Re: Subject: 6V to 12V

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

On Thu, 06 Aug 1998 06:48:24 -0600 Jim Shepherd <beltguy@domain.elided> writes:
>Lets take another look at this subject.  Some have used the equation
E=IR
>and have concluded that you get twice the amps.  That statement is true
for
>the maximum wattage *capability* for a given circuit.  However, whatever
is
>in the circuit has a wattage demand.  For example, a headlight might
take
>75 watts.  Here, the equation Watts=volts*amps applies.  Thus if you
double
>the voltage, you half the amps.  Obviously, with half the amps for every
>circuit, the wire is very well sized.  The only reason to change is if
the
>insulation has gone bad.
>
>
Time for a little basic physics to straighten this subject out.  I think
most everyone is basically correct in what they're trying to say;it's
just that not everyone is hearing the same thing as is being said or else
looking at only part of the situation.

I=V/R and Watts=V*I always apply in every case involving simple
resistance.  

If you double the voltage and change nothing else, you will get double
the current, at least momentarily until the resistance burns out.  You
will change your 75 watt bulb into a 150 watt flash bulb.  

If you change to a 12V, 75 W bulb, you will draw 1/2 of the previous
current (6.25 A vs. 12.5 A previously) because the 12 V bulb will have
double the resistance and the old wire used for 6V will be more than
adequate for the 12V circuit. 

 If you could just apply 12V without burning out the 6V bulb, the wire
might not be adequate to handle the doubled amps.

As Jim noted, some items (starter, horns) are rugged enough to withstand
the change to 12V, but this is partly because they are used
intermittently and don't get overheated by the extra current (and
power)(we hope).  I'm not sure, but I suspect the gauges worked
satisfactorily because they have a voltage regulator that is rugged
enough to handle the greater input voltage while still sending the proper
voltage to the gauges.  Here we have an active device so I=V/R doesn't
apply.  (That's why it's called a voltage regulator.)

As to Joel's question about keeping the power consumption constant, I'd
think you'd have to look at each item in the system and replace it with
one using the same power (or 1/2 the amps) on 12 V vs. the 6 V components
(doing this, I don't think you'll get brighter lights--a 75 W lamp puts
out the same amount of light whatever voltage it's DESIGNED for--even 110
V, but if you replace a 75 W, 6 V with a 85 W, 12 V .......)  

The resistance of the wire does remain constant and theoretically you'd
be wasting less power generating heat in the wire using it on 12 V vs. 6
V because you'd be carrying less current if you had the same power lamp,
but this effect would be so small relative to the power used in the loads
we can negelect it unless your wiring is hopelessly undersized.

Now class, any questions?  

	No? Good, close your books and we'll have a quiz.

Howard Pletcher
Howteron Products Scout Parts

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