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1. Junior Member
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Oct 2012
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Hello,

i tend to over think certain things and complicate matters.

Im going to explain my problem understanding this very basic situation regarding voltage current,resistance, in a resistive circuit and an inductive circuit. Now, I want to keep things very simple, getting into inductive reactance or impedance etc, shouldn’t need to be brought to the table unless completely necessary.

For what ever reason I’ve hit a road block, so here goes -

In a resistive circuit,in order to produce a wattage of 500W with 12VAC, current would be 41A and the resistance would need to be .292ohms. If the voltage was reduced even further the resistance would then need to be reduced in order to allow more current to flow to produce the same 500W.

To produce the same 500W with 600VAC .83A and the resistance would need to be 722ohms.

The reason we would need to reduce the resistance in the 12VAC circuit is because 12VAC simply isn’t enough voltage to push current through 722ohms and produce 500W. The same situation can be said for the 600VAC circuit, if applied to .292ohm the current would be greatly increased and their for wattage.
Im struggling with the relationships, high voltage low current same energy. Low voltage high current same energy.

Maybe this is because .83A flowing through a higher resistance of 722ohms would in fact cause 500W of heat, and 41A flowing through a much lower resistance of .292ohms is needed to produce 500w of heat.
More current is needed to flow in a lower resistance to produce the same heat as a lower current in a higher resistance? Is it just that simple?

Now, when talking about motors single phase squirrel cage induction motor, lets use the same values mentioned above. Wouldn’t the motor with 12V and 83A produce a might higher magnetic field on the stator then the 600V motor with .83A?

I’ve successfully confused the living hell out of myself.... sorry if this makes absolutely no sence to you guys.

2. Senior Member
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Apr 2009
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Williamsburg, VA
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I find it easier to work with whole numbers... So let's take a 120 watt incandescent light bulb. It draws 1 amp on a 120 volt circuit. (P=VI). It also has 120 ohms of resistance. (V=IR).

those two equations also lead to P=V2/R, and P= I2R.

The resistance of the filament of the light bulb will change some with the voltage applied, however consider it as an ideal resistor that does not change resistance based on voltage or temperature. If you drop the voltage by 10%, the wattage of light bulb decreases by the square of the drop. In this case, 21%... Which is fairly close to the wattage loss of using a 240-volt Appliance on a 208 volt circuit.

so, answer the first part of your question, I believe you on the right track. As to your motor question, I really have no idea.

3. gar
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181207-2023 EST

Phase:

The magnetic field intensity of a coil is B = K*N*i. K is some constant, N the number of turns, and i is the independent variable for current.

.

4. Senior Member
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NE (9.1 miles @5.07 Degrees from Winged Horses)
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Originally Posted by gar
181207-2023 EST

Phase:

The magnetic field intensity of a coil is B = K*N*i. K is some constant, N the number of turns, and i is the independent variable for current.

.
Wow, that brings back some vague memories of a long ago motors class. One I didn’t care about because I was never going to work with motors.

5. Senior Member
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Originally Posted by ptonsparky
Wow, that brings back some vague memories of a long ago motors class. One I didn’t care about because I was never going to work with motors.
Same here!

I get friends asking me to hook up their motors on well pumps or air compressors sometimes. I tell them flat out: I will hook it up, but I WILL NOT confirm if you've selected the right motor for the job.

In the HVAC world, of course, I swap motors everyday. And I have six motors on the truck so that I can replace like for like with fewer trips to the store. I don't need to know why it's a 1/3 HP for that configuration. I just need to know that the nameplate says it's a 1/3 HP motor in there, and sure enough, that's what I took out, so that's what goes back in.

6. Originally Posted by Phase
Maybe this is because .83A flowing through a higher resistance of 722ohms would in fact cause 500W of heat, and 41A flowing through a much lower resistance of .292ohms is needed to produce 500w of heat.
More current is needed to flow in a lower resistance to produce the same heat as a lower current in a higher resistance? Is it just that simple?
Yes. Remember that current is the result of a voltage applied across a resistance. One volt will push one amp through one ohm, and dissipate one watt. If you raise the voltage or lessen the resistance, current increases, resulting in greater power, and vice versa.

Power, in watts, is defined as voltage times current, so, in order to consume a given wattage (usually the design goal), increasing the resistance is the correct way to reduce the current when designing for a given power when connected to a greater voltage.

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