Copper Losses in an Electric Generator

In summary, the voltage and current of a generator are not directly related to V=I*R. The voltage and current depend on the load on the generator, and the generator's windings must be designed to handle the load.
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When I use Ohm's law and other related equations to calculate the heating losses in a generator I get the power output of the generator. I assume I'm using the equations wrong, and I was hoping that someone could help set me strait. I think what I would like to know is what determines the voltage and current of a generator and why it isn't directly related to V=I*R; but that might not be the right question.

Just as an example if I take this generator 350W at 24V gives us about 14.6A (P=I*V). That means the internal resistance of the motor is 1.6 Ohms (V=I*R). What I've always used for calculating power loss in a wire is P=V*I^2; which gives us 350W. That would mean that the efficiency of a generator is 50% but I've always understood that under optimal circumstances generators operate closer to 90% (just talking about the efficiency of mechanical rotation to electrical power, not including the efficiency of producing that mechanical rotation). I'm not surprised that the values are coming out the same since I'm using the same equations in reverse.

Thanks for the help.
 
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  • #2
The R you're calculating there is the load on the generator, not the generator's winding resistance. As if you had a 24V, 350W light bulb connected to the generator; then the light bulb would have those values. There is nothing in your problem description that gives us any information about generator's windings.
 
  • #3
Well first of all, every generator except the Faraday disc/homopolar one is an AC generator not DC. DC only arises after rectification and sufficient smoothing of the AC ripple voltage. All generators use rotating magnets and coils either way you slice them, that means you need to calculate the AC rms output which is the actual output of a generator since it outputs either a sine wave or in older DC style dynamos it outputs a pulsating DC which is essentially a half period sine.
P=I*V is only true for instantaneous power at any given moment and also true only if your voltage and current waveforms are in phase.
If their not in phase you have to additionally multiply by the cosine.

See the link
https://www.electronicshub.org/power-formula/

But overall there are two types of losses in an AC circuit, one is the same as in DC circuit and that is resistive loss , where you form a voltage drop across a piece of conductor that manifests as heat and the other loss is purely AC phenomena - reactive power/radiation.
Every time you have a time varying current you charge and discharge capacitance and inductance along the way , and every circuit has both to varying degrees.

Now in simple terms the way to know the resistive loss of a generator's winding is to measure the resistance of the winding and then to know the RMS voltage across it and out of that one can calculate how much power is wasted as heat. You have to use Ohm's law and get the current through the winding then use the voltage drop across it and multiply that by the current and you will get the power that the winding consumes as a resistor that manifests as heat

But a hint is this - it's a small amount compared to the total power output, otherwise your 350w generator would become a heating element and burn up quickly.
 
  • #4
mudmucker said:
Just as an example if I take this generator 350W at 24V gives us about 14.6A (P=I*V). That means the internal resistance of the motor is 1.6 Ohms (V=I*R). What I've always used for calculating power loss in a wire is P=V*I^2; which gives us 350W.
Your calculations assume the whole 24V is applied across the internal resistance. That is true only if the generator is short circuited. Instead, think of the circuit below. R1 is the internal resistance and R2 is the load resistance. Think of a light bulb for example. Now if R2 is much greater than R1, then most of the voltage drop and power consumption will be across R2, not R1.

Ideally, the internal resistance R1 would be zero and the efficiency of the generator would be 100%. But in the non-ideal case, we can make R1 as small as possible.
1667212891001.png
 
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  • #5
Thank you all! I knew that I was missing something.

The separation of load resistance from generator resistance was very helpful. I'm also thinking of voltage change in the generator as happening in two parts. There's the main voltage gain due to the generator operating, and then there is a small voltage drop in the generator due to resistive losses in the windings; with the sum of those two being the external voltage gain of the generator.
 
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  • #6
mudmucker said:
Thank you all! I knew that I was missing something.

The separation of load resistance from generator resistance was very helpful. I'm also thinking of voltage change in the generator as happening in two parts. There's the main voltage gain due to the generator operating, and then there is a small voltage drop in the generator due to resistive losses in the windings; with the sum of those two being the external voltage gain of the generator.
And just to add, the voltage that appears across any of the generator coils (for a free standalone generator not connected to grid) is dependent on the turns count of the coil, the max B field strength through the coil and the rate of change of that B field. The faster the field changes the higher the voltage will be.
That is why if you spin your generator faster you get higher output voltage as well as frequency.
 

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