A 12 kW, 120 V DC series generator has a series field resistance of 0.05 Ω. The generated EMF is 135 V when the generator delivers the rated load at its rated terminal voltage. The armature resistance is 0.1 Ω. If the generated EMF is 130 V and the generator is supplying 4.167 kW, what is the terminal voltage?
This question is from Bhag S. Guru, Hüseyin R. Hiziroglu: Electric Machinery and Transformers.
My solution:
Ea = 130 V
Ea - (0.1+0.05)IL - Vt = 0 where Vt = 4.167k/IL
Then I solve the quadratic equation for IL. The two solutions are 33.34 and 833.33.
IL must be 33.34 A because the rated load current is 100 A, and so the terminal voltage Vt must be 124.985 V.
Is this correct?
In contrast, here is my tutor's solution:
Ea - (0.1+0.05)IL - Vt = 0 where IL = 4.167k/120 and Ea=130 V Vt=124.79
But it doesn't make sense to me that IL = 4.167k/120. Isn't load current the output power (which is 4.167 kW in this case) divided by the terminal voltage (which is what we're trying to find)? Has my tutor made a mistake?
