# Solving for normalised current when given only normalised resistance?

1. Nov 4, 2011

### davidc95

For an experiment I am conducting in a grade 11 physics course, I am attempting to find the effects of resistivity on current density. To ensure accurate results, I attached, in seperate trials, the maximum length possible of 5 different types of wire (with varying material composition but constant thickness) to a circuit which could measure resultant voltage drop/gain and current.

As I used different lengths of wire for each trial there is the additon of an unwanted length variable, which to control, must be normalised. In doing this, I calculated the normalised resistance of each wire, assuming a constant length and thickness for the wire, with only resistivity varying.

In order to calculate current density, the current through the circuit is required, which, along with voltage, would have been affected when resistance was normalized. Is there any way to calculate normalised current when only normalised resistance has been determined?

Thanks.

2. Nov 4, 2011

### Simon Bridge

If I have understood your description, you are modelling a real circuit with real wires and whatnot by an ideal circuit with an ideal circuit ... where the effect of the real wires is represented by a small series resistor of resistance r (you have the dimensions and the resistivity right?. So you can find r.)

Your test resistor is one of a bunch of rods, with a resistance R. You have a voltmeter across R, and you have measured VR for different applied voltages V.

The current through each resistor is the same, given from Ohm's law taking into account the total resistance in the circuit.

The current density in the test resistance is the current divided by the cross-sectional area, which is known.
http://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity

3. Nov 4, 2011

### davidc95

That is correct. Multimeters were placed in such a way that they could measure both current and voltage. 4 trials were conducted, with a separate wire of varying length and resistivity used in each. I chose to use varying lengths as some wires with minimal length were producing imprecise readings so I decided to use for each the maximum amount of supplied wire possible to increase resistance so a more reliable reading from the multimeters could be read. Unfortunately, I did not consider that this would make it difficult to calculate normalised current, as right now I have only calculated normalised resistance and in the equation (I = V/R) both V normalised and I normalised are unknown.

4. Nov 4, 2011

### Simon Bridge

Are you using an AC voltage source or something?
Are the lines very long?

the normalized current is just the current you measured divided by the characteristic admittance - but your circuit has no reactive components so why would you not just use Ohm's Law on the things you've measured?

What is the aim of the experiment?

5. Nov 4, 2011

### davidc95

Ok, so you are suggesting that to find normalised current (which is all I need for the current density equation), I can just divide the original current by the length of wire it was transmitted across to find A/m (amperes per metre), which is normalised current that standardizes all values?

Also, here are my approximate wire lengths:

-21m
-2m
-6m
-4m

The aim of the experiment is to identify a relationship between resistivity and current density (which in my hypothesis I have already determined is unlikely).

6. Nov 4, 2011

### Simon Bridge

Don't you have a table of data for your test loads which includes a measurement of the voltage across each load and the current through them?

Note:

p=E/J

but E=V/L - eg. the volt-drop divided by the length
J = I/A - eg. current over area.
... both of these things you have.

What we'd normally do is have you plot V/L against I/A and find the slope.
You are saying you have to plot p against I/A=J ... presumably at the same volt-drop but different materials?
(What have you "determined is unlikely"?)

Did you have to design the experiment yourself, or were you given instructions.
Given that aim: I don't understand why you feel you need to normalize the current at all.

Last edited: Nov 4, 2011
7. Nov 4, 2011

### davidc95

OK, just to clarify, I will re-discuss the concept of my experiment.

In an Extended Experimental Investigation (EEI), I am investigating the effects that wires of various resistivity have on current density. Four different wires were used in four different trials, iron, tinned copper, constantan and nichrome, with the current and voltage drop in the circuit recorded for each. I attempted to retain the same length of wire for each separate trial, but the DC power pack I used for it would overload under lower wire lengths, so for each wire I used the maximum amount of length possible to prevent this. As such, a variable I want to control, wire length, had been manipulated.

As I am comparing resistivity against current density, wire length and thickness must be kept constant. To achieve this, I have attempted to normalise the resistance figures I calculated from (R=V/I) for a wire length of 10m and thickness of 1mm, leaving only resistivity as an manipulated variable. However, this new normalised resistance will have affected the values of V and I (in R(normalised) =V/I) and these new values are unknown.

Is there anyway to determine the normalised current from the normalised resistance, as more then two variables are unknown for this normalised Ohm's Law equation?

8. Nov 4, 2011

### Staff: Mentor

Suppose that you have some desired standard length L for the wire, and you've used a longer piece with excess length X. The total length of the wire is then L + X.

If you assume that the wire is uniform in cross section and material then L and X together will have some resistance R = k(L + X) for a linear resistivity constant k. Individually the sections of wire have resistances kL and kX.

Now, with voltage V applied across the full length of the wire the two resistances kL and kX will form a voltage divider. The voltage across the Standard Length will be given by:
$$V_L = V \, \frac{L}{L + X}$$
Since the L and X are in series the current is the same through each. So the resistance of the Standard Length of wire will be VL/I.

Does this diagram correspond to your setup?

File size:
2.5 KB
Views:
306
9. Nov 4, 2011

### Simon Bridge

OK - now I get you: you seem to be using the word "normalize" in a way I'm not used to.

Lets see if I have understood you ...

wire, length, diameter, voltage, current

wire type and diameter is read off the label while the rest are measured directly.
You need to know what the current would have been if the voltage was the same amount for each over a 10m length of the same wire. Is this correct?

You cannot do this without assuming a relationship between the current density and the resistivity. Your experiment is, as you suspected, broken.

You could have improved your original design by adding a voltage divider with a rheostat - you'd use the rheostat to set the voltage that was dropped across the test-load while a small known load stops the PSU overloading.

But one thing puzzles me - how were you expecting to measure the resistivity of the wires without assuming a relationship between resistivity and current-density?

If you were planning on using an ohmmeter, getting the resistance, then using p = RA/L to get resistivity, then you should check how an ohmmeter measures resistance: you'll find it uses the very relationship you are investigating. It's the same if you were to look the values up in a book of tables or online.

This may not be too bad though - by 11th grade you have probably confirmed Newton's force law somehow and that experiment has the same flaw.

You should repeat your experiment, collecting 10 pairs of V and I data for each wire and plot I/A vs V/L for each all on the same axis so you can compare them. Calculate the slope of each (with uncertainties if you know how to do that) and show that your result is (or is not) consistent with some official values of the resistivity.

That would be quick.
Or you could try investigating how resistivity changes with temperature.
That would be a very fine extended investigation.
And, of course, you could cheat :)

If you are short of time, what is the best you could do with your data without cheating?
Gneill's comments are apropos for this - though you still don't have a fixed electric field.
You could even pretend that was the aim all the time ... my cat insists that is not "cheating".

10. Nov 5, 2011

### davidc95

Ok, thanks for the feedback.

gneill's digaram is pretty much correct, except there is no resistor ('r' in the diagram). Are you suggesting that to find normalised voltage I can use the equation you stated, and then substitute that into I(normalised) = V(normalised)/R(normalised) to identify I(normalised)?

And Simon, yes, my experiment is flawed :) . I found that current density is dependant on more than just resistivity, somehow I'm hoping to justify and consider this in my analysis.

11. Nov 5, 2011

### Simon Bridge

You won't be able to get your original control back.
Note, if you "normalize" to 1m, the normalized voltage will be the electric field. You would then "normalize" your current by dividing it by the cross-sectional area of the test-wire... which would be the same as the current density.

There does not seem to be any point, in the context of your experiment, of normalizing to 10m by Gniell's suggestion because it's basically what I just outlined. (Remember, the current through all sections of the wire is the same?)

I suspect there is a deeper problem with your experiment - how do you intend to get the resistivity without exploiting the very relationship you want to test? That bit is missing from your description.

srsly: best practice from here is to write down what you've done and what you have discovered (that V and I are both important) and describe how you'd change things to do better next time. Hand that in (if you have time: redo, using your refined method).

I've graded thousands of these things. I cannot speak for your instructors but that would get high grades from me. A perfect paper could mean the author just got lucky... stuff-ups display your learning, reasoning, understanding and work. Well written it could be an A since failure is the most important part of science.

Anyway - I'll leave you to it.
Let us know how you did.

12. Nov 5, 2011

### Staff: Mentor

Resistor 'r' is meant to indicate that the circuit wiring itself will have some resistance. It won't play a significant role in your lab if the voltmeter is connected directly to, or at least very close to, the test-wire.

The current in a series circuit is everywhere the same. So your In will be the the same as your measured current. What is different is that the voltage across the standard length segment of wire isn't the end-to-end voltage that you actually measured, but a fraction of it as given by the formula. Note that this voltage is then a derived value, not a measured one; Its validity for your purposes depends upon making reasonable assumptions and arguments about Ohm's law, uniformity of the wire, symmetry, Kirchhoff's laws, etc.

You can always derive values from the measured data as part of your analysis provided that you can justify them. This may involve a short discussion and derivation in the Theory section of your lab write up.