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?

  • 6
    \$\begingroup\$ If you're about saving energy consider using LED technology. \$\endgroup\$ – Marcel Jan 23 '13 at 7:23
  • 7
    \$\begingroup\$ What sort of bulb? \$\endgroup\$ – Nick Johnson Jan 23 '13 at 10:32
  • 2
    \$\begingroup\$ I wonder if the answer is the same across all varieties of lights. I've seen tests of incandescent, florescent, and LEDs. But what about high pressure sodium, metal halide, mercury vapor, etc? \$\endgroup\$ – Grant Jan 24 '13 at 0:35
  • 3
    \$\begingroup\$ @marcel or really "anything but incandescent".. pretty decent 60w LED bulb equivalents are down to about $15 on Amazon at the time of this comment, and pull no more than 9w, often less. \$\endgroup\$ – Jeff Atwood Jan 24 '13 at 8:14
  • 2
    \$\begingroup\$ I always thought the myth was about the bulb lasting longer if left on not that you would save on power. The cycling of the power causes the filament to heat and cool and that causes it to wear out faster. I assumed it was the same thinking behind flourescent lighting being left on all the time. Things wear out faster if their state is changed to often. \$\endgroup\$ – Chef Flambe Jan 30 '13 at 2:16

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:

  1. this is for a fraction of a second
  2. getting it up to temperature doesn't take any more energy than it would have taken to leave it on to maintain that temperature
  3. 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.

  • 3
    \$\begingroup\$ One consideration is that some fluorescent bulbs don't warm up to full brightness until they've been on for a minute or more. They might turn on at only about a quarter of full brightness and rise slowly. This can be a nice feature to help your eyes adjust from a dark room to a light one, or annoying if moving from one lit room to another. One may wish to trade off a somewhat higher energy cost if this is an issue. \$\endgroup\$ – Matt B. Jan 23 '13 at 7:44
  • 4
    \$\begingroup\$ +1 for good answer. Do you happen to have a link to the report? Sounds very plausible, but it would be interesting to read (and always good to have the reference). \$\endgroup\$ – Leo Jan 23 '13 at 8:08
  • \$\begingroup\$ @Leo - just looking at the electrical properties of an incandescent bulb (resistance-temperature curve & time constant, etc.) should allow you to mathematically model it and convince yourself of the answer. \$\endgroup\$ – John U Jan 23 '13 at 9:14
  • 3
    \$\begingroup\$ 60 seconds sounds waayyy too long for a incandescent bulb. Consider they turn on pretty much instantaneously in human terms. Let's say 100 ms to be generous. Even if the cold current is 10x the hot current, that only accounts for 1 second of running time. \$\endgroup\$ – Olin Lathrop Jan 23 '13 at 13:28
  • 3
    \$\begingroup\$ For incandescent bulbs, anything that increases the power consumed by the bulb must increase the total amount of heat dissipated to the environment (via visible radiation, infrared radiation, conduction, convection, etc.) For normally-constructed bulbs, switching the bulb off for any interval will--so far as I can tell--unconditionally reduce the amount of heat dissipated, and must consequently reduce the amount of energy consumed. \$\endgroup\$ – supercat Jan 23 '13 at 15:57

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

Bulb Test Circuit

Here's the simulation, with the average power shown in the dialog box:

Bulb Test Simulation

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):

Bulb Test Closeup

And now, here is a simulation with the bulb switched on for the whole time, with average power shown:

Bulb Test On

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):

Bulb Test Quick Switch

  • 1
    \$\begingroup\$ +1 for graphs. Mythbusters proved the same, but they did show that fluorescent consumes significantly at startup. \$\endgroup\$ – Gustavo Litovsky Jan 23 '13 at 15:39
  • 1
    \$\begingroup\$ Yes, I think I recall seeing that show ages ago. I might take a look at the fluorescent bulb a little later and add it in, as I'm sure it will consume a lot more power on startup so it would be interesting to compare. \$\endgroup\$ – Oli Glaser Jan 23 '13 at 15:41
  • 1
    \$\begingroup\$ I don't think any duty cycle could increase power consumption in a conventionally-constructed incandescent bulb; perhaps you can read my answer and tell me if there's any flaw in my reasoning. \$\endgroup\$ – supercat Jan 23 '13 at 15:49
  • 1
    \$\begingroup\$ @supercat - probably not, due to the filament not cooling down enough between switching, which I realised just a little while ago. So I think you are quite right, and I'll add a note on this, and probably change that simulation later when I have a bit more time to look at the fluorescent bulb also. The main point (as you know) was to show how tiny the effect of the switching is overall. \$\endgroup\$ – Oli Glaser Jan 23 '13 at 15:57
  • \$\begingroup\$ @supercat - note that we are only taking the bulb itself into account here, and not the rest of the system. Looking at the wiring impedance and other non-ideal system factors might be interesting too (I just haven't time to do it justice now though) \$\endgroup\$ – Oli Glaser Jan 23 '13 at 19:30

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.

  • 1
    \$\begingroup\$ You need more energy to turn it on, but you must subtract the energy you save by turning it off. \$\endgroup\$ – Al Kepp Jan 23 '13 at 13:00
  • 1
    \$\begingroup\$ @AlKepp: This is all dependent on the "duty cycle" \$\endgroup\$ – Gustavo Litovsky Jan 23 '13 at 15:40

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.

  • 2
    \$\begingroup\$ The cost of the bulb is an excellent way to put a number to the transportation and manufacturing costs. People don't sell them to lose money, after all. You'd only have to worry about externalities like the environmental burden of manufacturing that isn't charged back to the manufacturer. But this is a diversion anyway, the question asks specifically about energy use, not cost. \$\endgroup\$ – Phil Frost Jan 23 '13 at 16:27
  • 1
    \$\begingroup\$ The idea would be that the cost constrains the manufacturing/transportation cost estimate and thus the energy consumed to accomplish that. But some externalities could internalize as energy usage, for example, processing a waste stream would likely consume energy (since the wastes involved here may not have potential as a fuel to drive the process) \$\endgroup\$ – Chris Stratton Jan 23 '13 at 16:31
  • 1
    \$\begingroup\$ +1 for another practical way to look at the problem. I think though it's a very simple question, you could write quite a lengthy (and math heavy) paper on the various factors hinted at in these answers. Just the impedance of the wiring, switch bounce, thermal characteristics of the bulb would be fun to start with... \$\endgroup\$ – Oli Glaser Jan 23 '13 at 19:18
  • 1
    \$\begingroup\$ According to the Mythbusters episode mentioned in another answer 'Furthermore, the wear and tear of turning the light on and off repeatedly did not reduce the bulb's total life expectancy enough to offset the increased electricity usage. Therefore, it is far more economical to turn a light off rather than leaving it on'. \$\endgroup\$ – Sarel Botha Jan 24 '13 at 14:22

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.

  • \$\begingroup\$ This reasoning makes an assumption of where you measure the energy consumption for the purposes of the original question: if you measure it electrically near the bulb socket this is correct. But it neglects the possibility of losses elsewhere due to the usage and pattern of usage, which would not show up as waste heat in the bulb itself but rather in the power plant, distribution grid, etc. The conclusion is probably unchanged though. \$\endgroup\$ – Chris Stratton Jan 23 '13 at 16:34
  • \$\begingroup\$ @ChrisStratton: If the infrastructure is modeled as resistive, that's similar to the situation with high- and low-temperature filaments; if the infrastructure is more "complicated", then anything's possible. Indeed, if a building which is served by long cables has a heavily-inductive load, switching on a bulb that was in series with a large cap could could--at least in theory--reduce the line losses by an amount much greater than the power consumed by the bulb. \$\endgroup\$ – supercat Jan 23 '13 at 19:38
  • \$\begingroup\$ "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." Don't those sentences completely contradict each other....? The only way to consume more power is more heat, but the one consuming more power gives off less heat.....? \$\endgroup\$ – Affe Jan 23 '13 at 20:22
  • 1
    \$\begingroup\$ @Affe - actually, they don't, but it's not a situation that can last. A hot bulb does dissipate more heat, because the dissipation rate is a function of temperature. However, the cold bulb's failure to dissipate as much will quickly result in it becoming a hot bulb. \$\endgroup\$ – Chris Stratton Jan 24 '13 at 19:03

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

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.