voltage drop

If you have a series circuit with two resistors on that circuit, and your power supply is a 6 volt battery, then given that both resistors are 1 ohm, you would end up
with 3 amps. This means that across each resistor, you would have a 3 volt drop. My question is this, the voltage drop represents some voltage lost, right? If that is the case, then by the time current has moved past the second resistor, you would have 0 volts! In which case, at this given point past the resistor, suppose you had another inch of wire for the free electrons to move across, they wouldn't be able to! Because we no longer have any voltage.
Is my thinking wrong? What is at conflict, is this. This is the model taught in my electronics class, yet realistically, this can't happen as those free electrons do find their way to the positive end of the battery so that the chemical reaction in the battery can take place again, to give up free electrons all over again. So what is the problem here?
-- conrad
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you are wrong for two reasons:
1) there is either a real, finite, resistance to the wire in which case you have additional "resistors" in your voltage divider affecting your calculation.
2) there is 0 resistance in the wire and therefore it takes 0 volts to move current (superconductor).

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Ah. So if after the two resistors, you experienced two voltage drops, both 3 volts, adding up to the total voltage of a 6 volt battery, then after the second resistor, when you have zero volts, and because the resistance is really low in the wire, current is still able to flow without the voltage?
-- conrad
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think of water pipes with a couple of restrictions; the restrictions represent the resistors. in the return leg of the pipes (downstream side) your pressure relative to the outside of the pipe is low or zero and yet the water still flows.
wrote:

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That is not what he said. In case 1, the wire adds a small resistance (let's say 0.1 Ohms for our example). Therefore, the current is now 6/2.1=2.857A, and the voltage across the resistors is 5.714V. The remaining voltage is across the wire.
In case 2, the wire is "ideal" with zero ohms resistance, and there is no voltage across it. It therefore doesn't exist, and the circuit is the same as the original. The current is 3A and the voltage across the resistors is 6V.
No matter what the wire resistance, by Ohm's law there will always be some voltage across it if current is flowing.
Ben Miller
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Benjamin D. Miller, PE
B. MILLER ENGINEERING
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True, but still convoluted. I suggest everyone goes back and reads Thevenin on equivalent generators and source impedance. Cheers, Roger
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----------------------------

-------------------------------------------- Go back further than Thevenin. What Ben stated has nothing to do with Thevenin as it is more basic than Thevenin
Thevenin is based on
a) Kirchoff's laws b)Linearity (which, strictly speaking, is what makes Ohm's Law true. )
a) is the main thing in consideration of series or parallel circuits -or any combination. Thevenin is secondary- a very useful circuit analysis approach but hardly fundamental and easily developed from a), b) above.
I would suggest that you, not Ben, are the one introducing unnecessary convolution.
Cheers
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Don Kelly snipped-for-privacy@shawcross.ca
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The problem with this whole silly thread is that the OP (and everbody else) keeps changing the simplifying assumptions.
The initial premise, that two 3 ohm resistors in series draw 2 amps at 6 volts assumes NO resistance in the connecting wires. Then he changes the premise and starts to argue about the effect of the wires. If they are real wires then the current simply can't be 2 amps and the voltage across the resistors can't be 3 volts.. If the wires are real and the current is 2 amps then the source can't be 6 volts.
And on and on it goes. This stuff was all figured out years ago, guys!!
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----------------------------

----------------------- You are right- it is a silly thread and I have joined in the silliness. Mea Culpa However, there are a lot if silly statements going around -possibly because of misinterpretation of what others have said. If I have misinterpreted- then I am sorry but I do find that calling on Thevenin is not improving the situation nor is calling on Ohms Law to explain that in a series circuit, the current is the same at evey point in the circuit.
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Don Kelly snipped-for-privacy@shawcross.ca
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On Sun, 29 Apr 2007 21:49:10 -0400, BFoelsch

Of course, my memory is twisted. Initial premise was that 2 one ohm resistors in series across 6 volts produce a current of 3 amps.
The rest of my criticism stands, however.
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On Sun, 29 Apr 2007 21:49:10 -0400, BFoelsch

And Ohm's law applies at any point, and across any given element in the circuit at ALL times!
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In a real-world circuit there is NEVER "no resistance" (0 Ohms). There is always a finite resistance, and an accompanying voltage drop. This may be micro-volts, pico-volts, or whatever, but as long as any current flows there will always be a voltage.
Zero Ohms is not practical. We can get rather close to it, but never zero. There is therfore always a voltage difference, as long as current flows. Electrons are also a lot more sensitive to voltage than a 1K-Ohm AVO. They still get the general idea as to where to go :-)
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