REFLECTOR:A (re)volting problem

Dave Black reflector@tvbf.org
Fri, 03 Oct 2003 15:51:54 -0400


Chuck,

> 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?  

No. This is what you'd need to do if you desire to hold power constant. In the
case at hand, power absolutely will NOT remain constant, because the
resistance of the pitot tube will not change. When you cut voltage in half,
you also cut current in half. Voltage x Current = Power, so power reduces to 1/4.

> So, an instrument rated for 10a at 24v is going to need a
> 20a circuit at 12v (actual draw is approximately 80%, or 16a)?

If you could reduce the pitot's resistance to maintain power this would be
true. But since you can't, current will go down with voltage, so you'll have
1/4 the power. A 5a breaker will do the job.
 
> Is that thinking correct?  Or is there some squaring and doubling going on?

Squaring. That's because when you double the voltage across a resistance, the
current doubles also. Power = voltage x current.

> If the circuit has 14 gauge wire, can a 20a push button breaker be safely
> used?

No. If I recall correctly, 14 gauge wire is rated at 15 amps. 
But that's fine -- all you'll need is 5 amps to run this pitot. Use a 5 amp
breaker. 

Dave Black
Shortwing RG