Re: Anyone know anything about electric heater wire

I hate people who understand this sort of sh*t

ken
:-)

"Johnny" <removethis.huuanito@xxxxxxxxxxx> wrote in message
news:%Tpmf.500$Ru.303@xxxxxxxxxxxxx
>
> "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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