220A, which puts the V across it as 5.28V, which is the same.

You use ohms law a lot, but with multiple sources the only way to do it is with Nodal or Loop analysis.

[Cazero]

Somehow, when I try to calculate V at R3, by using Ohms law to get I2 and I1, magically I get a higher voltage than I started with. I know that is impossible.

From the way you drawn the circuit, my bet would be on a sign error for I2.

[Anonymouswizard]

All currents into and out of a node must sum to zero, right? (just double checking I've remembered nodal analysis correctly)

So working off the top of my head, at N3 we get I_R1+I_R2=I_R3

V=IR, I=V/R -> (V_R1/10)+(V_R2/15)=V_R3/24

rearrange for V_R3 -> V_R3=24((V_R1/10)+(V_R2/15))

V_R3=24/30*(3V_R1+2V_R2)

At this point I get stuck remembering how to work out V_R1 or V_R2, because I was always stronger on electronics than electrics. There's some way to get the proportion of the P.D over each resistor compared to the emf of the relevant source out of the equation, but I can't quite remember it.

[Erloas]

All currents into and out of a node must sum to zero, right? (just double checking I've remembered nodal analysis correctly)

So working off the top of my head, at N3 we get I_R1+I_R2=I_R3

V=IR, I=V/R -> (V_R1/10)+(V_R2/15)=V_R3/24

rearrange for V_R3 -> V_R3=24((V_R1/10)+(V_R2/15))

V_R3=24/30*(3V_R1+2V_R2)

At this point I get stuck remembering how to work out V_R1 or V_R2, because I was always stronger on electronics than electrics. There's some way to get the proportion of the P.D over each resistor compared to the emf of the relevant source out of the equation, but I can't quite remember it.

You are correct in that they all have to add to zero.
The second link I posted has the method, which is a lot easier than me trying to explain it. I was good at it but it was so long ago since I've actually used it that I had to look up the specifics.

The problem is you've mixed together nodal analysis and loop analysis.