# Homework Help: Book wrong? Ohms law, Graph, wire, slope change question

1. Aug 16, 2015

### Barclay

1. The problem statement, all variables and given/known data
Hello.

The resistance of a wire increases when the temperature increases.
How would this cause the shape of the graph to change? Explain why. The graph shown is a straight line graph with the line travelling at 45 degrees path. The x-axis is VOLTS. The y-axis is CURRENT

2. Relevant equations
This is High School Physics and the chapter in the book is talking about Ohms law and V=IR

3. The attempt at a solution
In my opinion the gradient of the line should decrease. V = IR
So V is directly proportional to R. So as R increases V will increase too.
But I is indirectly proportional to R. So as R increases I will decrease.

So with V getting larger and I decreasing the slope will be shallower. Maybe even a downward curve.

The book answer says: "The current would rise ever more slowly as the potential difference increased. This is because increasing the current makes the wire hotter, increasing the resistance".

The BOOK ANSWER makes no sense to me. it doesn't even mention the slope of the graph.

Please advise. Thank you.

2. Aug 16, 2015

### Bystander

Reads pretty much in agreement with your summary/conclusion,

3. Aug 16, 2015

### SammyS

Staff Emeritus
Your book is correct. It just doesn't give a very detailed explanation.

You will no longer have a direct proportion.
Normally the direct proportion is stated as V is directly proportional to I, with R being the constant of proportionality.

If R is not constant, but varies with I, there is no longer a direct proportion. In this case R is the slope of the line tangent to the V vs. I curve, at any particular value of I
No. At any particular value of V, I is decreased. This moves the graph to the left from where it would be with constant R. That makes it steeper.
If you want to make yourself a more concrete example, Let R = R0 (1 + 0.1(I) ).

Then look at V = IR with this R.

4. Aug 16, 2015

### Barclay

Thanks for answers so far.

ByStander ... are you saying I'm correct or incorrect? Couldn't quite figure out your comment.

SammyS ... I knew R was no longer constant but it was all I could do to come to an answer. I shouldn't have done that.

Anyway I was thinking ... the question says "resistance of a wire increases when the temperature increases" so this means that there is resistance to the flow of current therefore the gradient of the slope will decrease. There will be less current able to flow through the wire and the hotter it gets the more the resistance will be rising and the less the current.

So really a downward slope. You said "steeper"

If I'm wrong its damaged all my fragile understanding of wires and electrons and coulombs and resistance.

5. Aug 16, 2015

### SammyS

Staff Emeritus
Current is plotted on the x-axis, Right?

So if you get the a given voltage with smaller current, then that moves the curve to the left. Therefore Steeper.

6. Aug 16, 2015

### Barclay

No SammyS the current is on the vertical y-axis and volts plotted on the horizontal x-axis.

So I assume that I'm correct in what I said ... that there will be a sloping downwards curve.

Please SammyS or someone confirm so I don't have a sleepless night

Last edited: Aug 16, 2015
7. Aug 16, 2015

### insightful

It's hard for me to see how the curve would ever go downward, i.e., an increase of applied volts will always cause an increase of amps.

8. Aug 16, 2015

### Bystander

... a decrease in slope .... It's still greater than zero.

9. Aug 16, 2015

### Barclay

The graph ORIGINALLY is a straight line graph with the line travelling at 45 degrees path.

So shall I just say a "decrease in the gradient of the line"?

10. Aug 16, 2015

### haruspex

I had the same intuitive response as all the other responders, that it cannot slope downwards. But I am not able to justify it by any simple argument. Certainly the power dissipated cannot decrease, as that would lead to a lower temperature and reduced resistance, but power is IV. That puts a limit on how fast the I/V graph could slope down, but doesn't rule it out.

11. Aug 16, 2015

### SammyS

Staff Emeritus
If that's the graph you're referring to, with the current on the vertical y-axis and volts plotted on the horizontal x-axis, then:
I = (1/R) V​

The slope is 1/R. The slope will decrease, but 1/R is still positive so the graph increases, just at a slower & slower rate.

12. Aug 16, 2015

### Barclay

So it is my understanding that the slope of the straight line graph will decrease as the resistance increases (with the rising temperature) BUT the decline will be slower and slower because it will mathematically (or physically) not be possible to curve downwards. In the end the wire will just melt when it gets too hot

13. Aug 16, 2015

### haruspex

I nearly fell into the same trap. The resistance is $\frac VI$, not $\frac{dV}{dI}$. If you join some point on the curve to the origin with a straight line then the slope of that is 1/R.
Did you read my post #10?

14. Aug 16, 2015

### insightful

So, what known "wire" would have a negative dI/dV?

15. Aug 16, 2015

### SammyS

Staff Emeritus
I read it -- not too carefully.

My post (#11) was an attempt to correct what I stated in my previous posts. I had originally misread OP and was thinking of the current as the independent quantity.

If R doesn't vary, then the slope for I = (1/R) V is 1/R .

However, if R does vary, then there are two standard ways to define resistance: Static_and_differential_resistance from Wikipedia

1. Static resistance ( chordal resistance ) is given by Rstatic = V/I

2. Differential resistance (dynamic or incremental resistance) is given by Rdifferential = dV/dI​

16. Aug 16, 2015

### haruspex

That's not how I read that reference. It defines differential resistance and points to circumstances in which that might be the more useful characteristic (note it does not list temperature dependence as such a circumstance), but I see no suggestion that this is an alternative definition of the unqualified term 'resistance'.

Now, there probably is no such material, but suppose the wire is insulated so that temperature is roughly proportional to power, and that (in some part of the curve) resistance rises according to some power > 1 of temperature. It would follow that increasing voltage would increase power but reduce current.

Edit: Just found this.... resistivity of tungsten rises as $T^{1.2}$
http://physics.info/electric-resistance/

Last edited: Aug 16, 2015
17. Aug 16, 2015

### haruspex

I'm not saying there is any such material, merely that there is no simple argument to rule it out. Pls see the theoretical example in my last post.

18. Aug 17, 2015

### Staff: Mentor

Impossible, since power = voltage x current

19. Aug 17, 2015

### SammyS

Staff Emeritus
That doesn't make it impossible for current to decrease while power and voltage increase. Remember, the ratio V/I is not constant here.

20. Aug 17, 2015

### Staff: Mentor

That's correct. With metals having a positive thermal coefficient of resistivity, the graph will maintain a positive slope, but the slope decreasing as current increases. (Meaning it will no longer be a straight line graph.)

21. Aug 17, 2015

### Staff: Mentor

But this homework question is seeking help with the case where "The resistance of a wire increases when the temperature increases."

Until the OP has got this clear, a parallel discussion concerning negative coefficients is just confusing the picture.

22. Aug 17, 2015

### SammyS

Staff Emeritus
I'm pretty sure that for a metallic wire, the current will not decrease with increasing voltage, but not for the reason you gave in post #18.

I also agree that this parallel discussion concerning negative coefficients is just confusing the picture.

23. Aug 17, 2015

### insightful

I find "there is no known material" to be a pretty simple argument.
Isn't a negative coefficient when the resistance decreases with increasing temperature? No one is saying that happens here.

24. Aug 17, 2015

### SammyS

Staff Emeritus
I think what was being discussed was the possibility that $\ dI/dV\$ could perhaps be negative, but in a way in which R is increasing. (1/R would decrease.)

As NascentOxygen points out, this parallel discussion concerning negative coefficients is confusing the the overall discussion (for the OP).

25. Aug 17, 2015

### haruspex

Consider $\frac VI = R = \rho \theta^{\mu}$, where theta is absolute temperature. If the wire is insulated then in steady state $\theta = \theta_0+cIV$ for some constant c. The condition for $\frac {dI}{dV}<0$ turns out to be $cIV(\mu-1) > \theta_0$. So it comes down to the question of whether mu can exceed 1. According to the link I posted, for Tungsten it is 1.2.

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