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When charges flow in a simple circuit, then if 5C of charge is transported by the battery(of 5V) in some time, then the work done/energy lost by the battery is 25Joules. During this time, the electrons collide with the lattice(atoms) of the conductor and heat it up.

This is what confuses me-

How is exactly all this energy converted to heat? I mean, what if the electrons collided with the lattice and lost 15J only in the collisions and still made it through the circuit?

(This might sound to be the stupidest question ever, but Im not able to create a perfect model of this topic in my mind and that is making me think all sort of weird things)

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    $\begingroup$ Then you have created a perpetual motion machine providing infinite energy. $\endgroup$
    – Jon Custer
    Commented Mar 23, 2023 at 15:57
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    $\begingroup$ Let's say we do a short circuit, then the 5C charge will happen/flow in a very short time .... all the heat will end up in the battery .... i.e. the electron having not encountered much resistance will just cause the atoms in the + electrode to vibrate, vibration is heat. (actually its the - electrode because physics got the convention wrong about 100 years ago!). $\endgroup$ Commented Mar 24, 2023 at 2:45

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The idea of electrons behaving as classical particles, bouncing with the nuclei and losing energy has its limitation as you see. Really, according to this description, the flow of charges (the current) could be split into electrons that collide, and electrons that don't collide.

The quantum mechanics picture of a metal is the band structure. Note that even without any battery attached, the electrons have a distribution of momentum and energy. But it is a stationary state like the energy levels of a single atom. They don't collide generating energy in spite of having momentum in this case.

Maybe a better way to understand qualitatively the situation is to compare with an electron that jumps to a higher energy level in a single atom when absorbs a photon. Here the battery plays the role of the photon and jumps a lot of electrons to a higher energy and momentum. They release energy when they fall back to the ground state. The transition must be very short, otherwise most of them would reach the battery without losing energy.

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How is exactly all this energy converted to heat? I mean, what if the electrons collided with the lattice and lost 15J only in the collisions and still made it through the circuit?

We know that current is constant through a resistor. This means the average kinetic energy associated with the electron flow through the resistor is constant.

The work energy theorem states that the net work done on an object equals its change in kinetic energy. This means the positive work done by the battery giving kinetic energy to the electrons, equals the negative work done by collisions with particles of the resistor, taking that kinetic energy away from the electrons and dissipating it as heat.

The overall result is the electrical potential energy given the electrons by the battery is dissipated as heat due to collisions in the resistor.

Hope this helps.

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You don't need a quantum mechanical picture of the metal (where electrons are still undergoing collisions by the way) to understand this. The Drude model is a simple model of conduction and suffices in this case. On average, electrons in a metal like copper collide roughly every $10^{-14}\text{ s}$. The drift velocity in copper is on the order of $1 \text{ mm/s}$, so between collisions electrons drift by roughly $10^{-17}\text{ m}$ against the electric field, on average.

The Drude model posits that electrons emerge from collisions with a random velocity distribution, with a mean of zero. So the electrons can be pictured as undergoing frequent collisions upon which they lose their kinetic energy, but inching forward in very tiny steps between collisions. Since the work done on each charge by the electric field is quickly lost as heat to the lattice, you can't really have a charge losing less energy than this work while going around a macroscopic loop.

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It seems like your mental picture is that electrons are given a large amount of kinetic energy by the battery, enough to propel them through the circuit and back to the battery.

First, if it did work that way, and the electron made it back with a significant amount of kinetic energy remaining, the battery would need to expend less energy accelerating it back to its maximum speed than if the electron had arrived back "exhausted". In other words, the leftover kinetic energy would not be lost. Only 15 J would be lost per cycle, and only 15 J of heat would be emitted. The battery would still run out of energy, all electrons would then slow to a stop, and 100% of the energy would have turned into heat.

Second, it doesn't work that way. Instead of ball bearings in a pinball machine, imagine ball bearings in a viscous liquid. They are nowhere close to having enough energy to make it back to the starting point of the circuit; if you stop pushing, they will stop more or less where they are. All of the work you put into pushing them becomes heat almost immediately.

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Your example assumes a conversion ratio of $\frac{15}{25}=0.6$. Let’s assume this holds for „1 loop“ of discharge current. As long as current is flowing, you‘ll loose this fraction.

Remember, these are all approximations and it depends on the given circumstances. Think of, e.g. Li-batteries, which go up in flames from mechanical damage.

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