I was wondering how they worked. A little bit of background info.
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I was wondering how they worked. A little bit of background info.
A high current passing through a thin wire (with high resistance) causes high heat (you can get a rough idea from the radiative power formula,).
This is essentially the same as how a filament lightbulb works.
Why did you disagree with me, our equations are identical. If you've ever passed a high current through a resistor or looked at a lightbulb, you'll know that in this case, the high current and the resistance are important to produce heat.
They might be identical equations, but they don't have identical conseqiences. R=pl/A or resistivity * length/(cross sectional area). In this case you will see that a smaller cross sectional area or thinner wire, means a HIGHER RESISTANCE, not lower as you stated.
We were designing a heating element for an electrical system whose voltage is fixed. Power will be another variable. Lets say the maximum is 120 W since a cigarette lighter typically has a maximum current of 15A, I'm using an easy number such as 10 Amps.
We then can select R
frm P=(V^2)/R or R=(V^2)/P we have (12)(12)/120W or 1.2 ohms would satisfy the design: 120W, 12V; we need a 1.2 ohm heating element.
Now lets design one for 60 W, 12V
(12)(12)/60 = 2.4 ohms
Thus a 2.4 ohm resistance would satiisfy our design of 60 W and 12V.
Is 60 watts less than 120 watts?
Is 2.4 ohms greater than 1.2 ohms?
Isn't Watts a measure of heat generated or dissapated?
If so, then high resistance causes low heat which is a directly opposite of your post.
We can plug it into your formula, only after we determine I which will be I=V/R, where V=12 and R is 1.2 and 2.4
I'll leave that as an exercise. But wait... You already told me the equations were the same.
PS: I know it doesn't help that a 31 AWG wire has a smaller cross secional area than a 0 AWG wire and that a 31 AWG is a "smaller" wire, but it's the same for drills and steel for that matter.
I quote: "a thin wire (with high resistance)". I never stated that a thinner wire would give a lower resistance. I said the opposite.Quote:
Originally Posted by KeepItSimpleStupid
P = I^2R is the equation that is used for radiated power. Do you disagree with this?
How about the following compromise:
Your explanation, if we are talking about fuses.
My explanation, if we are talking about cigarette lighters and resistance heaters.
The real explanation is somewhere else. It's the current required to raise the heating element above the ignition point of paper (451 deg F) and tobacco without cooling sufficiently or burning out the wire in the presence of air. Raising the temperature means lowering the resistance in a 12V system. Wires might be made of Ni-chrome which has a higher resistance than copper and will not oxidize or burn out in the presence of air.
I think It's perspective. I think Your correct and I'm correct.
KISS
... the wire is coiled, when heated its length would get longer, it have no room to expand; it POPOUT- ready for a mokeQuote:
Originally Posted by KeepItSimpleStupid
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