Why do current carrying wires with smaller area heat up more than those with larger area?

Question

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.

Answer 1

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!

Answer 2

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.