REFLECTOR:A (re)volting problem

John Dibble reflector@tvbf.org
Fri, 03 Oct 2003 10:34:44 -0400


For a given resistance, volts and amps change proportionally together. i.e. Higher
volts = higher amps and the power output increases by the square of the volt/amp
increase .  For a given power output, you can increase one and decrease the other, BUT
only if the resistance is changed to achieve the desired power output.  I would think
you would be drawing less than 10a unless, as was previously discussed, the resistance
is much less at the lower power output.

John

Chuck Jensen wrote:

> The plot thickens.  Yesterday evening, I went out to get my night landings
> in for currency and tested the pitot.  I pushed in the circuit breaker and
> turned on the pitot switch.  After 5-10 seconds, the "10" (amp, I assume)
> breaker popped.  I cycled it again, same results.  I checked and the voltage
> dropped from 12.8v to about 11.0v, then the breaker would pop after a few
> seconds.
>
> Based on my passing understanding of electricity (I once passed by the
> Electrical Engineering building), when you go up in volts, you go down in
> amps.  I assume the converse holds true; when the voltage is cut in half,
> the amps double?  So, an instrument rated for 10a at 24v is going to need a
> 20a circuit at 12v (actual draw is approximately 80%, or 16a)?
>
> Is that thinking correct?  Or is there some squaring and doubling going on?
> If the circuit has 14 gauge wire, can a 20a push button breaker be safely
> used or is that asking impolitely for trouble?
>
> Chuck, coming up short of a perfect solution.
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