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Thanks it was easier than expected!

Any Tips for the "PE Laws and Rules Exam" for those Applying to be a PE in CA? I was fortunate enough to pass the PE exam recently and just submitted by application here in CA. At the end of the application i see we need to take the "PE Laws and Rules Exam"? Is this one of those commonsense...

You know what i see it now. In the solution we find Scommon= Ic*Vout If we wanted Srated then Srated=Sout=I2*Vout or we use the relationship of the turns ratio provided. On a unrelated note just for clarity Nse-Series Winding turn ratio Nc=Common winding turn ratio How would you find the...

Thanks for the input Zach! So in such cases it seems its not straightforward to try to derive the rated power of the auto XFMR. It seems knowing the formula may be a good idea in this case. I'm curious if there any other alternative since in all NCEES and other practice auto XFMR materials i...

Sorry not following how this pertains to the question at hand

I made the same mistaken and overlooked the "No More Than.." text Thanks for the explanation Glad to be aware of this now in case this pops up on the exam!

Thanks for this clear explanation!!! I was having the same issue as @Sparky Bill PE

In the question below we are asked to find the voltage at panel A due to a voltage drop across the conductor Based on Table 9 I did the following: Zeff_ohms= (0.029 *PF + 0.048*sin(Arccos(PF)) ) * 250/1000 = 0.013 ohms So then i simply said VpanelA_LN= VLN_mainPAnel- Vdrop_Cable_LN=...

Thanks for the feedback this helps!

In the question below we are asked to determine the speed regulation of the induction motor. From what i recall and see in the NCEES handbook confirms that this is done using the following formula: slip=(Ns-N / NS )*100 For some reason the solution uses N instead of Ns as the term in the...

@main197 you happen to have an idea of how to solve this question ?

Thank you this makes complete sense now!

So in this question we assume we have converted Zdelta to Zy right? But since we already know the Line to Line voltage across the load we simply say if converted to Y then the LN voltage is VLL<-30deg /sqrt (3) is that right?

Agreed this is why i was confused it seems there something missing here @Zach Stone P.E. your feedback would be much appreciated

In the question below VAB across the load is given We also know the line current across the conductor In the solution the voltage across the load is divided by root three why is this the case isnt the typical rule of thumb VL=Vphase for a delta configuration?

Based on the references i saw S_Series=S_Common why is that we are saying this is not true? Based on Zachs video below thats what i gathered am i missing something here? Step DOWN Autotransformers Explained for the PE CBT Exam (including circuit analysis formulas) - YouTube

In the sample NCEES question below we are given the losses at 0% and 50% kVA I realize the 0% losses represent the no load losses which are constant Now the 50% loading value includes both the partial copper and the no load losses so subtracting the no load losses makes sense The part i dont...

I see so by using Isc= MVAsc/ (sqrt (3)*Vnom) what im doing is finding the ISc from the utility with a 40MVAsc assuming i want the ISC contribution to the node directly below it. this is not a relationship i can use to find the impedance However what we really want is the impedance which is...

Based on further review of the auto transformer theory i believe the following is true in this question: 1. Each 1P Auto Transformer is rated for 60MVA this means the power at the input and output is 60MVA 2. The question specifically calls for the power across the common winding therefore this...

For the Question below im having trouble understanding the simplifications made. So far i understand we are given 3P quantities and we are after the ratings of each 1P XFRM that make up the 3P auto XFMR where: 3Ph MVA= 180 Therefore, this means the 1P MVA= 60 Assuming this is an ideal...