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I passed the PE power exam with 91%.

Passed  - PE power in TX.

Yes - that's my understanding. Using trial and error method, we can determine the max load that can be supplied by the generator, without overloading either generator. The loads are shared according to the generator impedance.

The calculation results to problems 3-17 & 3-18 seem incorrect. I think the solution should be as follows: Problem 3-17: IN = - (IA + IB + IC) = - (54.8 <(0 – 36.9⁰) + 0.25<(120⁰-5⁰) + 96<(240⁰-36.8⁰)) = -(-44.52 - j70.49) = 44.52 + j70.49 = 83.37 <57.73⁰ Problem 3-18...

Actually I have the second edition and my concerns are regarding the answers in the second edition. The equation is correct, but the results seem to be wrong.

Have the errors in problems 4-62, 3-17 and 3-18 been corrected? The errata on your website does not list a revised solution. Can you please post the correct solution?

Hmm...maybe I did not have to sign up for the Testmasters course. I probably should have just studied with the Graffeo book and other references!

I am also confused with all the study material out there. I signed up for the Testmasters online course. So I have their course materials. Wondering what additional study material to buy. I already have the NEC 2014 handbook. I am planning to buy the Complex Imaginary and Spin-up tests. Not sure...