# Cram for the PE Sample Test 1 - Problem #13



## wiliki (Jan 3, 2020)

Hey guys, I know this is a simple problem, but can anyone provide their feedback on attaining the answer to this? The solutions manual says it's (D), but we had initially assumed the answer was (C) due to no ground.


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## Orchid PE (Jan 3, 2020)

C) doesn't have a ground because it's a DC source.

A MOV acts as a surge protector. It clamps voltage to a defined value when the voltage across its terminals gets too high.

All the connections are correct because:

A) Provides line to line protection

B) provides line to line and line to ground protection.

C) Provides positive to negative protection.


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## SCU2019 (Jan 3, 2020)

Hi Guys,

This is from the same set of problems but I dont understand why this is (B) and not (D) is this an error in the book. If not, please explain.


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## Orchid PE (Jan 4, 2020)

SCU2019 said:


> Hi Guys,
> 
> This is from the same set of problems but I dont understand why this is (B) and not (D) is this an error in the book. If not, please explain.
> 
> View attachment 15737


Since the question is asking which of the following is NOT true, we know the answer to be B.

This is because an inductor's reactance increases with frequently. The inductor is meant to block high frequencies, and the caps are meant to short high frequencies to ground. So this is a low pass filter.

We assume the AC ripple is a high frequency we are trying to filter out.

We know the inductor will increase efficiency, we know it will oppose current without generating heat, and we know it acts as a high reactance to high frequencies, so the answer that remains is not true.


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## wiliki (Jan 4, 2020)

Chattaneer PE said:


> Since the question is asking which of the following is NOT true, we know the answer to be B.
> This is because an inductor's reactance increases with frequently. The inductor is meant to block high frequencies, and the caps are meant to short high frequencies to ground. So this is a low pass filter.
> We assume the AC ripple is a high frequency we are trying to filter out.
> We know the inductor will increase efficiency, we know it will oppose current with generating heat, and we know it acts as a high reactance to high frequencies, so the answer that remains is not true.


Thanks so much for the detailed response on both problems!

Do you happen to have any recommendations on books on inductors, frequency, ripples, reactance, etc? You gave a great explanation, but we don’t know what to reference if we ever come across this problem (other than google).


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## Orchid PE (Jan 4, 2020)

wiliki said:


> Thanks so much for the detailed response on both problems!
> 
> Do you happen to have any recommendations on books on inductors, frequency, ripples, reactance, etc? You gave a great explanation, but we don’t know what to reference if we ever come across this problem (other than google).


For this type of stuff, probably just use whatever textbook you used for your circuits courses. That should cover filters and possibly MOVs.

I would suggest not planning to answer questions like the filter from reference material _during_ the exam. For _studying_, I would refresh with a circuits book. On the exam, try to be refreshed enough to answer these from your knowledge base.

Try to save as much time as possible for code questions.


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## Orchid PE (Jan 4, 2020)

I just realized that filter question is a VERY good example of answers on the exam.

Notice how B and D are opposites of each other. By inspection, we immediately know one of those _has_ to be the answer, because the inductor can't offer a high &amp; low reactance at the same time. Eliminating A and C now gives you a 50% chance of getting the right answer if you had to guess.


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## wiliki (Jan 6, 2020)

thanks again @Chattaneer PE ; this all makes sense now, appreciate you getting back to us so quickly


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## Orchid PE (Jan 6, 2020)

Keep the questions coming! It helps me stay refreshed on these topics.


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## wiliki (Jan 7, 2020)

hey @Chattaneer PE I've got another problem here from the Cram for PE test. I probably need to refresh myself on symmetrical components fault analysis after this...  

Why is the j0.07 reactance not included in the zero-sequence portion of the circuit? Also, is it standard to always assume V = 1 pu?


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## Orchid PE (Jan 7, 2020)

@wiliki

When I get home I'll type up a little more.

But for the V = 1 pu, yeah that's pretty standard. Since you get to pick which MVA and Voltage base you use, it makes things easier to just use whatever kV is given as the voltage base, making any voltage at that level = 1pu.

You could've selected any kV, but using 138kV is easiest.


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## Orchid PE (Jan 7, 2020)

Short answer: Because that's the zero sequence equivalent circuit for a delta-wye transformer. Since the generator impedance is connected to "L" in the case below, it has no connection in the equivalent circuit.


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## squaretaper LIT AF PE (Jan 7, 2020)

I'm not even gonna pretend like I know what any of this means.

...workin' on it.


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## wiliki (Jan 7, 2020)

Chattaneer PE said:


> Short answer: Because that's the zero sequence equivalent circuit for a delta-wye transformer. Since the generator impedance is connected to "L" in the case below, it has no connection in the equivalent circuit.
> 
> View attachment 15795


Please forgive my ignorance... but which one these is the delta-wye connection? The H represents the high side, and L represents low side?


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## Orchid PE (Jan 7, 2020)

wiliki said:


> Please forgive my ignorance... but which one these is the delta-wye connection? The H represents the high side, and L represents low side?


Correct. H = high, L = low, N = Neutral.

Delta = Triangle, Wye = Y-shaped one.

So connection (a) in the left-hand table.


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## Orchid PE (Jan 7, 2020)

So from the problem we know the connection of the transformer. All we have to do is flip the connections from the table of equivalent circuits.


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## wiliki (Jan 7, 2020)

Chattaneer PE said:


> Correct. H = high, L = low, N = Neutral.
> 
> Delta = Triangle, Wye = Y-shaped one.
> 
> So connection (a) in the left-hand table.


My initial understanding was that that (a) is a wye-delta connection? Or can I view these diagrams interchangeably for both the high and low side?


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## Orchid PE (Jan 7, 2020)

wiliki said:


> My initial understanding was that that (a) is a wye-delta connection? Or can I view these diagrams interchangeably for both the high and low side?


You can flip them to match.


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## Orchid PE (Jan 7, 2020)

Make sure to pay attention to the grounding of the Wye windings. They make a difference.


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## Orchid PE (Jan 8, 2020)

Here's a good explanation of why some equivalent circuits are open: https://circuitglobe.com/zero-sequence-current.html

In that explanation, it is determined there is no path for zero sequence currents to flow for ungrounded delta and wye (star) connections. So zero sequence current = 0. This is why there are equivalent circuits that shown open; because if I0 = 0, the return path has to show open.


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## wiliki (Jan 8, 2020)

Chattaneer PE said:


> Here's a good explanation of why some equivalent circuits are open: https://circuitglobe.com/zero-sequence-current.html
> In that explanation, it is determined there is no path for zero sequence currents to flow for ungrounded delta and wye (star) connections. So zero sequence current = 0. This is why there are equivalent circuits that shown open; because if I0 = 0, the return path has to show open.


Perfect, I’ll def print this out also for my references. Thanks so much again!! 


Sent from my iPhone using Tapatalk


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## wiliki (Jan 8, 2020)

Okay one more question here. The solutions to this says the answer is a) Profile 1 "for the given circumstances." 

We were a bit lost with this solution. If there is no load, shouldn't the voltage remain the same from send end to receiving end (Profile 2)? Why does voltage increase at the receiving end for Profile 1?


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## Orchid PE (Jan 8, 2020)

wiliki said:


> Okay one more question here. The solutions to this says the answer is a) Profile 1 "for the given circumstances."
> 
> We were a bit lost with this solution. If there is no load, shouldn't the voltage remain the same from send end to receiving end (Profile 2)? Why does voltage increase at the receiving end for Profile 1?


It is profile 1 because of the Ferranti effect.

A lightly loaded/line with no load will have a higher receiving end voltage than the sending end, because of the capacitive charging currents.


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## wiliki (Jan 9, 2020)

hello again! I'm here doing some problems on my lunch break and am a bit confused with this one.. 

So I see in the solutions that the positive, negative, and zero components were first derived. 

I am a bit confused with the -30*° *and 30*°*... I know that since this is a delta-wye transformer, there is a phase angle shift. But how would I go about designating the positive and negative signs for just phase A to the positive and negative components? I was referring the symmetrical components diagram below.


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## Orchid PE (Jan 9, 2020)

Maybe this helps?


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## Kris :) (Jan 19, 2020)

Hello ... i am actually was working on the same problem set and was wondering if you could help with the following question? 







Attached above is the question/answer.... i am having a hard time understanding where this formula comes from or how to derive it.

Also how would it change if its 3 phase, more than 2-pulse?, or not a fully controlled rectifier? I know .. a loaded question lol. Any help would be appreciated.

Thank you!


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## Orchid PE (Jan 19, 2020)

Kris :) said:


> Hello ... i am actually was working on the same problem set and was wondering if you could help with the following question?
> 
> Attached above is the question/answer.... i am having a hard time understanding where this formula comes from or how to derive it.
> 
> ...


To find the average value of a waveform we just integrate over the area of interest (finding the sum) and then divide by the period (finding the average). For the stuff below, we're integrating over the shaded area (in the case of a full wave rectifier, from theta to T/2). T=2π, w=2π/T, Vpk=Vrms.







The final formula is a tad bit different from what was used in the solution, but if you notice there are a few terms they could've canceled out.

EDIT: Typed this up for clarity.


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## Orchid PE (Jan 19, 2020)

The 2 pulse is simply because there are two thyristors that need to be turned on at the same time at any point.


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## Orchid PE (Jan 21, 2020)

Useful formulas for this: https://www.electronics-tutorials.ws/power/single-phase-rectification.html


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## Orchid PE (Jan 22, 2020)

@Kris :) The formula can be found on page 211, equation 8.6 of this book.


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## Orchid PE (Jan 22, 2020)

I drew two diagrams below; one for before the conductor break occurred and one for after. In the diagram before the break, the phase and neutral conductors are connected *directly to the load*. In the diagram after the break, the phase conductor is connected *directly to the enclosure*, and the neutral conductor connection is irrelevant (but it's still connected to the load). We don't need to worry about the neutral even though it's still connected to the load, because there is no power to the load since the phase conductor broke off. We only need to worry about the return path for current through the load's _enclosure_. So in the diagram after the break, the load inside the enclosure is drawn not connected. All current will flow through the enclosure. Since the enclosure is connected to ground, and the source's neutral is connected to ground, this will complete the circuit and will be the path for current. Once we calculate the current in the circuit, we can calculate the voltage at the enclosure by finding the resistance from enclosure-to-ground and multiplying it by the current.


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## DLD PE (Jan 22, 2020)

This is all great stuff!  Thanks everyone for posting (both the questions and answers).  

These qualitative questions are great.  I try, when I'm reviewing the more math-relative (quantitative) problems, how the NCEES might change it or ask a qualitative question on the same kind of problem.  For example, as Chattaneer wrote, losing the hot side of a WYE circuit.  What's the overall effect?  How does it affect the current/voltage though the other legs?  How is the load affect (before vs after)?


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## Orchid PE (Jan 23, 2020)

That'd be a nice zap if someone touched that enclosure without the 10ohm neutral grounding resistor installed!


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## wiliki (Jan 23, 2020)

Hello, we’re working on the same problem set, but from book #2 now. We’ve got a question for problem #19 below. 

The solution says that there is no neutral in the wye-wye transformer to carry harmonics. However, going back to the question.... we don’t see a mention of a wye with neutral. For problems like these, do we just assume no neutral? 

Also, if there was a neutral, does this help to mitigate the harmonics?


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## wiliki (Jan 23, 2020)

Chattaneer PE said:


> I drew two diagrams below; one for before the conductor break occurred and one for after. In the diagram before the break, the phase and neutral conductors are connected *directly to the load*. In the diagram after the break, the phase conductor is connected *directly to the enclosure*, and the neutral conductor connection is irrelevant (but it's still connected to the load). We don't need to worry about the neutral even though it's still connected to the load, because there is no power to the load since the phase conductor broke off. We only need to worry about the return path for current through the load's _enclosure_. So in the diagram after the break, the load inside the enclosure is drawn not connected. All current will flow through the enclosure. Since the enclosure is connected to ground, and the source's neutral is connected to ground, this will complete the circuit and will be the path for current. Once we calculate the current in the circuit, we can calculate the voltage at the enclosure by finding the resistance from enclosure-to-ground and multiplying it by the current.


Question on this one - Why is there a 0.1 ohm resistance in the equipment grounding conductor? Didn’t see it as a given in the problem - the problem specifically states “A single-phase load is connected phase-to-neutral and is feed from the source with each conductor, including the equipment grounding conductor, having a total resistance of 0.1 ohm” 

It says “total resistance”... so why does the hot have a resistance of 0.1 ohm, and why does the EGC have a resistance of 0.1 ohm?


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## Orchid PE (Jan 24, 2020)

wiliki said:


> Question on this one - Why is there a 0.1 ohm resistance in the equipment grounding conductor? Didn’t see it as a given in the problem - the problem specifically states “A single-phase load is connected phase-to-neutral and is feed from the source with each conductor, including the equipment grounding conductor, having a total resistance of 0.1 ohm”
> 
> It says “total resistance”... so why does the hot have a resistance of 0.1 ohm, and why does the EGC have a resistance of 0.1 ohm?


I think it makes more sense with a different wording of the question:

A single-phase load is connected phase-to-neutral. Each conductor has a total resistance of 0.1 ohms, including the equipment grounding conductor, .


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## Orchid PE (Jan 24, 2020)

wiliki said:


> Hello, we’re working on the same problem set, but from book #2 now. We’ve got a question for problem #19 below.
> 
> The solution says that there is no neutral in the wye-wye transformer to carry harmonics. However, going back to the question.... we don’t see a mention of a wye with neutral. For problems like these, do we just assume no neutral?
> 
> Also, if there was a neutral, does this help to mitigate the harmonics?


I honestly don't understand what he's trying to get at with this question.

We know delta-wye transformers are useful because the delta winding allows triplen harmonics to be contained within the transformer, reducing the harmonics on the wye side. Maybe he's assuming with a wye-wye transformer the harmonics are not contained within the transformer and are reproduced on the secondary side. I don't understand his solution of "because there is no neutral." Maybe he's implying that if the transformer had a neutral conductor the harmonics would flow in the neutral. However, I can't think of an instance where a wye connection would want to be used without a neutral. 

I don't like the question, but at least we can eliminated the a, b, and c answers.


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## DLD PE (Jan 24, 2020)

Chattaneer PE said:


> I honestly don't understand what he's trying to get at with this question.
> 
> We know delta-wye transformers are useful because the delta winding allows triplen harmonics to be contained within the transformer, reducing the harmonics on the wye side. Maybe he's assuming with a wye-wye transformer the harmonics are not contained within the transformer and are reproduced on the secondary side. I don't understand his solution of "because there is no neutral." Maybe he's implying that if the transformer had a neutral conductor the harmonics would flow in the neutral. However, I can't think of an instance where a wye connection would want to be used without a neutral.
> 
> I don't like the question, but at least we can eliminated the a, b, and c answers.


These are the type of practice exam questions I run into once in a while that I just can't seem to quite get.  My first thought was, is it not safe to assume the Wye has a neutral?  I don't remember seeing any kind of sample or practice exam problem showing a Wye without one.  I've seen one example of a transformer bank arranged in delta, and you lose one and it asks you how it can be connected (open delta).  

But examples like the one above I feel like I just have to move on, so I simply print it out with everyone's comments and put it in my binder and tag it in the rare event it might show up on the actual exam.


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## DilutedAr18_PE (Jan 24, 2020)

Chattaneer PE said:


> I honestly don't understand what he's trying to get at with this question.
> 
> We know delta-wye transformers are useful because the delta winding allows triplen harmonics to be contained within the transformer, reducing the harmonics on the wye side. Maybe he's assuming with a wye-wye transformer the harmonics are not contained within the transformer and are reproduced on the secondary side. I don't understand his solution of "because there is no neutral." Maybe he's implying that if the transformer had a neutral conductor the harmonics would flow in the neutral. However, I can't think of an instance where a wye connection would want to be used without a neutral.
> 
> I don't like the question, but at least we can eliminated the a, b, and c answers.


I agree that he’s trying to get at the harmonics reduction with a delta-wye transformer, but is doing it backwards by giving the example of the wye-wye. I do not understand the “there is no neutral” comment either. In the delta-wye, you don’t have a neutral on the delta side and create a separately derived system on the wye side, but that is not what the solution seems to be saying. 

I don't mind the question, but the solution explanation leaves a lot to be desired.


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## wiliki (Feb 5, 2020)

Another problem here, problem #21. We're at loss with the solution, as this is the first time we're seeing a relation between generator curves, to R-X diagrams, and to a distance relay. Does anyone understand the solution or is able to provide more clarity?


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## Orchid PE (Feb 5, 2020)

wiliki said:


> Another problem here, problem #21. We're at loss with the solution, as this is the first time we're seeing a relation between generator curves, to R-X diagrams, and to a distance relay. Does anyone understand the solution or is able to provide more clarity?


I'll check some of my protection books at home to see if they cover anything like this. This seems like a very specific question for protection engineers.


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## LyceeFruit PE (Feb 5, 2020)

This is from Elmore's Protective Relaying Theory &amp; Applications, sponsored by ABB.

The KLF &amp; KLF-1 relays are specific microprocessor relays from ABB, circa 2004.


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## LyceeFruit PE (Feb 5, 2020)

Also if you have 4th edition of Blackburn, check out page 262-266.


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## wiliki (Feb 5, 2020)

Thanks guys, that helps a lot  @Chattaneer PE @LyceeFruit PE

We're gonna have to purchase more books on protection over here. Heard about Blackburn, but have yet to look into it. Please let me know if you have any other good recommendations for protection references.


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## Dude99 (Feb 8, 2020)

My approach


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## Dude99 (Feb 8, 2020)

wiliki said:


> Question on this one - Why is there a 0.1 ohm resistance in the equipment grounding conductor? Didn’t see it as a given in the problem - the problem specifically states “A single-phase load is connected phase-to-neutral and is feed from the source with each conductor, including the equipment grounding conductor, having a total resistance of 0.1 ohm”
> 
> It says “total resistance”... so why does the hot have a resistance of 0.1 ohm, and why does the EGC have a resistance of 0.1 ohm?


NGR's are required by law in mining to limit frame V to &lt;40

An add on ?

what is the N-G V of the ungrounded phases during a gnd fault?

hint: what is the N V elevated to?


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## Cram For The PE (Feb 10, 2020)

wiliki said:


> Another problem here, problem #21. We're at loss with the solution, as this is the first time we're seeing a relation between generator curves, to R-X diagrams, and to a distance relay. Does anyone understand the solution or is able to provide more clarity?
> 
> View attachment 16328
> 
> ...




I posted my answer to this question here:

http://cramforthepe.com/index.php/2020/02/10/question-about-loss-of-excitation-protection/

If you have any other questions feel free to contact me.


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## BebeshKing PE (Feb 16, 2020)

wiliki said:


> Hey guys, I know this is a simple problem, but can anyone provide their feedback on attaining the answer to this? The solutions manual says it's (D), but we had initially assumed the answer was (C) due to no ground.
> 
> View attachment 15736


I believe letter C is still a proper connection. It still protect the load. MOVs are also used in DC applications ( electronics and semiconductors).


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## BebeshKing PE (Feb 16, 2020)

Chattaneer PE said:


> Maybe this helps?
> 
> View attachment 15873


Hi @Chattaneer PE, what if the connection will be High side wye and Low side delta? will the result for positive and negative sequence current change?

Will it become:

Ia1(wye) =Ia1 &lt;-30

Ia2(wye) = Ia2 &lt;30

Iao=0

???

I know the result will still be the same (1/square root of 3) since we are just adding the 3 components. But I am just curious with the result each components.

Thanks,


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## Orchid PE (Feb 17, 2020)

BebeshKing said:


> Hi @Chattaneer PE, what if the connection will be High side wye and Low side delta? will the result for positive and negative sequence current change?
> 
> Will it become:
> 
> ...


Yes it is dependent on connection (and technically phase sequence: abc vs acb, but most problems are assumed abc, read up on DAB vs DAC transformers).

The phase shift is either plus or minus 30 degrees from primary to secondary for *positive *sequence voltages and currents (essentially, positive sequence follows the phase shift of the transformer). *Negative *sequence shifts in the opposite direction from positive.

These two posts are worth reading, though they're not talking about symmetrical components:


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## BebeshKing PE (Feb 17, 2020)

Hi, I'm just going to use this thread for some of my* Cram Vol.1 practice exam* questions instead of creating a new one.

For *problem #2, *

If the efficiency lost is 20%, is that mean that the charging efficiency will be 80%?. And that being said, should the answer be

Time of charge= capacity/(charge rate current x efficiency) = 100Ah/(40A x 0.8) = 3.125 hours.???

[SIZE=11pt]Not sure where the 1.2 came from because 1/0.8 is 1.25, though.Could someone please tell me if my solution is correct?[/SIZE]

[SIZE=11pt]Thanks,[/SIZE]


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## Orchid PE (Feb 18, 2020)

BebeshKing said:


> Hi, I'm just going to use this thread for some of my* Cram Vol.1 practice exam* questions instead of creating a new one.
> 
> For *problem #2, *
> 
> ...


If there was no efficiency loss, the battery would take 100% of the time to charge (100 / 40 = 2.5hr).

But since there is an efficiency loss of 20%, that means it will take an _extra _20% of the time to charge. So 100% + 20% = 120% (or 1.2). We know to do this because an efficiency loss will increase the time it takes to charge the battery.

If, for whatever reason, we had an efficiency increase of 20%, we would multiply by 80% (because the total time to charge would be reduced).


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## Cram For The PE (Feb 18, 2020)

BebeshKing said:


> Hi, I'm just going to use this thread for some of my* Cram Vol.1 practice exam* questions instead of creating a new one.
> 
> For *problem #2, *
> 
> ...


A previous individual had the same question a while back. I answered it here:

http://cramforthepe.com/index.php/2019/07/27/batteries/


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## Dude99 (Mar 13, 2020)

wiliki said:


> hey @Chattaneer PE I've got another problem here from the Cram for PE test. I probably need to refresh myself on symmetrical components fault analysis after this...
> 
> Why is the j0.07 reactance not included in the zero-sequence portion of the circuit? Also, is it standard to always assume V = 1 pu?
> 
> ...


This is a good example: THINK before grabbing the pencil and calculator.

can't be 0 unless it is the faulted line or inf bus no and no Z.

can't be 13800 unless NO Z except a hi NGR.

same for 12090, nrg must be &gt;&gt;&gt;&gt; X1

that leaves 6990, which makes sense since L-N is ~7970
 

imo the point of a question like this is to make you think about basics and give exercise on SC's, but I personally spend time on it during testing.  No if the range of answers was: 6900, 6990, 7100, 7600, maybe.  It can be done quickly using only Z and arrive at 6900.  Don't always jump into the most complicated method, think basic concepts first.


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