# Electrical PE Review Practice Exam



## Byk (Jan 3, 2021)

I was wondering if anyone went trough this exam and found any errors. There are few answers that simply do not make sense.

Also, am I the only one who feels that the exam was written with a "gotcha" intent?

For the reference I was working with version 1.0.5


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## akyip (Jan 3, 2021)

I have the exam, though it is an earlier version.

The author of the exam also visits and peruses this board. I confirmed with him that there are no major changes between the early version I have, and the latest version.

Which specific questions are you having trouble with on the Electrical PE Review exam?


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## Byk (Jan 3, 2021)

akyip said:


> I have the exam, though it is an earlier version.
> 
> The author of the exam also visits and peruses this board. I confirmed with him that there are no major changes between the early version I have, and the latest version.
> 
> Which specific questions are you having trouble with on the Electrical PE Review exam?


I see, thank you for the info.

I do not understand question 32 when it asks to calculate the percent impedance of the generator.

I do not get why are we to use 15kV as a new Vbase.


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## akyip (Jan 3, 2021)

Byk said:


> I see, thank you for the info.
> 
> I do not understand question 32 when it asks to calculate the percent impedance of the generator.
> 
> I do not get why are we to use 15kV as a new Vbase.


I have it in front of me and am looking at 32.

The question states that the system voltage base at the distribution bus (all the way downstream) is equal to the secondary voltage rating of transformer T2, the most downstream transformer.

So then you have to work upstream converting the voltages via the turn ratios to get the system generator base voltage:

V dist bus, base = 69 KV

V transmission, base = 69 KV * 230/69 = 230 KV

V gen, base = 230 KV * 15/230 = 15 KV

This 15 KV also represents the new voltage base for the generator, when you are asked to convert the generator's p.u. impedance to the system base.

The generator's old base voltage is the rated voltage, which is the 13.8 KV given at the generator.

The formula for converting p.u. impedance to a new system base is:

Z pu new = Z pu old * (S base new / S base old) * (V base old / V base new)^2

So the (Vbase old / V base new)^2 = (13.8 KV / 15 KV)^2

Hope this helps.


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## akyip (Jan 3, 2021)

Also I think I should mention the rules for setting a system base.

When you are given a base power (VA) value to use, this base power applies to the entire system (regardless of transformers in the system).

When you are given a base voltage value and there are transformers in the system, the base voltage will be transformed via the turns ratio of the transformer.

So in problem 32, the distribution zone's base voltage is given as 69 KV, and calculating with the given transformer turns ratios, the transmission zone's (middle zone's) base voltage is 230 KV, and the generator zone's base voltage is 15 KV.


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## Byk (Jan 3, 2021)

akyip said:


> I have it in front of me and am looking at 32.
> 
> The question states that the system voltage base at the distribution bus (all the way downstream) is equal to the secondary voltage rating of transformer T2, the most downstream transformer.
> 
> ...


So we cannot use Zpu new=Zp old * (Snew/Sold) because of zone 2?

On the other hand I also tough of converting pu to actual value of the generator and then back to pu using Zbase at the T2 but that did not work either.


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## Byk (Jan 3, 2021)

akyip said:


> Also I think I should mention the rules for setting a system base.
> 
> When you are given a base power (VA) value to use, this base power applies to the entire system (regardless of transformers in the system).
> 
> ...


Why wouldn't you use 13.8kV as Vbase for the Generator?


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## akyip (Jan 3, 2021)

Byk said:


> Why wouldn't you use 13.8kV as Vbase for the Generator?


The problem statement specifically stated to use another voltage value as the voltage base - the 69 KV voltage as the base on the distribution zone, then converted through the two transformers' turn ratios to get the 15 KV base voltage at the generator that you're supposed to use for the generator system base.

Basically the problem is telling you to use another base value as the system base value, instead of the 13.8 KV given at the generator.


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## akyip (Jan 3, 2021)

Byk said:


> So we cannot use Zpu new=Zp old * (Snew/Sold) because of zone 2?
> 
> On the other hand I also tough of converting pu to actual value of the generator and then back to pu using Zbase at the T2 but that did not work either.


You can still (and you are supposed to) use Z pu new = Z pu old * (S base new/S base old) * (Vbase old/V base new)^2. You just need to make sure you are using the correct base voltage value, keeping in mind that each transformer changes the base voltage value between different zones.


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## Byk (Jan 3, 2021)

akyip said:


> You can still (and you are supposed to) use Z pu new = Z pu old * (S base new/S base old) * (Vbase old/V base new)^2. You just need to make sure you are using the correct base voltage value, keeping in mind that each transformer changes the base voltage value between different zones.


My problem was that I ignored zone 2 and how that will effect the base values.

Thanks for the help @akyip.


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## jd5191 (Jan 3, 2021)

Byk said:


> Also, am I the only one who feels that the exam was written with a "gotcha" intent?


I totally get where you're coming from. I haven't taken Zach's course myself, but due to the overwhelmingly positive reviews of Zach's course and exam, I saved this exam as one of the last unseen exams I was going to time myself on. Half way through the morning section, I realized I was burning through a lot of time reading and re-reading questions just trying to understand the wording and how I was going to setup the problem, maybe this is partially because I didn't take his course and had to really parse out the sentences in each problem. I feel some of the problems requiring the NEC code required way more time than I would've expected. I found myself wondering if the nuances the exam gets into are helpful or distracting, but he eventually does a good job explaining things in the solutions that I walk away thinking this will probably be helpful to know, even if I did sink way more time than I wanted to on the problem. I eventually stopped digging into every problem I was confused about and thought I would research later if I had time and maybe email Zach about them later or post here. Since you made the post, I'll post a few things I was wondering about here:

Problem #3: the integration uses both a starting and ending angle. Is this unique to IGBTs because SCRs conduct until the end of the cycle so ending angle is always 180 degrees

Problem #6: Do all inverse time curves look like this? Google search brought up some inverse time curves that look like below, but I can't tell if its for different pickup settings or not.


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## akyip (Jan 3, 2021)

jd5191 said:


> I totally get where you're coming from. I haven't taken Zach's course myself, but due to the overwhelmingly positive reviews of Zach's course and exam, I saved this exam as one of the last unseen exams I was going to time myself on. Half way through the morning section, I realized I was burning through a lot of time reading and re-reading questions just trying to understand the wording and how I was going to setup the problem, maybe this is partially because I didn't take his course and had to really parse out the sentences in each problem. I feel some of the problems requiring the NEC code required way more time than I would've expected. I found myself wondering if the nuances the exam gets into are helpful or distracting, but he eventually does a good job explaining things in the solutions that I walk away thinking this will probably be helpful to know, even if I did sink way more time than I wanted to on the problem. I eventually stopped digging into every problem I was confused about and thought I would research later if I had time and maybe email Zach about them later or post here. Since you made the post, I'll post a few things I was wondering about here:
> 
> Problem #3: the integration uses both a starting and ending angle. Is this unique to IGBTs because SCRs conduct until the end of the cycle so ending angle is always 180 degrees
> 
> ...


I can't really answer for 6, but I will answer for 3.

Yes, the integration from theta 1 to theta 2 is unique/exclusive to IGBTs. They have both a firing angle for starting conducting and a firing angle for stopping conducting. It's a step up for controlling firing, from SCRs which only control starting the conducting.


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## Byk (Jan 3, 2021)

jd5191 said:


> I totally get where you're coming from. I haven't taken Zach's course myself, but due to the overwhelmingly positive reviews of Zach's course and exam, I saved this exam as one of the last unseen exams I was going to time myself on. Half way through the morning section, I realized I was burning through a lot of time reading and re-reading questions just trying to understand the wording and how I was going to setup the problem, maybe this is partially because I didn't take his course and had to really parse out the sentences in each problem. I feel some of the problems requiring the NEC code required way more time than I would've expected. I found myself wondering if the nuances the exam gets into are helpful or distracting, but he eventually does a good job explaining things in the solutions that I walk away thinking this will probably be helpful to know, even if I did sink way more time than I wanted to on the problem. I eventually stopped digging into every problem I was confused about and thought I would research later if I had time and maybe email Zach about them later or post here. Since you made the post, I'll post a few things I was wondering about here:
> 
> Problem #3: the integration uses both a starting and ending angle. Is this unique to IGBTs because SCRs conduct until the end of the cycle so ending angle is always 180 degrees
> 
> ...


I think overall exam is very good, it makes you go a couple steps further then actually needed for the actual exam. But I think it's good since you get a lot more practice.

Regrading question 6:

I want to say that the graph will vary from manufacturer to manufacture but the overall principal stays the same.

I believe that Zack was going here more for the basics which is that  CB will have less time delay when going from inverse to very inverse.

Here is what I printed from ABB (it might help to compare the graphs) 

https://sertecrelays.net/wp-content/uploads/2019/02/41-100.1B.pdf


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## jd5191 (Jan 3, 2021)

Byk said:


> I think overall exam is very good, it makes you go a couple steps further then actually needed for the actual exam. But I think it's good since you get a lot more practice.


Agree 100%


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## Dothracki PE (Jan 4, 2021)

Byk said:


> I was wondering if anyone went trough this exam and found any errors. There are few answers that simply do not make sense.
> 
> Also, am I the only one who feels that the exam was written with a "gotcha" intent?
> 
> For the reference I was working with version 1.0.5


I believe @Zach Stone, P.E. does tend to write many of his questions with a few tricks thrown in to better prepare you for the exam. It is possible that NCEES could do the same with certain problems by giving you extra information in the problem statement or giving incorrect solutions with common errors like an incorrect sign or using line voltage value that should be phase voltage. It is really testing your understanding of the fundamentals of each topic, which is really the basis of the exam.


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## jd5191 (Jan 4, 2021)

Another question, on Problem #53:

The correct answer includes "a portable GFCI cord" according to 590.6(A)(1). Below is this section




My initial reading of this section made it made me think the outlet itself must have a mandatory GFCI, the portable stuff is optional. Anyone else read it that way?


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## Byk (Jan 4, 2021)

jd5191 said:


> Another question, on Problem #53:
> 
> The correct answer includes "a portable GFCI cord" according to 590.6(A)(1). Below is this section
> 
> ...


Portable cord is just a part of the answer. If you read the whole answer it says GFCI receptacle first.


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## jd5191 (Jan 5, 2021)

Byk said:


> Portable cord is just a part of the answer. If you read the whole answer it says GFCI receptacle first.


If portable cord is permitted "in addition" to the other mandatory GFCI requirements, then isn't the answer (B) and not (A)? The only difference between (B) and (A) is the portable cord.


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## Dothracki PE (Jan 5, 2021)

I work in construction, so I am very familiar with this section. Note this is only for temporary use on construction sites and does not meet the requirements for permanent GFCI protection of receptacles required in 210.8.

The last part of 590.6(A)(1) &amp; (2) states that listed cord sets or devices incorporating listed GFCI protection shall be permitted to meet the GFCI requirement such as the one shown below. The three other methods that I have seen, and allowed by this code, is possible such as using GFCI-type receptacles as the temporary receptacle, installing GFCI modules at some point in the line (shown in the second picture below), or using GFCI type breakers at the panel. Based on this, I agree that the correct answer is (A).


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## jd5191 (Jan 5, 2021)

Thanks @Dothracki PE I guess I'm continuing to read the code wrong. I pulled open my NEC handbook and saw that this part of the code had changed so I thought perhaps the solution was a remnant of a previous code version. My reading is that the 2014 code allows for portable cords only, and the 2017 code allows for portable cords in addition to other requirements, which to me means you need other GFCI whether you have portable cords or not, which would mean Solution (A) is incorrect. Thanks for the expert input  I think I've spent enough time on this problem lol


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## Dothracki PE (Jan 5, 2021)

jd5191 said:


> Thanks @Dothracki PE I guess I'm continuing to read the code wrong. I pulled open my NEC handbook and saw that this part of the code had changed so I thought perhaps the solution was a remnant of a previous code version. My reading is that the 2014 code allows for portable cords only, and the 2017 code allows for portable cords in addition to other requirements, which to me means you need other GFCI whether you have portable cords or not, which would mean Solution (A) is incorrect. Thanks for the expert input  I think I've spent enough time on this problem lol


Ahh yes, need to be careful with the correct version of the electrical code. There are not many drastic changes between 2014/2017, but this is one example that the changes have a difference in the answer. I am on the 2008 NEC in New York City so I probably would have made that mistake as well if I did not read the 2017 NEC so much for the PE exam.


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## Zach Stone P.E. (Jan 5, 2021)

There is a lot of great discussion in this thread. I'll try to address all of it in one post to help reduce clutter.



Byk said:


> I was wondering if anyone went trough this exam and found any errors. There are few answers that simply do not make sense.
> 
> Also, am I the only one who feels that the exam was written with a "gotcha" intent?
> 
> For the reference I was working with version 1.0.5


Hi @Byk!

Yes, our practice exam was written intentionally with lots of "gotchas". A lot of practice exams available are very similar and only offer a slight variety of problems. In my experience this can lead to a false sense of confidence.

Our practice exam (Electrical Engineering PE Practice Exam and Technical Study Guide) was made to serve first as a learning tool. A lot of feedback I've received over the years is that practice exams are helpful to learn how to solve specific problems from specific topics, but left engineers vulnerable to not being able to solve every type of problem like it, and from a variety of different perspectives and possible "gotchas", similar to the tougher questions you'll see on the PE exam.

The problems in our practice exam are intended to be challenging in order to expose any blind spots, with extremely detailed explanations to help fill in any gaps in your understanding regardless of what they may be. Blindspots being defined as gaps in your knowledge that you don't yet know that you have. For the PE exam, blindspots are extremely dangerous.

You'll generally find that once you take the time to work through each problem and solution in our practice exam, your overall scores in other practice exams will dramatically increase. Each problem is designed to help you solve *every* problem in that specific topic.

With this practice exam out of the way, future practice exams we publish will be simpler and designed to serve as more of a benchmark tool than a learning and competency tool.



Byk said:


> So we cannot use Zpu new=Zp old * (Snew/Sold) because of zone 2?
> 
> On the other hand I also tough of converting pu to actual value of the generator and then back to pu using Zbase at the T2 but that did not work either.


I enjoy reading your thought process, these are the types of blindspots that the practice exam hopes to bring to light and address.

Have you worked through the per unit article on our website yet?

It clears up just about every aspect of the per unit system. It's a long article but just like the questions in the practice exam, it will help you solve just about every per unit problem you can expect to face. Here's a link:

Per Unit Example – How To, Tips, Tricks, and What to Watch Out for on the Electrical PE Exam
 



jd5191 said:


> Problem #3: the integration uses both a starting and ending angle. Is this unique to IGBTs because SCRs conduct until the end of the cycle so ending angle is always 180 degrees
> 
> Problem #6: Do all inverse time curves look like this? Google search brought up some inverse time curves that look like below, but I can't tell if its for different pickup settings or not.
> 
> View attachment 20496


*Problem #3* - Yes, unlike thyristors (SCR) that can only be controlled when they  turn "on" (close), IGBTs can be controlled both when they turn "on" (close) and when they turn "off" (open).

For thyristors (SCR), your upper and lower limits should be the firing angle when the gate pulse starts (the thyristor will close) - and the when the voltage across it reaches zero (the thyristor will open).

For IGBTs, your upper and lower limits should be the firing angle when the gate pulse starts (the IGBT will close) and the firing angle when the gate pulse stops (the IGBT will open).

*Problem #6 *- Curves depend on the type of relay or breaker. For example, consider the different curves for various CO relays from page 8 to 14 on this ABB paper.

Problem #6 asks for the change in trip characteristics after the time delay setting has been changed to be more inverse *compared to the previous setting with no change to the pick up*. 

In general, as the time delay for the TCC trip characteristic for a device is increased to be more time inverse with an increase in time dial setting, the breaker will take longer to operate for the same level of fault current. The more time inverse, the longer the time the breaker or device will take to operate. This is the purpose of time delay.



Byk said:


> I think overall exam is very good, it makes you go a couple steps further then actually needed for the actual exam. But I think it's good since you get a lot more practice.




Glad to hear it, this was the intent behind our practice exam - to be much more useful for increasing your overall exam score in all areas compared to just being another practice exam to add to the pile to work through.

We receive a lot of feedback that you can find answers to a lot of the questions you are still left with after working through other practice exams by looking at the solution steps to similar questions in our practice exam. 



Dothracki PE said:


> I believe @Zach Stone, P.E. does tend to write many of his questions with a few tricks thrown in to better prepare you for the exam. It is possible that NCEES could do the same with certain problems by giving you extra information in the problem statement or giving incorrect solutions with common errors like an incorrect sign or using line voltage value that should be phase voltage. It is really testing your understanding of the fundamentals of each topic, which is really the basis of the exam.


Exactly.

It's great being able to solve the easier, quicker problems from most subjects. But what happens when variables are changed that carry extra nuance to how the problem must be set up to solve correctly that may not be obvious at first? If you can solve these problems, you can solve any problem.


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## Byk (Jan 5, 2021)

Zach Stone said:


> There is a lot of great discussion in this thread. I'll try to address all of it in one post to help reduce clutter.
> 
> Hi @Byk!
> 
> ...


@Zach Stone, P.E. thank you so much for taking time to respond it is much appreciated.

I am glad to hear that the exam is structured that way because it took me whole 8 hours to go through each problem. 

I also found your exam to be the most helpful as it covers a lot of details.



Zach Stone said:


> Have you worked through the per unit article on our website yet?
> 
> It clears up just about every aspect of the per unit system. It's a long article but just like the questions in the practice exam, it will help you solve just about every per unit problem you can expect to face. Here's a link:
> 
> Per Unit Example – How To, Tips, Tricks, and What to Watch Out for on the Electrical PE Exam


I went through the the article couple of times. What threw me off is part that says "the new base will be the new bases we have chosen".

That worked for the example in the article because T2 was in Zone 2 (between 2 zones).

However, the part that I did not consider is that the base value will change when going from Zone 1 to Zone 3 or vice versa.

I also want to take a moment and say *THANK YOU* for everything you do. I cannot express how thankful I am for all of the free content that you provide.


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## Byk (Jan 5, 2021)

jd5191 said:


> Thanks @Dothracki PE I guess I'm continuing to read the code wrong. I pulled open my NEC handbook and saw that this part of the code had changed so I thought perhaps the solution was a remnant of a previous code version. My reading is that the 2014 code allows for portable cords only, and the 2017 code allows for portable cords in addition to other requirements, which to me means you need other GFCI whether you have portable cords or not, which would mean Solution (A) is incorrect. Thanks for the expert input  I think I've spent enough time on this problem lol


I would suggest to use actual code not the handbook as handbook will not be available on the test


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## Zach Stone P.E. (Jan 5, 2021)

Byk said:


> I also want to take a moment and say *THANK YOU* for everything you do. I cannot express how thankful I am for all of the free content that you provide.


My pleasure.

I'm glad you've found it helpful. Helping engineers pass the PE exam and learn more of the technical aspects of each subject that often get overlooked, ignored, missed, or misunderstood is what I really enjoy doing.


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## lost4ever P.E. (Jan 7, 2021)

Anyone can help me please

Question 69 (Electrical PE Preview Technical Guide)

Z0+Z1+Z2 = 10 + 7j ohm

I0 = 646.6A&lt;-35.8

SLG fault

*My approach :*

For SLG fault I0=I1=I2

Ia = I0+I1+I2=3*646.6&lt;-35.8 = 1939.8&lt;-35.8

Vph = 1939.8&lt;-35.8 * (10+7j) =23.898&lt;-0.1 kV

Vline = 23.898*1.732 = 41.39kV

What I am missing here?


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## DarkLegion PE (Jan 7, 2021)

lost4ever said:


> Anyone can help me please
> 
> Question 69 (Electrical PE Preview Technical Guide)
> 
> ...


I think it's because that's the ground fault value, not the impedance value of the system


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## COJeff (Jan 7, 2021)

Read the question and look at the NCEES manual chapter 5 page 66.

Dont want to give too much away.


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## COJeff (Jan 7, 2021)

DarkLegion said:


> I think it's because that's the ground fault value, not the impedance value of the system


I reworked this problem and I think I misunderstood it as well as I got zero.


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## lost4ever P.E. (Jan 7, 2021)

DarkLegion said:


> I think it's because that's the ground fault value, not the impedance value of the system


@DarkLegion Can you show me your attempt? I tried again and getting similar results.


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## COJeff (Jan 8, 2021)

I re-did this but with out showing the numbers to make it easier to understand.


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## Byk (Jan 8, 2021)

COJeff said:


> I re-did this but with out showing the numbers to make it easier to understand.
> 
> View attachment 20527


I feel like you are making in more complicated than it is.

Two important things to notice here are:

1) Question is asking for line voltage and you solving for line to neutral.

2) All the currents are same in the circuit (I0 = I1 = I2 = Ia1) therefore you can simply combine all of the impedance together and solve it for Vp

You had your whole solution in the first line. Vp= I0 * Zt. From there you just convert form Vln to Vl.

The rest of your work was simply deriving the same equation again.


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## akyip (Jan 15, 2021)

Hey guys,

Responding to the questions about the single-line-to-ground fault from the Electrical PE Review exam.

See the attached circuit diagram analysis. I should note the following:


For a single line to ground fault with Phase A faulted, Van = 0 at the location of the fault. Ib = Ic = 0
Here, we are solving for the voltage from the source. First, I am solving for the Van voltage from the source. Then, we can multiply the magnitude by sqrt(3) to get the line-to-line voltage.


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## lost4ever P.E. (Jan 15, 2021)

Thanks all appreciate your support. It is clear to me now.


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