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I have read that the electric energy flows from the battery into the circuit through electric field created outside the wires into the light bulb (or any other resistor). I have also read that the friction caused in the movement of electrons cause light bulb to heat up. So which is the correct explanation?


marked as duplicate by ACuriousMind, Kyle Oman, John Rennie, Kyle Kanos, Chris Mueller Apr 24 '15 at 12:02

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The electric field that is applied to the wire (it doesn't matter if you consider it to be outside or inside the wire for the purpose of this explanation) causes the electrons that are indeed inside the wire to move. This movement involves the "friction" you heard of ( electrical resistance), which in turn causes heat (and emission of photons).

The important thing is that it is not the field that's flowing. The field causes a flow and that flow causes the heat.

  • $\begingroup$ Thanks a lot both of you. It was useful for improving my understanding. I have a follow - up question. I have seen the diagrams of electric circuits in which the electric field develops between two HALVES of the circuits. But that doesn't make sense. This is because, as per my understanding, the electric field should develop between two atoms of the conductor at any given part of a closed circuit. So how does the two HALVES scenario come into the picture ?? $\endgroup$ – Abhishek Apr 27 '15 at 16:35
  • $\begingroup$ But do the electrons from wire come back to their original "positions" in atoms of the wire? $\endgroup$ – ado sar Jan 5 at 2:12

I have read that the electric energy flows from the battery into the circuit through electric field

simple circuit

This is a simple circuit with a battery a switch and a light bulb.

The battery has the energy in chemical form, i.e. there are atoms and molecules separated into ions that generate a voltage drop, (an electric field ) accross the switch. No current flows as long as the switch is open, and no energy is transfered.

Once the switch is closed a current develops in the wires, electrons in the conduction band move through the wires from the positive to the negative pole through the light bulb. The electrons have an average drift velocity, carrying kinetic energy. In the wire itself they do not lose energy. In the resistance of the lamp they lose energy transferring it to the molecules of the resistor. The way electrons lose energy when decelerated is through photons, in the case of the light bulb infrared photons which make the lamp incandescent and then excite the electrons of the molecules of the resistor to high levels, enough to start deexciting in visible light photon transitions.

created OUTSIDE THE WIRES into the light bulb ( or any other resistor). this is wrong.

I have also read that the friction caused in the movement of electrons cause light bulb to heat up.

See my description above . It is nor really friction, as friction is a macroscopic effect. It is electromagnetic interactions.

  • 1
    $\begingroup$ There are some confusion here. The way electrons lose energy is not "through photons", but by collisions with impurities in the resistor. (by impurity I mean anything that disrupts perfect crystalline order of the ion cores: atoms of different materials, randomly oriented domains, vibrations, ...). In a collision, the electron does work on the impurity (applies a force and displaces it). The displaced ions accelerate, and can start vibrating. Thus, the resistor heats up. This is how the electron loses energy. "Photons" (radiation) is a consequence of the motion of the ion cores. $\endgroup$ – garyp Apr 23 '15 at 17:00
  • $\begingroup$ @garyp in my view at the level of electrons everything happens with the exchanges of photons, either virtual or real. Heat is infrared photons, light has higher frequency and comes from excited levels due to multiple infrared absorptions $\endgroup$ – anna v Apr 23 '15 at 17:19
  • $\begingroup$ No heat is NOT infrared photons, heat is random motion of electrons and vibration of atoms. Motion of particles has nothing to boil down to further. $\endgroup$ – Milind R Sep 25 '15 at 14:01

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