I-V graph of two wires and temperature

[shihab-kol]

Homework Statement

Homework Equations

## V=IR##

The Attempt at a Solution

This is an I-V graph so the slope must represent ##\frac{1}{R} ##
Now,
$$ R∝Temperature$$ {This is assuming that the wires are of metal}

Now , since the graph is of ##\frac{1}{R} ## , the steeper the slope the higher I and lower R

Thus, wire 2 has a higher temperature.

Is my solution all right?

 

[Asymptotic]

What is the definition of electrical power?
How does this relate to wire temperature?

 

[berkeman]

Thus, wire 2 has a higher temperature.

Is my solution all right?

Looks right to me!

 

[haruspex]

Thus, wire 2 has a higher temperature.

No. You have only shown it has the higher resistance. Why would that mean it has the higher temperature?
Are you sure you have provided all the information in the question?
Try to answer @Asymptotic's questions.

 

[berkeman]

Why would that mean it has the higher temperature?

R∝Temperature

Are you thinking that the question is more complicated than just 2 similar wires at different temperatures?

What is the definition of electrical power?
How does this relate to wire temperature?

What if the wires are in two different ovens at different ambient temperatures, and that's why they are at different temperatures? Maybe the OP needs to tell us if there is more information supplied. My assumption is that the 2 wires are at different temperatures for reasons other than just self-heating...

 

[haruspex]

My assumption is that the 2 wires are at different temperatures for reasons other than just self-heating...

Maybe I'm too influenced by a recent question @shihab-kol posted. It dealt with very elementary relationships between P, I, V and R. So I don't read it as anything to do with the way resistance varies with temperature.
Whatever, more information is needed to answer the question.

Edit:
If it is to do with the increase of resistance with temperature,

1. We should be told that the wires are identical.
2. The graphs should not be straight lines.

If it is to do with the increase in temperature with power generated,

1. We should be told that the ambient conditions are the same.
2. (Writing out my second point might give too much away)

 

[shihab-kol]

No,no the question is reproduced exactly, nothing is left out.
Well, sorry for the delay (I have my exams)

I think I get it.
Yes, I have shown the difference in resistance without thinking about its implications.
Well, resistance is a constant here since both are straight slopes.
So, in
$$H^2=I^2RT $$
##R## is a constant for both the wires
Thus, for the same time ##T##
$$H^2∝I^2 ~ ⇒~ H∝I$$
Now, since ##I_1\gt I_2##
Then, ##H_1\gt H_2##
So,now is it all right??

 

[kuruman]

This problem does not specify the conditions for the comparison. Clearly the resistances are different, but what about the voltage across and the current through each? The power dissipated is P = IV. I see three cases.

Case I, the voltage is the same.
Consider a point on line 1. Drop a perpendicular to the voltage axis. It intersects line 2 at a point below. You get two right triangles, the area of each being half the power dissipated by each resistor. Clearly, resistor 2 dissipates less power (its base is the same - its height is smaller) and might be assumed to be at a lower temperature than resistor 1.

Case II, the current is the same.
Consider again a point on line 1. Now draw a perpendicular to the current axis that intersects line 2 at a point to the right. Again two right triangles are formed, the area of each being half the power dissipated by each resistor. Here resistor 1 dissipates more power (its height is the same - its base is larger) and might be assumed to be at a higher temperature than resistor 1.

Case III, neither the current nor the voltage are the same.
In this case anything goes.

 

[shihab-kol]

This problem does not specify the conditions for the comparison. Clearly the resistances are different, but what about the voltage across and the current through each? The power dissipated is P = IV. I see three cases.

Case I, the voltage is the same.
Consider a point on line 1. Drop a perpendicular to the voltage axis. It intersects line 2 at a point below. You get two right triangles, the area of each being half the power dissipated by each resistor. Clearly, resistor 2 dissipates less power (its base is the same - its height is smaller) and might be assumed to be at a lower temperature than resistor 1.

Case II, the current is the same.
Consider again a point on line 1. Now draw a perpendicular to the current axis that intersects line 2 at a point to the right. Again two right triangles are formed, the area of each being half the power dissipated by each resistor. Here resistor 1 dissipates more power (its height is the same - its base is larger) and might be assumed to be at a higher temperature than resistor 1.

Case III, neither the current nor the voltage are the same.
In this case anything goes.

Perhaps you are getting to ahead .
Actually we have not yet gone into all those details.
Also the heat formula in my previous post is wrong. It should be ##H## not ##H^2##

 

[kuruman]

Actually we have not done gone into all those details.

What details? It seems you already know that
1. H = V²T /R
2. H = I²RT
3. V = IR
if you replace R = V/I from (3) into either (1) or (2), you get H = IVT.

 

[Windadct]

This problem does not specify the conditions for the comparison. Clearly the resistances are different, but what about the voltage across and the current through each? The power dissipated is P = IV. I see three cases.

Case I, the voltage is the same.
Consider a point on line 1. Drop a perpendicular to the voltage axis. It intersects line 2 at a point below. You get two right triangles, the area of each being half the power dissipated by each resistor. Clearly, resistor 2 dissipates less power (its base is the same - its height is smaller) and might be assumed to be at a lower temperature than resistor 1.

Case II, the current is the same.
Consider again a point on line 1. Now draw a perpendicular to the current axis that intersects line 2 at a point to the right. Again two right triangles are formed, the area of each being half the power dissipated by each resistor. Here resistor 1 dissipates more power (its height is the same - its base is larger) and might be assumed to be at a higher temperature than resistor 1.

Case III, neither the current nor the voltage are the same.
In this case anything goes.

If this "instructor" is implying a voltage perspective, Voltage is the same, I would like to take have words with them...

 

[haruspex]

$$H^2∝I^2 ~ ⇒~ H∝I$$

That is true for each wire individually, but the constant of proportionality will be different.
You cannot use this to compare the wires.

Now, since ##I_1\gt I_2##

You have no information relating the current in one to the current in the other.
The graphs show a range of currents and voltages for each.
Can you find a point on the upper line and a point on the lower line such that the power dissipated is greater for the upper line? Is there anything in the question statement that rules out this combination of points?
Can you find a point on the upper line and a point on the lower line such that the power dissipated is greater for the lower line? Is there anything in the question statement that rules out that combination of points?

 

[shihab-kol]

I am sorry guys . It seems the person who have me this question had got something wrong.
Actually,the wire is the same, there has been no change in length or area.
So the graphs are of the same wire of same length,same area except temperature.
In that case,perhaps the first post should do.
I am pretty sorry for all this trouble.

 

[haruspex]

I am sorry guys . It seems the person who have me this question had got something wrong.
Actually,the wire is the same, there has been no change in length or area.
So the graphs are of the same wire of same length,same area except temperature.
In that case,perhaps the first post should do.
I am pretty sorry for all this trouble.

Ok, except that as I wrote in post #6 the graphs should not be straight. As the current increases more power will be dissipated, raising the temperature of the wire for the same ambient temperature.

 

[Windadct]

"the wire is the same, there has been no change in length or area." Now this is simple...

Conventional wire has Positive Temperature Coefficient - so wire with higher temp has higher resistance, #2.

A sufficiently large wire could negligible heating due to current over this range - so without additional data IMO heating due to the current should not be included in this analysis.

 

[David Lewis]

As the current increases more power will be dissipated, raising the temperature of the wire for the same ambient temperature.

Good point but sometimes these graphs ignore the time dimension and just show the static resistance. That is, if you apply a voltage to a wire of a given temperature, what is the initial current before the current heats up the wire further?

 

[kuruman]

I think folks read into this question more than there is. I agree that in general the graphs should not be straight if a wide range of currents and voltages are shown on the graph. However, the scale is not shown so the graphs could be straight, if say the voltage range is 0 - 1 mV. I think the point of the problem is, "given that the graphs are straight lines, how do the temperatures of the wires compare?" The answer is it depends: If the wires are connected to the same battery in parallel, then the wire with the lower resistance dissipates more power (P = V²/R); if they are connected to the same battery in series, then the wire with higher resistance dissipates more power (P = I ²R); if they are connected separately to different batteries, anything goes.