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@zgsharonI did it on NEC so I am sure it can be used.

@akyip They have a line tool which you can use to draw a straight line from end to end. And it never goes away, so if you want to recheck your values anytime, you can do that easily as well.

Just came in!!. Huge shout out to this forum for all the help!! Also, a big thank you to Zach's, Engineering Pro Guides and Grafeo's practice tests. And not to forget, the awesome videos on Zach's youtube channel. All of these were monumental in getting me through this exam!! I found the...

Sorry to hear @DuranDuran. I hope you pass on Jan 7th! In the diagnosis, do we get to know how many questions we did right? Can you let us know how many you got or what you think the passing score might be? Just trying to figure out if the passing score has lowered now in CBT format.

Yes that is correct. As far as I understand, when we use the Zeff formula, Zeff is already at the pf angle, -36.86. If we again multiply it by [email protected] degrees, the Vd will be at an angle of - 72.73 which will be incorrect.  May be my understanding is incorrect, but that's how I thought it works.

The extra 90 is to take into account for the j in 0.9j which is the synchronous reactance.

In my opinion, the angle is already taken care of when using the Zeff formula. We can either use the pf in Zeff calculation and mulitply with I mag OR use (R+jX) and then use I at an angle. 

This is from my notes when I solved it in the exam. Its a bit messy so let me know if you cant read it and I will try to write it up on a clean sheet.

Alternate method being the alternate solution they mentioned above. I calculated the Zeff based on the formula on NEC chapter 9 Table 9. Then multiplied it by the current. The same formula is also given in the handbook. NEC -    Handbook - 

From the alternate method, I am getting Vd of 5.2V. Now given Vline is 480 and the Vd is per phase, so converting it to Vphase, we get V=277.128. Vphase - Vd= 277.128 - 5.2 = 271.92. Converting it back to line value, Vline=sq.rt 3 * 271.92 = 470.99 Let me know if there is something wrong with...

Apart from the angles, shouldn't Ib in the addition for In be negative? If we do KCL, Ia and Ic will be leaving the neutral point N and Ib will be entering it. Or am I missing something?

1. Equivalent Z = 50.225 2. Transferring to the primary side, Zp= 50.225 * 7/2 * 7/2 since transformers have same polarity, the ratio is 7/2, while transferring impedance, we multiply by square of the ratio 3. I = 240/Zp = 0.3900 A

Thanks for putting out all the helpful stuff Zach! Can't wait for the lectures to start! Would love to receive a free copy! Thank you!

Hi Zach, your online lectures and tips are really helpful. I would love to get a copy of this, please!