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Re: Lights out!!

To: glenel@infobahnos.com
Subject: Re: Lights out!!
From: jurrasm@genesis.torrco.com (Mark Jurras)
Date: Mon, 17 Jun 1996 09:38:17 -0400
> Reply-To: glenel@infobahnos.com
> 
> Mark et al.,   From the depths of my rusty store of electrical formulae, 
> I seem to remember that watts divided by volts equals amps.  Given that 
> is so, if the voltage drops for a given wattage of lamp, that SHOULD 
> mean that the amperage goes up.  I don't know how much current that 
> headlight circuit is fused for, but a drop of 4 volts in a 12 volt 
> circuit means an increase of 50% in the amperage flowing through it.  
> Somewhere, something has got to give; let's hope it doesn't require a 
> fire extinguisher! :-)

Let's take a walk back to physics class. Three things you need to
remember are:  
1) F=MA 
2) V=IR 
3) You can't push with a rope.

>From equation 2 it is intuitively obvious (I always loved that one)
that Power = Volts * Amps. For a 100 Watt 12 Volt bulb that would be
about 8.3 amps. The resistance of the bulb is around (R=V/I) 12/8.3=1.4
Ohms. If you were to reduce the voltage to say 4 volts the current
would be (I=V/R) 4/1.4=2.8 amps. Because the resistance doesn't change
when you reduce the voltage the current also drops, therefore the power
drops.

I hope this jogs a few memory cells. 

- -Mark   = =o&o

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