Cram for the PE Sample Test 1 - Problem #13

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wiliki

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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. 

Problem 13.JPG

 
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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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.

Capture1.PNG

 
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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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).
 
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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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 & 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.

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

 
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? 

Cram for PE Test 1 - #7 Problem.PNG

Cram for PE Test 1 - #7 Solution.PNG

 
@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.

 
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.

Transformer_sequence_networks.jpg

 
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? 

 
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.

 
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.

Untitled.png

 
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? 

 
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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