All 20 amp breakers

[Robert Meier]

Originally Posted by tclayton

Hello David A. Andersen, I am not picking on you just asking if you have any IR pictures depicting the elevated IR signature of the wiring concealed with in the wall. I understand if there is a load applied to the circuit you may see a elevated temp due to the equipment drawing additional current because of the reduced voltage. The comments you posted just may not have given all of the info I needed to understand it. Thanks I am just trying to learn.

Be careful when you say that a reduced voltage will increase current. That is not true when dealing with resistive loads. If you reduce the voltage to a heater by half you will reduce the current by half as well.

[Roy D. Cooke, Sr]

Originally Posted by Robert Meier

Be careful when you say that a reduced voltage will increase current. That is not true when dealing with resistive loads. If you reduce the voltage to a heater by half you will reduce the current by half as well.

I am many years out of school but I think if you half the voltage you quarter the load ... Roy

[Robert Meier]

Originally Posted by rcooke

I am many years out of school but I think if you half the voltage you quarter the load ... Roy

That depends on how you define your use of the word load. It terms of current, the voltage and current are directly proportional so they would increase or decrease at the same proportion, decrease the voltage by half and the current decreases by half too. But it terms of output in watts you are correct, decrease the voltage by half and the wattage drops to 1/4.

[Robert Meier]

Originally Posted by rcooke

I think you need to talk to your apprentice and get him to explain it to you .

Are you saying that my post was incorrect? Then maybe it's true that you've been out of school for too long. For the record I said:

-Resistive load
-Reduce voltage by 1/2
-Current is reduced by 1/2
-Wattage (output) is reduced to 1/4

[Roy D. Cooke, Sr]

Originally Posted by Robert Meier

Are you saying that my post was incorrect? Then maybe it's true that you've been out of school for too long. For the record I said:

-Resistive load
-Reduce voltage by 1/2
-Current is reduced by 1/2
-Wattage (output) is reduced to 1/4

Sorry Robert My mind is tired , Thanks for bearing with me ... Roy

Trying to Follow our Federal election and this Forum is not a smart thing to do .

[Robert Meier]

Originally Posted by rcooke

Sorry Robert My mind is tired , Thanks for bearing with me ... Roy

No problem. Your one-liner was pretty funny.

[Terry Clayton]

Robert I apologize for my statement, I was actually thinking incorrectly while typing, thinking along the lines of the difference 120v vs 240v sorry. But I was just curious mostly about his pictures, thanks fro your response.

[Robert Meier]

Originally Posted by tclayton

Robert I apologize for my statement, I was actually thinking incorrectly while typing, thinking along the lines of the difference 120v vs 240v sorry. But I was just curious mostly about his pictures, thanks fro your response.

No need to apologize. We're all here to share ideas and maybe even learn a thing or two.

[David A. Andersen]

Originally Posted by tclayton

Hello David A. Andersen, I am not picking on you just asking if you have any IR pictures depicting the elevated IR signature of the wiring concealed with in the wall. I understand if there is a load applied to the circuit you may see a elevated temp due to the equipment drawing additional current because of the reduced voltage. The comments you posted just may not have given all of the info I needed to understand it. Thanks I am just trying to learn.

I'm not sure I understand what it is you're looking for or asking.
I have most of my IR scans on another computer/server. I don't really take pictures of hot wires inside the wall, though I do have some somewhere. I usually try for a more direct temperature measurement by removing an electrical component on the circuit as the direct temperature of the conductors are more important than sheet rock wall temperature.
　



The circuit that this outlet was on had a voltage drop. This was the hottest outlet on the circuit and was the source of resistance on the circuit.

The voltage drop occurs downstream of this outlet, not upstream.
We see this temperature rise with the thermal camera due to conduction of heat energy down the electrical wire. It is not necessarily because there is an increase in current through the conductor.

If you have a very high resistance load on the circuit which is at close to full capacity, you may see upstream conductors warmer than downstream because the current passing through the upstream conductor is equal to the load on the circuit plus the high resistance loss occurring on the circuit.

It is unlikely you will see this with most thermal cameras.
You can see that the upper wires on this plug are warmer than the lower wires, but this is more likely due to convection within the electrical box.
　

[Allen L. Catterton]

So good to see Marc Shunk posting here again. I have always enjoyed his informative and specific statements regarding electrical issues.