Or do they just capture the incoming energy and turn it into heat at the unit and save you nothing?

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    $\begingroup$ I'm voting to close this question as off-topic because it is about the features of a device which is unrelated to physics. $\endgroup$ – Danu Mar 16 '16 at 14:01
  • $\begingroup$ A light dimmer is not a resistor. It works by switching the power on and off. In the the max-brightness setting, the power will be on 100% of the time. In the min-brightness setting, the power will be on about 50% of the time. $\endgroup$ – Solomon Slow Mar 16 '16 at 15:29
  • $\begingroup$ I think they save energy. A good inductor can reduce current without dissipation of energy, and hence reducing power consumption. $\endgroup$ – Anubhav Goel Mar 16 '16 at 16:26
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    $\begingroup$ In the old days dimmers were rheostats, adjustable resistors, so they dimmed the lights but did not save any energy. Today they do both, and they are most often implemented via pulse width modulation. $\endgroup$ – Peter Diehr Mar 16 '16 at 17:25
  • $\begingroup$ I understood that pulse width modulation was quite destructive for various sort of light sources (but the regular filament bulb) ? $\endgroup$ – Fabrice NEYRET Mar 17 '16 at 4:48

Yes, dimmers save energy. Modern dimmers save more energy than older (rheostat) dimmers, but both save energy.

Modern dimmers use a "triac", which is an inexpensive way of reducing power by pulsing the current. These pulses result in lower average current going to the bulb, and therefore lower average power. They are very efficient, so providing current 50% of the time implies using about 50% less current, or 50% less energy overall.

Older dimmers used rheostats, which are basically just resistors. These are less efficient but reduce energy. Consider the case of a bulb with resistance $R_B$ in series with a rheostat of variable resistance $R_V$. The total resistance is then $R = R_B + R_V$, and the total power is $P = {{V^2}\over{R}} = {{V^2}\over{R_B + R_V}}$. So if the rheostat is set to $0\ \Omega$, then $P = {{V^2}\over{R_B + R_V}} = {{V^2}\over{R_B}}$, and if it's set to the same value as $R_B$, then $P = {{V^2}\over{R_B + R_V}} = {{V^2}\over{2R_B}}$, or half as much power into the combined resistors. Note though, that while the total power goes down by two when dimmed (with these values), so dimming uses less power overall, the power to the light goes down by four, because half of the total power is lost in the rheostat... and that's why rheostats are inefficient.

A further complication to this with incandescent bulbs is that they are less efficient at producing visible light when operating out of their designed range since a larger proportion of energy ends up in the infrared range. That is, if you dim a single bulb to get less light it uses less energy than undimmed; but even with a 100% efficient dimmer, if you combining multiple dimmed bulbs they will use more energy than a single undimmed bulb, for the same amount of visible light. Or said another way, if you want less light, it's more efficient to use a bulb designed for less light -- and this will always be true due to the usual inefficiencies, but is exaggerated for incandescent bulbs.


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