Let's say I have a 60W bulb in a lamp in my bedroom. If I kept the lamp on for 2 hours straight but the next day, I switched it on and off 10 times in intervals of 5 minutes. Which scenario would use more energy?
Leaving it on would use more energy, absolutely. Sometimes, people try to convince themselves that turning a light on and off uses more energy because there is some high inrush current, or some such thing.
Firstly, incandescent lights hardly have any inrush current, because they don't have any capacitors to charge, and they need not strike an arc in the bulb. The current is initially higher because the filament resistance is lower, but:
- this is for a fraction of a second
- getting it up to temperature doesn't take any more energy than it would have taken to leave it on to maintain that temperature
- even though the current may be higher, it's not that much higher. Do all the other lights in your house dim temporarily when you turn one on?
Secondly, if you take a fluorescent bulb, which may have capacitors, and thus may require some inrush current, it doesn't begin to make up for the cost of leaving the light on. Consider again how short the turn-on period is relative to the leaving-on period. Even if you consider the wear-and-tear on the bulb and the starter and the fixture, it's almost always more economical to turn the bulb off. I read a report by someone who bothered to do all the math, and they concluded that if you intend to leave the light off for more than about 60 seconds, it's more economical to do so.
Okay, let's set up a simple simulation:
According to the Wiki page on incandescent bulbs, for a 100W, 120V bulb, the cold resistance is ~9.5Ω, and the hot resistance ~144Ω. It takes around 100ms for the bulb to reach the hot resistance on turning on.
So armed with this info, we can simulate and prove the initial surge would be absolutely insignificant if we switched the bulb every 5 minutes. We don't really need to run the simulation for 2 hours to prove this, but we will. I have even extended the "warm up" time to 300ms.
Here's our SPICE circuit, the bulb is represented by a switch which gradually changes resistance from 9.5Ω to 144Ω over the control signal rise (300ms) The light switch is represented by another switch, which just changes from 1mΩ to 10MΩ
Here's the simulation, with the average power shown in the dialog box:
Here is a close up of the switching, with the bulb resistance shown(don't worry about the resistance being negative, that's simply because SPICE calculated it that way using the current flow - it's still a real positive resistance):
And now, here is a simulation with the bulb switched on for the whole time, with average power shown:
You can see that the average power is 95.659W, which is only slightly less than if we doubled the initial 5 minutes on, 5 minutes off test value of 48.2W (48.2" * 2 = 96.4W) so the difference the switching made is tiny.
How quick would you need to switch for it to be worse?
It's probably not possible to make it worse as Supercat rightly notes, since the filament will not cool enough between switching. So take the graph underneath as the worst case scenario (e.g. the bulb is blasted with freezing gas between switching or something :-) Note that this would be adding another source of energy to the system though, so would obviously be cheating) Just how fast it cools down and the effect would be interesting to look at though, and if time permits I'll add some more on this.
So, assuming the above, pretty fast, around once every 2 seconds according to the exaggerated simulation above (in reality, probably about once a second) Here's two minutes worth of switching once evry two seconds, and the average power is just over 100W (~104W):
According to a Mythbusters episode summary on Wikipedia:
" The MythBusters calculated that the power surge from turning on a light would only consume as much power as leaving it on for a fraction of a second (except for fluorescent tube lights; the startup consumed about 23 seconds' worth of power)".
So in fact it is possible that on/off would consume more power if fluorescent was constantly being turned on and off.
The constantly on setting would consume more energy powering the bulb.
A possible counter-argument would be that the turn-on/turn-off cycling would shorten the bulb life, and thus the energy cost of manufacturing, transporting, and disposing of it would be amortized over fewer service hours. But without digging up actual numbers, my gut feeling is that this is unlikely to exceed the operational energy. One plausible way to bound an estimate is to compare the cost of the bulb itself to the cost of powering it.
All of the energy that goes into an incandescent bulb will get converted into heat, which must then be dissipated somehow. Some of that heat will then be radiated off in the form of light, but the energy must start out as heat. Therefore, the only way an incandescent bulb can use more power is for it to dissipate more heat. A bulb which is cold consumes more electrical power than one which is hot, but also dissipates less heat. If a bulb which is powered at a stable temperature is switched off at time T1, cools down somewhat, is switched back on, and has returned to its earlier temperature by time T2, the total energy consumed between time T1 and T2 must be the total amount of heat dissipated, and that is going to be less than the amount of heat which would have been dissipated had the bulb been on continuously.
The only scenario in which an incandescent bulb could use more power when cycled than when operated continuously would be if the bulb had different filament sections which were wired in series and operated at different temperatures (some projector bulbs are constructed like that). In that scenario, cycling the bulb would cause the high-temperature portion to radiate less, but under some duty-cycle conditions would cause the low-temperature portion to radiate more. It would be possible to construct the bulb in such a fashion that the increase in dissipation from the low-temperature portion exceeded the reduction in dissipation from the high-temperature portion, thus increasing overall energy usage; I'm not sure if such conditions would ever apply to any "practical" bulb designs, though.
Leaving a light on uses more power. Switching a light off saves power.
Just assume the light takes zero power when off (POWER_OFF=0), and 100W or whatever when on (POWER_ON=100).
Total power in Watt hours is equal to: POWER_ON * TIME_ON + POWER_OFF * TIME_OFF.
Notice that since POWER_OFF=0, the total power is determined solely by the TIME_ON term.
protected by W5VO♦ Jan 29 '13 at 18:13
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