Why Does Increasing the Cross-Sectional Area of a Wire Reduce Heat Dissipation?

[PFuser1232]

The power dissipated in a resistor of resistance $R$ with current $I$ passing through it is $I^2 R$. We can write $R$ as $\rho \frac{L}{A}$ where $L$ is the length of the wire and $A$ is the cross sectional area of the wire. Clearly, increasing $A$ decreases $R$. Why is less heat dissipated in thicker wires, though? I thought changing the resistance of the wire changes the current passing through it as well. How can we guarantee from the equation $P = \frac{\rho L I^2}{A}$ that increasing $A$ decreases $P$? Isn't $I$ functionally dependent on $R$?

 

[Averagesupernova]

You are not applying ohms law correctly. Given a current of 1 ampere through a resistor of 1 ohm will cause a dissipation of 1 watt. There will also be a voltage drop across the resistor of 1 volt. Nothing in ohms law says that says that decreasing the resistance forces the current to increase unless the voltage is fixed. Solve some problems with a voltage source and multiple series resistors. Make some of the resistors several thousand times the value of the smallest resistor which would represent the wire. Changing the smallest resistor slightly will not affect the current in the circuit very much at all. You are not the first poster here who has had similar questions. Strangely enough though, the first time I saw this type of question asked was here on physics forums.

 

[phyzguy]

It depends on whether you are in a constant current or a constant voltage situation. If the current I is held constant, then everything is as you have written it, and the current is independent of the area A. If you apply a constant voltage to the wire, then the current through the wire is \frac {V}{R}, and the power dissipated is \frac{V^2}{R}. In this case, increasing A will cause R to decrease, I to increase, and the power dissipation to increase.

 

[PFuser1232]

You are not applying ohms law correctly. Given a current of 1 ampere through a resistor of 1 ohm will cause a dissipation of 1 watt. There will also be a voltage drop across the resistor of 1 volt. Nothing in ohms law says that says that decreasing the resistance forces the current to increase unless the voltage is fixed. Solve some problems with a voltage source and multiple series resistors. Make some of the resistors several thousand times the value of the smallest resistor which would represent the wire. Changing the smallest resistor slightly will not affect the current in the circuit very much at all. You are not the first poster here who has had similar questions. Strangely enough though, the first time I saw this type of question asked was here on physics forums.

You are not applying ohms law correctly. Given a current of 1 ampere through a resistor of 1 ohm will cause a dissipation of 1 watt. There will also be a voltage drop across the resistor of 1 volt. Nothing in ohms law says that says that decreasing the resistance forces the current to increase unless the voltage is fixed. Solve some problems with a voltage source and multiple series resistors. Make some of the resistors several thousand times the value of the smallest resistor which would represent the wire. Changing the smallest resistor slightly will not affect the current in the circuit very much at all. You are not the first poster here who has had similar questions. Strangely enough though, the first time I saw this type of question asked was here on physics forums.

So the effect of cross sectional area on the dissipated heat I described above only applies to wires which typically have a much smaller resistance than all other components in a circuit. Increasing the cross sectional area of a resistor with significant resistance might not necessarily decrease the heat dissipated in the resistor, right?

 

[jim hardy]

How can we guarantee from the equation P=ρLI2AP = \frac{\rho L I^2}{A} that increasing A decreases P ? Isn't I functionally dependent on R ?

You cannot guarantee it.

As previous posters pointed out, this is not a physics problem but an algebra problem.

If I is a function of A and you wish to vary A , your equation ought to express I in terms of A

 

[Averagesupernova]

So the effect of cross sectional area on the dissipated heat I described above only applies to wires which typically have a much smaller resistance than all other components in a circuit. Increasing the cross sectional area of a resistor with significant resistance might not necessarily decrease the heat dissipated in the resistor, right?

Jim Hardy pointed it out quite well. It's a math problem. In a resistor any time you double the current through said resistor you will double the power it dissipates. But that scenario isn't exactly what you questioned in your original post. At least through my interpretation. Don't give up. We're here to help.