Re: Anyone know anything about electric heater wire


"Jack Denver" <nunuvyer@xxxxxxxxxxxx> wrote in message
news:eP2dnUnKLoqjTATeRVn-qA@xxxxxxxxxxxxxx
> You would want to use them in series for greatest resistance - you
wouldn't
> get much light though. I could see buying a cheap fixture like a six bulb
> bathroom light bar and rewiring it in series with 6 100 watt bulbs. If you
> created some kind of simple switch network you could even vary the
> resistance .
>
That's not correct.
The bulbs need to be in parallel and here's why:
For a 1500W element the on resistance at full power is R=V^2/P =
115*115/1500 = 8.8 ohms
If we want to drop down to say 1150 W over the lights + heater then its
11.5 ohms total (neglecting where the heat from the extra 2.7 ohms will go).
Now the _cold_ resistance of a 100W light bulb is of the order of 10 ohms
so already even before you pass current through it it's too much resistance
since we only need an extra 2.7 ohms. To get a lower resistance the bulbs
need to be in parallel.
Based on the _cold_ resistance you would need 4x100W bulbs in parallel to
get close at 2.5ohms.
However as soon as any appreciable current passes through the bulbs their
resistance will go up making the need for more than 5 bulbs.
I suspect that you might need about 20 bulbs to get those 2.7 ohms. And
then bulbs would then be dissipating 1150*2.7/11.5 = 270W total or 13.5 W
apiece leaving only 880W for the heater.
Conversely even two bulbs in series when cold would cut the power down to
115x115/(10+10+8.8)= 460W most of which (approx 2/3) would be dissipated in
the light bulbs.





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