Timeline for How to calculate the effect of different voltages on an incandescent bulb?
Current License: CC BY-SA 4.0
8 events
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| Nov 7, 2023 at 8:06 | comment | added | Russell McMahon♦ | @Neil_UK Yes. Even a few percent works woners. Lonnnnng ago people used to make startup surge reducers. Memory said that just those made a very great difference. || Meaurement is what I'd usually do :-) . | |
| Nov 7, 2023 at 6:22 | comment | added | Neil_UK | @RussellMcMahon Lots of mechanisms, so measure it, and fit it. The main takeaways are that the current exponent is a long way from linear, and the lifetime exponent is huge. | |
| Nov 7, 2023 at 1:05 | comment | added | Russell McMahon♦ | @Mattman944 Without vast thinking (rushing): Sounds right. Rear of brain niggles. Changing temperature of filament affects resistance (of course). Power loss through enevelope is 4th power of delta t related, and so temperature and so resistance vary with wattage with differing laws. Then there is envelope to ambient, which is less variable. (Convection also matters in m=both cases, almost certainly to a lesser extent). There are so many mechanisms with different exponents and laws that I suspect that "it's complex" :-). Maybe not. | |
| Nov 6, 2023 at 14:32 | comment | added | Mattman944 | @RussellMcMahon - Note that the current exponent and the power exponent should differ by exactly 1. They are both approximations, and apparently two people came to slightly different conclusions. See my answer. | |
| Nov 6, 2023 at 8:58 | history | edited | Neil_UK | CC BY-SA 4.0 |
added 117 characters in body
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| Nov 5, 2023 at 12:22 | comment | added | Russell McMahon♦ | Nice reference. From that page this is useful. Note that that they say Power is proportional to V^1.6 BUT they also say " ... The relationships above are valid for only a few percent change of voltage around standard rated conditions, ...". | |
| Nov 5, 2023 at 12:01 | vote | accept | Dennis | ||
| Nov 5, 2023 at 10:16 | history | answered | Neil_UK | CC BY-SA 4.0 |