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[SOLVED] rotary pot power handling

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Zak28

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Suppose a rotary pot rated to 200mW is across a 39v power supply capable of 1.17w the wiper turned to no resistance or its very minimal resistance in few milliohms range, supposing the pot is at 20mΩ the current would be 30mA and power dissipation would roughly be 30mA^2*39v yeilding 0.0351W, safely below the pots power rating. There has to be a mistake in this since supply is 1.17w capable exceeding the pots power handling, I believe there is a mistake somewere.
 

Hi,

to me ... all your values make no sense.

Show a circuit. Show your resistor values and your calculations.

Klaus
 
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    Zak28

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A very confusing question but I think you are asking why a supply rated at 1.17W is only producing 0.0351W of heat. If my assumption is correct, you are confusing the rated power with actual power. The rated power is the MAXIMUM the supply can produce but not necessarily what is actually being utilized. Think of it like a car, it might be capable of driving at 100 MPH (160Km/H) at full throttle but most of the time (hopefully!) you only drive it at lower speed.

Brian.
 
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    Zak28

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Hi,

The rated power is the MAXIMUM the supply can produce
and you (by the circuit) have to take care not to exceed this specification.

Klaus
 
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    Zak28

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30mA^2*39v yeilding 0.0351W

Up to now, power formula has been P=I*V, e.g. 30mA*39V = 1,17W

What's the potentiometer resistance value?
 
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    Zak28

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Hi,

Again: show your circuit. Values alone is not enough.

Klaus
 
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    Zak28

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power dissipation would roughly be 30mA^2*39v yeilding 0.0351W...

Power is I*V=(I^2)*R=(V^2)/R

Your calculation is faulty (unless you are trying to say something else...)
 
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    Zak28

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show your circuit
We have:

voltage source
three potentiometer terminals
some load (if any)

How these are connected?
 
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    Zak28

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