Book wrong? Ohms law, Graph, wire, slope change question

[Barclay]

Homework Statement

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 traveling at 45 degrees path. The x-axis is VOLTS. The y-axis is CURRENT

Homework Equations

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

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.

 

[Bystander]

doesn't even mention the slope of the graph

book answer says: "The current would rise ever more slowly

Reads pretty much in agreement with your summary/conclusion,

V getting larger and I decreasing the slope will be shallower

 

[SammyS]

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

Homework Statement

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 traveling at 45 degrees path. The x-axis is VOLTS. The y-axis is CURRENT

Homework Equations

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

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.

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

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

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.

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.

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.

 

[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.

 

[SammyS]

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

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.

 

[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

 

[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.

 

[Bystander]

a sloping downwards curve.

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

 

[Barclay]

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

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

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

 

[haruspex]

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

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.

 

[SammyS]

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

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.

 

[Barclay]

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.

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

 

[haruspex]

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.

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?

 

[insightful]

But I am not able to justify it by any simple argument.

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

 

[SammyS]

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?

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 R_static = V/I

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

 

[haruspex]

there are two standard ways to define reistance: Static_and_differential_resistance from Wikipedia

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

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

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/

 

[haruspex]

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

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.

 

[NascentOxygen]

It would follow that increasing voltage would increase power but reduce current.

Impossible, since power = voltage x current

 

[SammyS]

Impossible, since power = voltage x current

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

 

[NascentOxygen]

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

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.)

 

[NascentOxygen]

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

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.

 

[SammyS]

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.

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.

 

[insightful]

I find "there is no known material" to be a pretty simple argument.

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

Isn't a negative coefficient when the resistance decreases with increasing temperature? No one is saying that happens here.

 

[SammyS]

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.

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).

​

 

[haruspex]

Impossible, since power = voltage x current

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.

 

[haruspex]

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)

Isn't a negative coefficient when the resistance decreases with increasing temperature? No one is saying that happens here.

Quite so. Seems to me that NascentO has misinterpreted the discussion as concerning negative coefficients, which had not been mentioned at all.
And this discussion is absolutely relevant to the OP. Barclay wrote:

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

This triggered a torrent of replies saying the curve cannot go downward. I wrote one myself before realising my error and deleting it. As my post just above shows, it can go downward, and without involving any exotic materials.

 

[insightful]

Here's some data on tungsten filaments at a range of volts and amps:

http://donklipstein.com/incchart.html

And attached is a plot of the 100W bulb data.

 

[haruspex]

Here's some data on tungsten filaments at a range of volts and amps:

http://donklipstein.com/incchart.html

And attached is a plot of the 100W bulb data.

I was careful to specify an insulated filament. A filament in a lightbulb will radiate according to a fourth power law, easily overwhelming the effect of the power law of the resistance.

 

[insightful]

Yeah, I saw that, but we're only talking about temperature, right? At 160 volts, the nominal 750-hour filament life is down to less than 20 hours. The 3210K temperature is still quite a ways from the 3700K melting point; however, I found it interesting that the curve straightens out at the higher temperatures.

 

[NascentOxygen]

Quite so. Seems to me that NascentO has misinterpreted the discussion as concerning negative coefficients, which had not been mentioned at all.

Negative coefficient was not correct, it's the region of negative gradient.

A material where an applied voltage could correspond to either of two currents, depending on its recent electrical history? If there were such a material, that region beyond the current peak would amount to a negative resistance (i.e., a negative incremental resistance).

Complicating this discussion with thermal time constants by introducing the metal into essentially an oven with non-ideal thermal insulation is not appropriate in this thread.

 

[haruspex]

A material where an applied voltage could correspond to either of two currents, depending on its recent electrical history? If there were such a material, that region beyond the current peak would amount to a negative resistance (i.e., a negative incremental resistance).

Complicating this discussion with thermal time constants by introducing the metal into essentially an oven with non-ideal thermal insulation is not appropriate in this thread.

The question is about the effect of heating on the voltage/current relationship. If you eliminate the history then there's no heating. You have to assume steady state, that's all.
The question does not specify whether the filament is exposed, so radiating, or insulated, so losing heat in a manner proportional to temperature, or something between the two.
The OP is interested in what could happen, and the answer is that with an insulated filament of tungsten at a high temperature the gradient could go negative. The intuitive response we all made that it was impossible turned out to be wrong. Personally, I find this very interesting.

 

[Barclay]

Thanks for advice. I was lost when people started using differentiation maths. I picked bits out of posts that made sense and concluded "The gradient of the slope decreases".

 

[haruspex]

Thanks for advice. I was lost when people started using differentiation maths. I picked bits out of posts that made sense and concluded "The gradient of the slope decreases".

That's a good answer, leaving open the question of whether it could go negative. Saying that it could go negative could lose you marks, even though it turns out to be true.

 

[Barclay]

That's a good answer, leaving open the question of whether it could go negative. Saying that it could go negative could lose you marks, even though it turns out to be true.

That's such a relief. People talk in riddles so much. With every question I ask there's a riddle to the answer. When I reply with my answer they give me further riddles to tell me what they think of my answer.

 

[haruspex]

That's such a relief. People talk in riddles so much. With every question I ask there's a riddle to the answer. When I reply with my answer they give me further riddles to tell me what they think of my answer.

That's inevitable in a forum where the objective of the responders is to help you develop more ways of thinking about a problem.