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In a text book, when they calculate power dissipated in a closed circuit, they use circuit with a battery and one resistor. Idealized positive charge travels from + terminal (point B) of the battery then through resistor, then to the - terminal (point A) of the battery. Then battery pushes the charge back to the + terminal and the cycle begins again. The calculation based on the assumption that as the charge travels from point A back to point A, its net gain of energy is 0, that is the potential energy given to it by the battery is totally spent by the resistance in the circuit (in particular at the resistor if considering idealized wire).

Somehow I fail to see this as obvious. Yes, when the charge arrives back to point A it will have the same potential energy as when it started (in the example in the book point A is grounded so potential is 0). But what prevents it from having more kinetic energy then when it started? What prevents the charges from being accelerated?

I understand that resistance acts against acceleration, like friction force. But what makes those forces (electrical and resistance) and corresponding energy gains/losses exactly balanced?

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  • $\begingroup$ The "change" in potential energy will be zero because electrostatic force is conservative. $\endgroup$
    – An_Elephant
    Commented Sep 24, 2023 at 12:26
  • $\begingroup$ The current in the wire is electrons, while the current through the battery is ions. From the particle point of view, the circuit isn't closed. $\endgroup$
    – John Doty
    Commented Sep 24, 2023 at 13:25

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what prevents it from having more kinetic energy then when it started? What prevents the charges from being accelerated?

Let’s suppose that the KE of the charge carriers does increase around the circuit. That implies that the velocity increases. Since the current is proportional to the velocity of the charge carriers, that implies that the current increases. Since the current increases, by Ohms law the energy lost in the resistor increases. This uses up more of the PE reduces the amount of excess energy available for KE to increase further. This process continues until a steady state is reached where all of the energy is lost in the resistance.

So what you suggest in fact is a transient state which leads to the steady state analyzed by the text. It happens every time you close a switch.

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    $\begingroup$ It's worth also pointing out that the increase in the velocity of the electrons in the transient state is extremely small. Free electrons accelerated across a ~10 V potential would achieve speeds of ~1000 km/s, but electrons in a wire don't come close to that. Typical drift speeds associated with electric currents are ~0.1 mm/s (even less than speeds associated with thermal motion, ~100 km/s). $\endgroup$
    – FTT
    Commented Sep 24, 2023 at 13:09
  • $\begingroup$ @FTT yes, and the time to reach the steady state is correspondingly minuscule. On the order of the size of the circuit divided by the speed of light. $\endgroup$
    – Dale
    Commented Sep 24, 2023 at 13:18
  • $\begingroup$ This actually raises another question, if the time to reach steady state is so short, how do electrons "know" to gain just enough speed to lose all potential energy in the resistance in the circuit, before they even traveled any considerable portion of the circuit? Or i guess at any point any excess kinetic energy gained above the steady state gets consumed by the resistance? $\endgroup$
    – Yevgeniy P
    Commented Sep 25, 2023 at 12:17
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The only thing that is important in conservative forces is the final and initial position.Since the final and initial position are the same there isnt any change in the potential energy of a test particle driven by a conservative force.

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