V, I, R in series circuit confusion

1. Nov 25, 2012

JoeN

I'm doing coursework on a circuit experiment we performed in class. I'm pretty sure the results are wrong or flawed, but we were testing the effect of increasing the length of wire on the resistance. I had a voltmeter about the wire, a cell, an ammeter, and completed the circuit by attaching the croc. clips to the wire...

Of course, the resistance should increase and the current decrease, but should the voltage increase, as it did in my experiment? And should it increase enough to also make the current increase despite the higher resistance? I know that more work is needed to push the charge through the wire if you increase its length, so in theory the wire's potential difference should increase (?) but I'm not so sure the current should increase... This didn't occur in the preliminary data I carried out either.

I'm not sure about the voltage of the cell. Well, I know it was 1.5V but it had been used enough to make that irrelevant.

Oops, wrong forum. Sorry about that

2. Nov 25, 2012

K^2

It should increase slightly, yes.

No. Current should decrease.

Can you post your measurements? These could help. I can think of several reasons for your measurements, but I might be able to eliminate some of them if I saw your data.

Now, why should the voltage increase as you increase the resistance? Imagine that you have 2 resistors of 10 Ohm each in series with 3V applied across. You have 20 Ohms of total resistance, giving you current of 0.15A. And that gives you 1.5V across each resistor. Now say, I increase resistance of one of these to 20 Ohm. That means total is now 30 Ohm, current drops to 0.1A, and the 20 Ohm resistor has 2V across it now.

Wire you are measuring across isn't the only thing with resistance. There are other wires and battery's own resistance. So when you increase resistance of the wire, more of the battery's applied voltage is dropped across the wire. Hence, higher voltage measurement. But the current should only decrease. Unless, something else is going on.

3. Nov 28, 2012

JoeN

My preliminary data:
length of wire V A1 A2 mean ohms
10 0.28 1.20 1.25 1.23 0.23
20 0.24 1.21 1.22 1.21 0.20
30 0.32 1.10 1.16 1.13 0.28
40 0.30 1.11 1.13 1.12 0.27
50 0.39 0.98 0.97 0.98 0.40

This experiment followed my expectations: V increased slightly, A decreased, R increased (although I think the R and A should have increased and decreased a lot more?). And thanks for that explanation.
L.O.W V A1 A2 mean ohms
10 0.06 0.64 0.68 0.66 0.09
20 0.08 0.62 0.65 0.64 0.13
30 0.10 0.70 0.72 0.71 0.14
40 0.12 0.72 0.71 0.72 0.17
50 0.14 0.77 0.75 0.76 0.18

And this is where it messes up. I performed both experiments in the same way, so my initial conclusion was a mistake in the circuit or faulty equipment. Is there another conclusion?

4. Nov 28, 2012

Drakkith

Staff Emeritus
Shouldn't your voltage equal the applied voltage from your cell? Or are you not able to develop voltage very well across a conductor?

5. Nov 29, 2012

davenn

true, you cant get more voltage out than what you put in to a passive circuit like that

Dave

6. Nov 29, 2012

K^2

Both sets look very strange.

First of all, it looks like the battery might be dying. That alone could explain the increases in current. If you give a dying battery a bit of time, ti recovers a bit. So the internal resistance can decrease a bit. That could explain your data.

Second part, I don't think your contacts were very good. You simply shouldn't be getting that kind of resistance dependence on length. However, if there was significant additional resistance on clips, it fits.

There are two things you can do to significantly improve this experiment.

1) Put a 5-10 ohm resistor in series with the wire. This will reduce the current through the whole thing, and tax the battery a lot less.

2) Get more data points with different lengths. The range of 10-50 (mm?) is fine. Then plot all these points resistance vs length. Try to fit a curve that looks like R = k*L + R0 through these points for some constants k and R[/sub]0/sub]. Assume that this R0 is due to bad contacts. This way, you can predict the "real" dependence of resistance on length via the remaining terms, R = k*L. Alternatively, you could just subtract off this R0 from your data at this point.

7. Dec 1, 2012

JoeN

I'm not in a position to perform the experiments again, unfortunately, I just have to evaluate the data.

1) Battery dying is likely. They have probably been used many times before I did the experiment. So a dying battery can eventually recover some voltage and decrease in resistance for a higher current?
2) "You simply shouldn't be getting that kind of resistance dependence on length" - you mean the resistance is too great a change, or too small a change? Is the resistance not proportional to the length of the wire?

8. Dec 1, 2012

K^2

Yes, that's what bothering me about resistance measurements. There could be any number of reasons for this, but I suspect contact resistance. In which case, you have a somewhat noisy bias to your resistance measurement.

Linear regression might still give you a reasonable fit. Though, more data points would help a lot.

Do you by any chance know the material and diameter of the wire? And the lengths, are they in mm? It'd be easy enough to compare results to expected value if these things are known.