1
\$\begingroup\$

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?

\$\endgroup\$

1 Answer 1

0
\$\begingroup\$

Your solution appears to be correct.

Your tutor appears to have made an error. He/she assumed the output voltage is 120 volts in order to calculate output current and then, concluded that the output voltage is 124.79 volts. This method would need several more iterations to get accuracy.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.