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Why does the temperature of a current carrying wire increase? watch

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    Quick question: Why does the temperature of a current carrying wire increase?

    I am asking this because I am doing an experiment where I increase the temperature of a copper wire by increasing the voltage I pass through it. Except I don't fully understand the reason why the copper wires temperature increases.

    I first thought it was due to the resistance of the wire, but shouldn't the resistance of the wire remain constant throughout the experiment? Which would mean the temperature would also remain constant, which is not the case.

    Any help would be much appreciated!

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    (Original post by YdLeet)
    Quick question: Why does the temperature of a current carrying wire increase?

    I am asking this because I am doing an experiment where I increase the temperature of a copper wire by increasing the voltage I pass through it. Except I don't fully understand the reason why the copper wires temperature increases.

    I first thought it was due to the resistance of the wire, but shouldn't the resistance of the wire remain constant throughout the experiment? Which would mean the temperature would also remain constant, which is not the case.

    Any help would be much appreciated!

    Thanks
    Let's not consider the change in resistance of the wire due to change in temperature for a while; we'll deal with it in a moment.

    Power dissipated in a wire is I^2R=V^2/R. If you increase voltage, higher current goes through. This means that more power (heat) is dissipated. Every second the surroundings "takes" a certain amount of heat from the wire - this amount is proportional to difference between the wire's temperature and temperature of the surroundings. This means that the temperature of the wire will stabilise at such a level that the amount of heat dissipated (per second) because of the flow of current and the amount of heat "taken" (per second) by the surroundings will be equal. So the greater current passes through wire, the more heat it can give out, and the higher temperature will be settled.

    Now we can move back to the change in resistance due to change in temperature. For metals, resistance increases along with temperature. This means that every time the temperature has increased by some amount, the resistance would increase as well. Therefore less heat would be dissipated than in the case when we assumed resistance to be constant.

    So in both cases temperature of the wire would increase along with the increase in voltage. Only in the second case, when resistance increases, the same increase in voltage would cause smaller increase in temperature than if resistance didn't increase.

    I'm sorry if I haven't explained it clearly. If so, let me know, I'll try again
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    (Original post by jaroc)
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    Thank you!

    So does the second explanation comply with V=IR? Voltage will increase and resistance will increase. Why exactly is heat given out due to the current and the resistance of the wire? more electron colliding with the atoms in the wire causing more vibrations and more heat?
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    (Original post by YdLeet)
    Thank you!

    So does the second explanation comply with V=IR? Voltage will increase and resistance will increase. Why exactly is heat given out due to the current and the resistance of the wire? more electron colliding with the atoms in the wire causing more vibrations and more heat?
    From what I know you're right, heat is dissipated due to collisions with atoms (or some elements of atoms - I don't know exactly) .

    Two things can happen to dissipated heat - either it will be taken by the surroundings or it can cause the wire's temperature t_w to increase. If the difference between the t_w and surrounding temperature t_s is not high enough, some part of the heat will be given out to the surroundings, and some will contribute to increase in wire's temperature. When the difference in temperature \Delta t=t_w-t_s has reached value high enough, thermodynamical equilibrium is reached, so that the temperature of the wire remains constant from this point on.

    If you now increase voltage, more heat is dissipated. This "extra heat" cannot be taken by the surroundings because the difference in temperature is too small, so it "stays within the wire" and causes t_w to increase. When t_w has increased by an appropriate value, new thermodynamical equilibrium is reached, and a new, higher temperature t_w is settled.

    This explanation isn't influenced by the fact of increase in resistance qualitatively, but only quantitatively. However, quantitative treatment of the problem would be a bit complicated.

    I'm not sure if I've managed to make it clearer, but I hope so As to whether it complies to V=IR - yes, it does, but as I said mathematical treatment would be complicated. If you can specify your doubts about its complying with V=IR more precisely, maybe then I could be of more help.
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    (Original post by jaroc)
    From what I know you're right, heat is dissipated due to collisions with atoms (or some elements of atoms - I don't know exactly) .

    Two things can happen to dissipated heat - either it will be taken by the surroundings or it can cause the wire's temperature t_w to increase. If the difference between the t_w and surrounding temperature t_s is not high enough, some part of the heat will be given out to the surroundings, and some will contribute to increase in wire's temperature. When the difference in temperature \Delta t=t_w-t_s has reached value high enough, thermodynamical equilibrium is reached, so that the temperature of the wire remains constant from this point on.

    If you now increase voltage, more heat is dissipated. This "extra heat" cannot be taken by the surroundings because the difference in temperature is too small, so it "stays within the wire" and causes t_w to increase. When t_w has increased by an appropriate value, new thermodynamical equilibrium is reached, and a new, higher temperature t_w is settled.

    This explanation isn't influenced by the fact of increase in resistance qualitatively, but only quantitatively. However, quantitative treatment of the problem would be a bit complicated.

    I'm not sure if I've managed to make it clearer, but I hope so As to whether it complies to V=IR - yes, it does, but as I said mathematical treatment would be complicated. If you can specify your doubts about its complying with V=IR more precisely, maybe then I could be of more help.
    Well throughout my experiment the copper did not reach a constant temperature, it kept rising until I chose to end it. Therefore I think the first part of your reply doesn't exactly apply here?

    I see where you are getting at with the quantitative explanation, but all I'm more interested in the qualitative explanation, which is as I increase voltage, the resistance the electrons passing through the copper experience is increased and as a result temperature increases due to collisions. V=IR, was the only formula I could find that fit my explanation for the increase in temperature.

    I was just confused as to whether or not the value for resistance stays the same. I now know that it increase, am I right?

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    [QUOTE=YdLeet;30546251]Well throughout my experiment the copper did not reach a constant temperature, it kept rising until I chose to end it. Therefore I think the first part of your reply doesn't exactly apply here?[/latex]

    It didn't stop rising for given voltage? This probably means that you hadn't waited long enough


    I see where you are getting at with the quantitative explanation, but all I'm more interested in the qualitative explanation, which is as I increase voltage, the resistance the electrons passing through the copper experience is increased and as a result temperature increases due to collisions. V=IR, was the only formula I could find that fit my explanation for the increase in temperature.

    I was just confused as to whether or not the value for resistance stays the same. I now know that it increase, am I right?

    Thanks
    All right, I now see what you mean

    When you increase voltage, both current and resistance increase (although the increase of resistance is not a direct consequence of increasing voltage - it's because of the rise in temperature). So when you increase voltage, more power is dissipated because of the increase of the current but not because of the increase of resistance - if resistance didn't increase, even more power would be dissipated! It's not so easily seen from the formula V=IR, because I is dependent on R. I suggest that you look at P=V^2/R. The increase in dissipated power is mostly due to the increase in current - electrons have greater velocities, so when they collide, more energy is dissipated.
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    [QUOTE=jaroc;30546496]
    (Original post by YdLeet)
    Well throughout my experiment the copper did not reach a constant temperature, it kept rising until I chose to end it. Therefore I think the first part of your reply doesn't exactly apply here?[/latex]

    It didn't stop rising for given voltage? This probably means that you hadn't waited long enough




    All right, I now see what you mean

    When you increase voltage, both current and resistance increase (although the increase of resistance is not a direct consequence of increasing voltage - it's because of the rise in temperature). So when you increase voltage, more power is dissipated because of the increase of the current but not because of the increase of resistance - if resistance didn't increase, even more power would be dissipated! It's not so easily seen from the formula V=IR, because I is dependent on R. I suggest that you look at P=V^2/R. The increase in dissipated power is mostly due to the increase in current - electrons have greater velocities, so when they collide, more energy is dissipated.
    Ah, wow okay, it took a while but I've got it now. Thanks a lot for your help!!!
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    (Original post by YdLeet)
    Ah, wow okay, it took a while but I've got it now. Thanks a lot for your help!!!
    Sorry for the confusion

    And to make sure I haven't caused misunderstanding: there are two factors that contribute to the amount of dissipated heat - current and resistance. The higher the current is, the faster electrons are, and more energy is dissipated. The higher resistance is, the more frequent collisions are. However, higher resistance not always means more heat dissipated. Let's consider two cases: (1) constant current, (2) constant voltage, and see how the increase of resistance affects dissipated power in each of them.

    (1) Constant current.

    P=I^2R.

    If we can ensure a constant current regardless of resistance, it easily follows from the above formula that the higher resistance is, the more heat is dissipated.

    (2) Constant voltage.

    The formula P=I^2R is not too helpful here. If R increases, would P increase? We can't tell at once because I is a function of R! (Note - in case (1) current was independent of resistance) But we can rearrange the formula to show explicitly the relation between power and resistance: P=I^2R=IR\times IR/R=V^2/R.

    So if we increase resistance, the current decreases. This can be explained this way: although the collisions are more frequent, the electrons are slower, and overall less power is dissipated.

    You were investigating the second case, because you were increasing voltage, and didn't know what current was passing through.


    I hope I'm not causing any more confusion by writing this post I really do. However, if this is the case, and you're patient enough, I'll still be willing to help
 
 
 
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