What is the Full-Load Voltage of a DC Generator at 800 rpm?

[Katsmed23]

Homework Statement

A dc generator is rated at 10 kW, 208 V, and 50 A when it rotates at 1000 rpm. The resistance of the armature windings is 0.56 . Assume that the magnitude of the magnetic field remains constant.
(a) Calculate the "no-load" voltage at 1000 rpm (when there is no circuit hooked up to the generator).

(b) Calculate the full-load voltage (i.e. at 50 A) when the generator is run at 800 rpm

Homework Equations

V-Vw - Vl = 0
Vw = IRw
Faraday's Law (where EMF = the magnetic flux with respect to time ->the derivative)

The Attempt at a Solution

I got part a because i know that the no load voltage supplied by the genrator is the voltage of the idealized generator. Thus 208V + (.56 x 50A) = 236 Volts = CORRECT

Part B I can't seem to find! I know we're looking for the voltage load, and that the voltage is poportional to faradays law, where (BxA) x w = radians / sec..

BAsically if V = 236 V at 1000rpm, vL = ? at 800 rpm? I even tried figuring that the rpm frequency is 20% less, so maybe the voltage would be 20% less? (188.8) = wrong

 

[Merlin3189]

Your idea sounds right to me. So what answer does it give?
Uload <> 188.8 because it is the generator voltage, U, that is proportional to speed.
So at 800 rpm U = 188.8 V
Then Uload = U - Uw as you said.
So what is the given answer for (b)?