# Resistance of a loop of wire: confusing!

1. Jan 12, 2009

### mklein

Dear physics forums

I am an A-level physics teacher and I have built, admittedly, a VERY simple circuit. The problem is that I cannot explain its behaviour. It involves a metre long wire, twisted at the ends to form a loop, and incorporated into a circuit as shown below: (or attached)

http://mpklein.co.uk/loop_circuit.jpg

What I cannot understand is that as I move the slider to the right, the voltage over the length "L" decreases and the total current drawn from the power-pack increases.

Is it not true that the circuit could effectively be redrawn as a parallel circuit? I have always taught in lessons that the voltage over any branch in a parallel circuit remains the same as the supply? In this parallel circuit I would expect lengthening one branch to reduce the length of the other. Therefore I would expect different currents through the branches e.g. if the top branch is shorter than the bottom, then I would expect a larger current through the top branch and smaller current through the bottom. But I would expect these currents to add up to some total which stays the same. So I would have expected my ammeter reading to stay the same!

I have built and tested the circuit yet it continues to confuse me. Please help as I am feeling very useless and not worthy of teaching at A-level !

Many thanks

Matt Klein

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2. Jan 12, 2009

### Staff: Mentor

Sure.
The voltage across each branch must be the same.
Don't guess, calculate. Does the total resistance of the parallel circuit element (your wire loop) change as you move the slider? (Treat the branches as resistors in parallel.)

3. Jan 12, 2009

### mgb_phys

Quick thought experiment.
If you have 2 identical resistors (R) in parallel the total resistance is R/2
Now make one resistance halve the other

edit - Doc Al's suggestions (below) is better

Last edited: Jan 12, 2009
4. Jan 12, 2009

### Staff: Mentor

I would revise the thought experiment slightly to compare the total resistance of:
- two indentical resistors (R) in parallel
- a 3R/2 and a R/2 resistor in parallel

5. Jan 12, 2009

### mklein

Thank you for your reply Doc Al. I have tried your suggestion and I arrived at a total resistance of 3R/8.

So this shows us that adjusting the distribution of resistance in each branch DOES affect the total resistance.

But you did agree that the voltage over each and every branch in a parallel circuit should be the same (and equal to the supply voltage). So why was it that moving the sliding croc clip changes the voltage over the section "L"? Shouldn't the voltage over section "L" just equal the supply voltage? (if this is just a parallel circuit as we said). When I built the circuit the voltage over section "L" changed and did not equal the supply voltage !

Sorry to be dense about this, but I think I need people to spell it out to me very simply!

6. Jan 12, 2009

### mgb_phys

The voltage should equal the supply voltage for ideal components, with simple teaching lab moving coil meters it might not.
Can you try with a DVM instead , or just short out the Ammeter.

7. Jan 12, 2009

### mklein

Hi mgb_phys

I used two digital multimeters - one for the ammeter and one for the voltmeter

I am wondering if the change in voltage was caused by internal resistance. That is, as we draw different currents (by changing the combined resistance) we change the volts lost to internal resistance.

The circuit is supposed to be used as a good method for finding the resistance or resistivity of a wire (vary L, calculate the resistance in each case and then plot R/L against L). Could this method still be used even if the voltage is changing? (for whatever reason !)

8. Jan 12, 2009

### stewartcs

What values did you get? I suspect it would be a rather small difference but essentially equal to the source voltage.

CS

9. Jan 12, 2009

### mklein

Hi Stewart

I think it varied between about 0.7V and 1.1V. I don't really know how significant this is...

Does anybody have any thoughts on how such a circuit could be used to measure the resistance of the wire? (i.e. by producing a graph etc)

10. Jan 12, 2009

### skeptic2

There are several things you should do. The first is to measure the battery (or source) voltage with the tap at the different points on L. The voltage may drop slightly with increased current and if you are using a battery the voltage may even decay somewhat with time.

The second is to measure the voltage drop of the rest of the circuit including the ammeter. The small amount of resistance in the rest of the circuit may be affecting your results.

Yes this method can be used even if the voltage is changing but you will need to measure the battery voltage and the voltage drop (or I x R if you know the resistance) of all the other components simultaneously.

11. Jan 12, 2009

### skeptic2

What were the readings on the ammeter for the respective voltage readings?

12. Jan 12, 2009

### mklein

Hi Skeptic

Thanks for joining in on this thread!

Assuming then that the circuit is not overly affected by the internal resistance, how would I go about using this method to determine the resistance of the wire using a graphical method?

For example, if it was one straight piece of wire, being trimmed shorter and shorter, I could plot R against L and the gradient would be equal to p/A (where p=restivity and A= area, from the resistivity equation).

However I cannot visualise how this would be done with this particular circuit. But apparently it can !

13. Jan 12, 2009

### aniketp

I have simplified the circuit here. (forgive me, im bad at paint ;))
Let the net resistance be Rnet

(Rnet)-1 = (R1)-1 + (R2)-1

The net resistance will thus change if R1 and R2 change, though the sum remains constant. R1 and R2 change with the length as
R is proportional to L.

14. Jan 12, 2009

### mgb_phys

Doesn't it just add unnecessary complexity (using a loop in the first place to determine resistivity)?
Assume the resistance is R ohms/m and the total length of the loop is 1m
The resistance of the two arms is LR (top) and R+(1-L)R = R(2-L) bottom

So Rtotal = RtopRbottom / ( Rtop+Rbottom )
= LR R(2-L) / ( LR + R(2-L) ) = 2LR2 - L2R2 / ( RL+2R-RL )
Rtotal = (2LR - L2R )/2 = R(2L-L2)/2

edit - aniketp, that's not quite the same thing.

Last edited: Jan 12, 2009
15. Jan 12, 2009

### aniketp

This i guess is the net result, which is to be used in plotting the graph to find out the resistance

EDIT: at the risk of sounding stupid, what is not quite the same thing??

16. Jan 12, 2009

### mklein

Hello again folks, I have found out from a colleague, who knows less about this than myself, that the equation relating the resistance of the wire to its length, in terms of RESISTIVITY (p) is as follows:

R/L = -kL + p/A

where "R" is total resistance and "k" is a constant. This looks similar in ways to mgb_phys's work above. I think you might be on to something there...

17. Jan 12, 2009

### mgb_phys

Sorry my mistake, I thought your V was in a different place in the circuit, but redrawing it I see.

18. Jan 12, 2009

### mklein

Hi Aniketp

If anyones stupid here its me!

What I meant was that with a single straight piece of wire L refers to its whole length and the voltage is measured across its whole length. The current is flowing through that whole length so it is easy to see the relationship with "L"

This circuit confused me because the current is not even measured inside "L" !

19. Jan 12, 2009

### skeptic2

I would move the tap as far to the left as possible and measure the voltage. The resistance is voltage divided by current. Each leg of the loop would be twice that.
[Suppose you measure the resistance at 1 ohm, each leg would be 2 ohms and the total resistance of the loop would be 4 ohms.]
Since the resultant resistance for two resistors in parallel is Rr = R1*R2/(R1 + R2) and since as you move the tap you can calculate the resultant resistance from v / i, and since you know that Rr = R1 + R2 (4 ohms in the example above), R1 can be calculated from R1 = -R2/(1-R2/Rr).

20. Jan 12, 2009

### mklein

mgb_phys, what do you mean?

Do you mean that your maths earlier does not apply?

Does your maths match the equation that I posted two posts back?