Question about Dimming Light

EKKC

Diamond Member
May 31, 2005
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Does a dimmed light use the same amount of energy than if it is maxed out?

say i have a 100w bulb, would it use the same if I dim it at say 25% than if I turned it all the way up?
 

Mark R

Diamond Member
Oct 9, 1999
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Nope, the lamp will use less energy when dimmed.

However, as you reduce power to the lamp it will get less efficient. So you might be using 75% of the power to get 25% of the light.
 

RaynorWolfcastle

Diamond Member
Feb 8, 2001
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How long before someone posts an equation involving blackbody radiation in this thread?

edit: From an electrical point of view, P = V^2/R_sys where P is power, V is voltage, and R is the system's resistance.

If we ignore all this blackbody radiation mumbo jumbo to determine the brightness vs power, we can get an idea of what's going on.

The easiest (and most commonly used) method to create a dimmer is to put a potentiometer (the dimmer) in series with the light such that R_sys = R_bulb + R_dim. The voltage supplied is constant at 120Vrms.

So the power used by the system is V^2/(R_sys+R_dim). Clearly any value of R_dim will reduce the power consumption. Also, by voltage division, we know that the voltage povided to the bulb will be V*R_bulb/(R_bulb + R_dim). However, as others have said, the relation between brightness and voltage is not linear so it's difficult to tell exactly how much power you save by dimming the light.
 

bsobel

Moderator Emeritus<br>Elite Member
Dec 9, 2001
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Originally posted by: Mark R
Nope, the lamp will use less energy when dimmed.

However, as you reduce power to the lamp it will get less efficient. So you might be using 75% of the power to get 25% of the light.

Its not nearly that bad, it's fairly close to linear (not exact obviously) from what I recall. Also, a normal bulb ran at about 90% will last up to 3x as long as the same bulb at 100% (todays trivia fact)
 

Heisenberg

Lifer
Dec 21, 2001
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Originally posted by: RaynorWolfcastle
How long before someone posts an equation involving blackbody radiation in this thread?
Power/unit area = a*sigma*Temp^4

Happy now? :p