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Consider two wires of length $L$ and resistivity $\rho$, and consider a fixed voltage source with voltage $V$. The first wire has cross sectional area $A_1$ and the second wire has cross sectional area $A_2$ such that $A_1 > A_2$.

If we connect the first wire across $V$, Ohm's law gives:

$$V = I_1\rho\dfrac{L}{A_1}$$

(where $I$1 is the current corresponding to wire $1$ ) which gives:

$$\frac{I_1}{A_1} = \frac{V}{\rho L}$$

This means that for a given voltage, $V$, the current density for a given length $L$ of wire with resistivity $\rho$ does not depend on its area:

$$\frac{I_1}{A_1} = \frac{I_2}{A_2} = \frac{V}{\rho L}$$

Similarly, using $P = IV$, we can see that:

$$\frac{P_1}{A_1} = \frac{P_1}{A_1} = \frac{V^2}{\rho L}$$

So the power dissipated per unit area does not depend on the cross sectional area of the wire.

Since this is the case, what is the explanation for the everyday phenomenon that wires of smaller cross sectional area heat up more when the same voltage is applied? On a per unit area basis, they dissipate the same power, so intuitively it does not seem that the smaller wire should experience higher temperatures.

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  • $\begingroup$ Are you sure that "wires of smaller cross sectional area heat up more when the same voltage is applied?" This thing is definitely true when the same current is applied, how are you so sure about the voltage? $\endgroup$ May 25, 2020 at 17:49
  • $\begingroup$ Yes I definitely knew this was true about fixed current, but what motivated me to ask was primarily 4:00 - 4:10 in this practical engineering video about transmission lines: youtube.com/watch?v=qjY31x0m3d8&t=261s that show smaller wires melting when they were put in a 120v circuit with a hairdryer. $\endgroup$
    – Tim Clark
    May 25, 2020 at 18:40
  • $\begingroup$ I don't know much about Electromagnetism above the standard High School level. But one thing that I definitely know is that standard Ohm's law are not applicable on "Transmission Lines". It requires a very different kind of analysis. $\endgroup$ May 25, 2020 at 18:45
  • $\begingroup$ You should use mathjax for writing comments. $\endgroup$ May 25, 2020 at 18:50
  • $\begingroup$ Well the whole explanation in the video, as well as every explanation I've ever seen for why they use high voltages for transmission lines uses the assumption that P = IV = I^2R which is based on Ohm's Law. $\endgroup$
    – Tim Clark
    May 25, 2020 at 18:51

2 Answers 2

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You need also to consider how the wire gives heat to its surroundings.

You've shown that for a given voltage the power dissipation in a wire per unit length is proportional to its cross-sectional area, A. The wire will get hot and give out heat at a rate proportional to its surface area, which for a given length of wire, is proportional to $\sqrt A$. So, using your notation, if $A_1 = 4A_2$, the first wire will generate 4 times as much thermal power per unit length as the second, but will have to get hotter in order to give out this amount of power through a surface area only twice as big, so as to reach thermal equilibrium.

For a given applied voltage, the fatter wire gets hotter!

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  • $\begingroup$ That's interesting! That contradicts everything I thought i.e. that smaller wires would get hotter when exposed to the same voltage. I did not think to use the Stefan - Boltzmann law. In the case of constant current, similar analysis using the 4th power temperature / emittance relationship shows that the 4th power of the temperature is proportional to the negative 3/2 power of the surface area, correctly predicting that the wires heat up with smaller diameter. $\endgroup$
    – Tim Clark
    May 25, 2020 at 21:02
  • $\begingroup$ Yes, the cases are quite different! Incidentally, I made $T^4$ proportional to $(\text{surface area})^{-3}$ for the constant current case. $\endgroup$ May 25, 2020 at 21:26
  • $\begingroup$ Sorry wrote that wrong. I have it proportional to the negative 3/2 power of cross sectional area which means it's proportional to the negative 3rd power of surface area. $\endgroup$
    – Tim Clark
    May 25, 2020 at 22:29
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From Ohm's law

$$V=IR$$

and

$$R=\frac{ρL}{A}$$

For fixed $L$ and $ρ$, if you double $A$, you halve $R$. If the voltage across $R$ is fixed, you double $I$. But the power dissipated in the resistor is

$$P=I^{2}R$$

So the power dissipated in the resistor is squared if the cross sectional area is halved. So more power is dissipated in the larger cross section conductor than the smaller for a given material, length and voltage. But how does this translate in terms of the temperature of the conductor?

To determine that we need to also consider the rate at which heat transfers from the interior of the conductor to its surroundings. All other things being equal, the heat transfer rate is proportional to area across which the transfer occurs. That surface area is the surface area of a cylinder (minus the end caps), or

$$A_{S}=πDL$$

The cross sectional area of a conductor, as a function of its diameter, is

$$A_{X}=\frac{πD^2}{4}$$

For a given length, the surface area and thus the heat transfer rate is proportional to the diameter. However, the cross sectional area goes up as the square of the diameter, so the rate of heat generation goes up as the square of the diameter. In effect, the larger the diameter the smaller the ratio of the surface area to volume of the conductor, and the greater retention of heat.

Bottom line: The temperature rise of of the larger size (diameter) wire should be greater than the smaller size (diameter) wire.

Hope this helps.

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