## Circuits with two voltage sources

 Quote by gneill Vr is the potential across the 2Ω resistor in the complete circuit (with both voltage sources operating). So the top of the 2Ω resistor is at Vr above the ground potential. You can find the currents in the branches by doing a "KVL walk" from ground to the node where Vr sits. For example, for the left branch you have +3V - I1*(4Ω) = Vr. Solve for I1.
So what I did was I1 = (3V - 19/17V)/4Ω = 8/17A

I2 = (19/17V - 2V)/10Ω = -3/34A

I'm assuming Vr is universal from 2Ω to 4Ω and 10Ω. If not, how would I find Vr for the other resistors? using the same method just add them up?

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 Quote by Chaso So what I did was I1 = (3V - 19/17V)/4Ω = 8/17A I2 = (19/17V - 2V)/10Ω = -3/34A
Those currents look good.
 I'm assuming Vr is universal from 2Ω to 4Ω and 10Ω. If not, how would I find Vr for the other resistors? using the same method just add them up?
Vr is the potential across the 2Ω resistor only. To find the others, use KVL since you know the potential at ground is 0 and that at the top node is Vr. The potential across the given resistor in the path must be such that it makes sum of potentials along the path equal this total. Another method is to use Ohm's law. If you know the current through the 4Ω resistor is I1, then the potential drop across it must be I1*4Ω.

 Quote by gneill Those currents look good. Vr is the potential across the 2Ω resistor only. To find the others, use KVL since you know the potential at ground is 0 and that at the top node is Vr. The potential across the given resistor in the path must be such that it makes sum of potentials along the path equal this total. Another method is to use Ohm's law. If you know the current through the 4Ω resistor is I1, then the potential drop across it must be I1*4Ω.
Hm okay that makes sense but why are we using Vr for the 2Ω to solve for the current of the other 2 resistors?

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 Quote by Chaso Hm okay that makes sense but why are we using Vr for the 2Ω to solve for the current of the other 2 resistors?
Well, rather than use the currents that you've already calculated, this provides a separate check on your results (you were looking for ways to check your results). You can also compare these currents with the the currents through these resistors found by superposition.
 Hmm okay that makes sense so how would I solve for I1 and I2 without Vr?

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 Quote by Chaso Hmm okay that makes sense so how would I solve for I1 and I2 without Vr?
Well, that's what you were doing before .

There are a number of circuit analysis techniques that can be applied. You chose superposition. You could also have applied source transformations, mesh analysis, or nodal analysis...
 Hmm okay thanks for the wonderful help. I think I got the hang of this now. Hmm quick question would a fan be equivalent to a resistor?

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 Quote by Chaso Hmm okay thanks for the wonderful help. I think I got the hang of this now. Hmm quick question would a fan be equivalent to a resistor?
A motorized fan? No, not really. It'll most likely look like a combination of resistance (small) and inductance. Is it a DC motor an AC motor?