# If energy delivered to the resistor is not the kinetic energy of the electrons then how is it delivered? [duplicate]

In a help session by Walter Lewin he says that the energy delivered to the resistor is not the kinetic energy of the electrons. If not how is energy delivered.

and the time into the video is about 4:00 minutes.

## marked as duplicate by ZeroTheHero, Jon Custer, Community♦Jan 22 at 8:11

• A while back someone posted a link to an excellent article that bridged the concepts and phenomena of current electricity with motion of electrons and the building up of static charges etc. I wish I could find the link. I think it would help answer this questions. – M. Enns Nov 11 '18 at 3:09

A full description of forces between particles the size of electrons requires quantum mechanics. But it is possible to state some results as a few simple rules. They don't tell you everything, but they are good enough to be useful here.

• In a solid, nuclei and most electrons play a background role. They form neutral atoms rigidly held in fixed positions by electric forces. The forces can be quite large.

• In an insulator, electrons are stuck to individual atoms.

• In a metal, some electrons can flow around very freely in the metal. Relatively small electric forces can make them move. Never the less, they cannot easily leave the metal.

In some ways you can think of the electrons in metal as little balls and springs. They push each other apart as much as possible without leaving the metal. Be careful with this picture. It is useful to visualize some ideas about energy, but it is wrong in other ways.

• It is possible to add or remove electrons. This leaves the metal charged. For a negative charge, the electrons are crowded together a little, compressing their springs. Compressed springs store potential energy.

• For a positive change, the electrons are a little spread out, leaving a positive charge from excess nuclei. The analogy of springs between nuclei and electrons is a little awkward because electrons can move around freely, and springs would tie them down. But you can still sort of think of it that way. Stretched springs store potential energy.

• In general, if positive and negative charges are close together, it takes energy to separate them. For example, it takes energy to pull an electron off one piece of metal, and put it on another. This leaves one positively charged and the other negative.

• It takes less energy if the separation is small. A capacitor is two plates very close together. It takes energy to charge a capacitor. But less than the same charge on well separated objects. So a capacitor stores potential energy in stretched springs. It isn't really springs. It is potential energy from having moved charges against the electric forces of repulsion and attraction.

• If you connect the plates of a capacitor with a wire, the compressed springs on the negative plate will push electrons through the wire to the positive plate. As they flow, they more or less bump into electrons and make them vibrate around their fixed positions. The way electrons interact with atoms isn't exactly "bump into," but it will do. Potential energy turns to heat.

• A resistor is like a wire, except electrons bump into atoms more. It impedes their flow. The capacitor discharges slower.

• It isn't true that the compressed springs accelerate the electrons really fast, so they have lots of kinetic energy and bump into atoms hard.

• A circuit is often powered by a battery or generator instead of a capacitor. A battery pushes electrons out the negative terminal and receives them in the positive terminal. The energy to do the pushing comes from chemical reactions in the battery. A generator uses magnetic fields to push electrons.

• But this still does not answer how energy is transferred to the resistor. – Allen Aug 13 '17 at 14:20
• @Allen - I added a brief section on a resistor. – mmesser314 Aug 13 '17 at 16:53
• So are you saying that it is indeed the kinetic energy that is converted to heat @mmesser314 – Allen Aug 14 '17 at 16:28
• If you like. But electric forces continually push electrons. Without that push, a resistor wouldn't produce heat. – mmesser314 Aug 15 '17 at 5:49

Consider a simple circuit of a resistor of resistance $R$ connected to a battery of emf $\gamma$. When the battery is connected, it produces an electric field $\textbf E$. The wires of the circuit are obviously made of conducting material so that most of the electrons are in conduction band and are loosely bound. The electric field acts on these loosely bound electrons and causes an acceleration,
$$\textbf a=-\frac{e\textbf E}{m_{e}}$$ The negative sign implies that electrons move in direction opposite to the field, as the direction of the field is defined conventionally for unit positive charge and the electrons are negatively charged. The electrons that are now under motion, will have some kinetic energy and will in due time lose the energy as $\textbf{heat}$ when it collides with the kernals(metal ions) of the resistor. But the electrons after collision will get re-accelerated due to the field, so an overall current, say, $I$ flows through the circuit. Simultaneously, the electrons keep extracting kinetic energy from the field to continue their motion and keep giving off the energy in the form of heat in collisions. So, ultimately, $\textbf{all}$ of the energy extracted from the field will be lost as heat when the battery is switched off or the charge is drained out. Let the energy that is lost as heat be $Q$.

Assuming that the connecting wires are ideal and offer no resistance, then the emf of the battery, $\gamma$, will be equal to the potential drop across the resistor, say, $V$. Thus $\gamma=V$. By Ohm's Law, we have
$$V=IR$$ so $\gamma=IR$
$\implies\gamma I=I^2R$
Now $\gamma$ is defined as the work done by the battery or the electric field in circulating $1C$ of charge across the circuit and $I$ is the rate of flow of charges(current). So the quantity $\gamma I$ denotes the rate of doing work by the battery or the field. If after time $t$, all the charge of the battery is drained out. Then work done by the battery is:
$$W=\gamma It=I^2Rt$$ The work done by the battery is the energy lost by the battery or the energy that the electrons extract from the field. We know that the energy lost as heat after the battery is drained of charges is $Q$. So, we have,
$$W=\gamma It=I^2Rt=Q$$
So in fact, all of the energy lost by the battery, ultimately appears in the form of heat.

• "lose the energy as heat when it collides with the kernals(metal ions) of the resistor." No, heat is the transfer of energy due to differences in temperature. Heat is not due to collision of individual particles. It is a statistical process. The transfer of energy does result in the increase of temperature, but the transfer is not a heat transfer. – Bill N Aug 13 '17 at 18:46
• If the collisions lead to increase in temperature, then won't energy flow in the form of heat from higher temperature (of the resistor) to the lower temperature of environment? Correct me: how is the kinetic energy converted to heat otherwise? – Nishant Garg Aug 13 '17 at 19:37
• Or do you mean I should first talk about the temperature difference due to collision and then talk about heat transfer due to difference in temperature? @BillN – Nishant Garg Aug 13 '17 at 19:40
• Why even mention heat? An electron doesn't have temperature and can't lose energy via heat. It will lose or gain kinetic energy, and it can be part of a system of increasing or decreasing potential energy. You're correct in the first comment regarding the resistor getting warmer and losing heat to the surroundings, but the electrons aren't losing energy via heat. – Bill N Aug 14 '17 at 0:49
• Heat is the term used for the process of transferring kinetic energy of particles @NishantGarg – Allen Aug 14 '17 at 16:30