Re: ignition and calculations

[n9ads@domain.elided (Howard R Pletcher)]

On Sun, 30 Aug 1998 21:03:21 -0400 (EDT) cookiedan@domain.elided (Daniel
Nees) writes:
>John,
>   I screwed up and didn't explain myself properly. The calc that I was
>getting at was the amount of power the resister uses. Ex. 1 12V*.43
>ohms=5.6V Now in your example of the 12V*2ohms=24V is corect as far as
>the math goes, but; wouldn't work.................. 

Back to the books, Dan.  I'll vote for John's analysis of the situation. 


I''m not sure what your V*R calculates, but the voltage drop across a
resistor is I*R and the power is I^2*R  (Current squared * resistance).

And current will still continue to flow through a 100W light bult with 12
V applied, just not enough to make it give off light.  A 100 watt bulb
has approximately 144 ohms resistance--Power=100 Watts = I^2*R  which is
also = V^2/R  so 100=120^2/R  or
R= 120 * 120  / 100  =  144.    And the current flowing through it is
120/144 = .833 amps.

 At 12 volts,  current =  V  /  R  =  12/144  =  .0833 amps, or 1/10 as
much as could be expected from applying 1/10 as much voltage.  (Actual
test results may vary as I believe the resistance of the filiament does
not remain constant, but increases as it reaches the temperatures that
make it white hot.)

Howard Pletcher
Howteron Products Scout Parts


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