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From what I understand, the lower the resistance of an object in a circuit, the more it produces heat (all else kept equal): P = V^2 / R

It's deeply bothering me because I don't understand why that wouldn't also mean that in a circuit most of the heat would be given off by the most conductive parts of the circuit (i.e. the power cord would heat up, not the resistor).

Would anyone help liberate my mind from this torture?

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(all else kept equal): P = V^2 / R

This formula doesn't apply to the case of "all else kept equal", it applies to the case of "voltage across the resistor kept equal".

There's another common case: Current through the resistor kept equal. In this case the formula is $$P = I^2 R.$$ You can see that this has a very different effect: higher resistance leads to higher power dissipation rather than lower.

It's deeply bothering me because I don't understand why that wouldn't also mean that in a circuit most of the heat would be given off by the most conductive parts of the circuit

Often when you change the resistance of some element of a circuit, you aren't keeping the voltage across the element equal, because of the interaction with the other elements. You might have to analyze the whole circuit to determine the effect on the power dissipation of some specific element.

the power cord would heat up, not the resistor

If the power cord is well chosen, the current through it should be determined by the load, not by the properties of the power cord itself. This means the power cord should be analyzed by the formula for the case where the current is kept equal, not the voltage. And then you get the result that reducing the resistance of the power cord reduces its power dissipation, as you expect.

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  • $\begingroup$ @The_Photon: Thank you! 🙏 $\endgroup$
    – Hans
    Jul 14, 2021 at 18:37

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