The battery provides a potential difference between its terminals that allow acceleration of the electrons in the circuit, i.e the potential energy gets converted to kinetic energy, these electrons have a certain kinetic energy that depends on the voltage of the battery.

Now what exactly happens once the electrons reach the light bulb? Why does the light bulb need some voltage limit or higher to emit light? Maybe I don't understand what voltage drop really means, I searched throughout the site but no luck.

  • $\begingroup$ You mean a tungsten filament lamp? $\endgroup$ Feb 15, 2019 at 13:58
  • $\begingroup$ Maybe, anything that demands some voltage to work really $\endgroup$
    – khaled014z
    Feb 15, 2019 at 14:08
  • $\begingroup$ So you do not understand Ohm's law and the theory of electrical conductivity? $\endgroup$ Feb 15, 2019 at 14:30
  • $\begingroup$ I don't understand what is happening microscopically when electrons flow through the light bulb, what makes it emit light? $\endgroup$
    – khaled014z
    Feb 15, 2019 at 14:40
  • $\begingroup$ It depends on the design of the lamp. $\endgroup$ Feb 15, 2019 at 14:58

1 Answer 1


Let's consider an old-fashioned filament bulb. The potential difference of the battery will cause electrons to move through the circuit. Part of that circuit is a thin piece of tungsten metal. The tungsten is not a perfect conductor and in a good bulb design it is the worst conductor in the circuit. The moving electrons in the tungsten interact with vibrational modes in the metal called phonons which creates resistance in the filament. If the voltage is too low then very little energy is dissipated in the filament and it does not get very hot. At higher voltages more electrons flow with the current, I, being roughly governed by the equation V = IR. (This isn't quite so simple because R increases with filament temperature.) If the voltage is too high then so much current flows that the filament heats up and melts. At just the right voltage, however, the filament will heat up to about 2700 to 3000 degrees and give off the black body radiation that we know and love as the phonon vibrations couple with the photon field in and around the filament.

  • $\begingroup$ And if it is a lamp with LEDs? :) $\endgroup$ Feb 15, 2019 at 15:01
  • 1
    $\begingroup$ hey, I don't see the tag semiconductor physics on this one ;-) $\endgroup$
    – Paul Young
    Feb 15, 2019 at 16:08

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