DC- Generator

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Eduardo Maun

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A DC generator has a no-load output voltage rating of 120v. It's armature ckt. resistance is 0.9 ohms and it's field coils are separately. If load rated @ 2 Kw, @ 115 volts is connected across the output. Neglect all the losses.

What power would be absorbed by the load?

My calculation(s) gives me 1,665 watts.

Solution(s):

Rl = V square / Pr = 6.6125 ohms

Il = E / Ra+Rl = 120 / 0.9+6.6125 = 15.87A

Pl = I square * Rl = 1,665 watts

My question(s) is if I P=IV then;

Pl = 15.87 * 115 = 1,825 watts

1,665 watts is not equal to 1,825 watts; so what happened to the equation? something wrong here? which is the correct answer?

Thank you!!!

Ed...
 
Ed - This is a trick question - right?

Eduardo Maun said:
...My question(s) is if I P=IV then;

Pl = 15.87 * 115 = 1,825 watts...
Click to expand...

Where did you get the 115V for this equation? I get 105.6V

carl
 
DC-Generator

DC-Generator

I'm not sure if this a trick question or not, but this is the given voltage in the problem 115, I'm just wondering the solution presented and the answer is not the same with the equation P=IV, what power absorbed by the load which is the correct answer? Is it the 1,665 watts or 1,825 watts? Or something went wrong here?

Thank you!

Ed
 
The load was specified as using 2KW when supplied with 115V.

But your generator is not supplying 115V to this load.

You presumed that the load was a resistor and solved for its resistance. You than placed the resistance in series with the generator resistance, and supplied the combined resistance with the 120V produced by the generator when open circuit. All of these steps are approximations, but they imply that the voltage actually applied to the load is about 106V, not 115V.

-Jon
 
DC-Generator

DC-Generator

Are you saying the correct answer is 1,665 watts not the 1,825 watts? So the power absorbed by the load is equal to 1,665 watts right?

Again thank you for your big help I really appreciate it.

Ed.
 
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