voltage drop

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).

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

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.

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

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 :-)

True, but still convoluted. I suggest everyone goes back and reads Thevenin on equivalent generators and source impedance. Cheers, Roger

-------------------------------------------- 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

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!!

----------------------- 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.

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.

On Sun, 29 Apr 2007 21:49:10 -0400, BFoelsch Gave us:

And Ohm's law applies at any point, and across any given element in the circuit at ALL times!