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That is not correct. All codebooks will be provided. You will be able to use the search function as well. And you don't need to memorize codes, but you need to know where to look at. For example, if the question asks what the width of the working space in front of the electrical equipment is...

Someone I know said it took them about six months. This was about 8-10 months ago. Not sure if it is better now.

@Zach Stone P.E. explains this in his class. I am not sure if that particular video is available for free. Don't take it the wrong way but looking at your posts, it seems like you are struggling with the basics. If you want to pass the exam, you need to know the basics. I highly recommend you to...

That's because the power loss is given in Watts, which is the real power loss. Real power loss per phase= I^2*R. Power Loss per phase= 100W. R is given. So, 100=(I^2)*3, which gives I= 5.77A. Here phase current=line current.

Looks like it comes from IEEE C57.13 permissible error. Here, the graph shows 0.6% accuracy region for currents less than 100% for 0.3Bxx CT with 4.0 rating factor. "The limit of permissible error in a current transformer accuracy class has one value at 100% rated current and allows twice that...

You can't go wrong with one of these: 1. Electrical PE Review 2. Engineering Pro Guides

I have some materials I am looking to sell.

80 questions are divided into two sections. Usually it's not 40-40, rather 41-39 or 42-38. The 8-hour timer will start once you go to the first question. Once you finish the first 41 questions, you will have the option to take a scheduled 50-minute break. Once you submit the first section, you...

When you solve these kind of circuits, always draw the per phase equivalent circuit. The impedance of line are also per phase values. So, you calculate the phase values and change it to line values. If the load is in delta, convert it into Y. For balanced load, Z_Y=Z_delta/3

Copper loss at full load is Pfl= (I_fl)^2*R At 50% load, I_50= 0.5* I_fl This gives P50= (I_50)^2*R= (0.5*I_fl)^2*R= 0.25 *(I_fl)^2*R= 0.25*Pfl So, P50= 0.25*Pfl. And Pfl= P50/0.25= 4*P50

Interesting. If we use Zach's equation, Sc=Sse= 16.5 MVA. This gives Srated= Sc+Sse= 16.5+16.5= 33MVA. They should add up to 60MVA, no? Srated is 60 MVA. I am not sure if I am missing something. Maybe @Zach Stone P.E. would like to chime in.

For step up auto transformer, Scond=V1*I2 and Sind= V1*Ic

1. Correct 2. Correct 3. Sc= V2*Ic, Sseries= V2*I1 or you if you have Sc, Sseries= Srated-Sc= 60MVA-16.5MVA= 43.5 MVA Srated= V1*I1 or V2*I2

Do not use that equation to calculate the pu value. That is the actual value of short circuit current. I_base= VA_base/sqrt(3)*V_base and Isc_actual= Isc_pu*I_base Use this equation to find the impedance, S=V^2/Z, so Z=V^2/S. To make it easier to memorize, remember the Ohm's law (V=IR)...

Autotransformers are different from a two-winding transformer. In a two-winding transformer, power is transferred inductively. But, in an autotransformer, power is transferred both inductively (at common winding) and conductively (series winding). There is no electrical isolation between primary...

See this: NCEES Problem #70

Select transformer's primary as base kV. Now the Vpu of source will not be 1, it will be Vpu=Actual Voltage/Base Voltage=20kV/22kV=0.9 pu Calculate the impedance up to fault point only. If the fault is at VT, you do not need to calculate the impedance beyond that fault point.

Can you upload the diagram?

1. Select transformer's voltage rating as base values. In above example, the faulted bus is on the transformer secondary. So, Vbase at that zone is 230kV. There are some cases where source voltage does not match the transformer's voltage. For e.g. if the source was 20 kV and the transformer was...

If you use transformer's MVA as base MVA, for transformer, Znew=Zold=0.15 pu (both kV and MVA ratio is 1). For generator, Znew= 0.23*(22/22)^2*(933/834) (because kV_new=kV_old=22kV)