# NCEES #519



## robertplant22 (Feb 10, 2012)

I do not understand why the answer is (D). If the circuit is functioning as a single phase full wave rectifier, the output current would be similar to that of the voltage output waveform; which is, waveforms with only positive values as shown below.

For anyone who is wondering where the picture came form; it came from the book Electrical Machines, Drives, and Power Systems pg. 482


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## knight1fox3 (Feb 10, 2012)

robertplant22 said:


> I do not understand why the answer is (D). If the circuit is functioning as a single phase full wave rectifier, the output current would be similar to that of the voltage output waveform; which is, waveforms with only positive values as shown below.
> 
> For anyone who is wondering where the picture came form; it came from the book Electrical Machines, Drives, and Power Systems pg. 482


Have a look at this thread and see if it helps.


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## robertplant22 (Feb 10, 2012)

Not sure why the picture did not upload, but here it is. I read the thread knight1fox3 pointed to; but I'm still confused. In the graph below, the current never goes below 0; were as in the answer given in the NCEES book it does.


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## knight1fox3 (Feb 14, 2012)

For this particular problem, I don't think that graph is helpful.

To quote another poster in that thread, the key here is understanding what is actually being asked.



pelaw said:


> I think the key to this problem is understanding what is asked. The problem asks *ONLY for PHASE A*. Not for the circuit.
> 
> Second part is understanding that PHASE A will (1) have one peak over a cycle, 1/60s, *and (2) one negative peak* as it will carry the return current from phase B.


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## ecbahr (Oct 20, 2012)

Can someone explain why it is D and not B? Why is the waveform not sinusoidal?


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## Silkworm (Oct 20, 2012)

Since the motor load is nonlinear, the phase current waveform will not be sinusoidal. So you can quickly discount B.

For one cycle, diodes connected to Phase A will only conduct between 0 and 60 degs (+ve) and 120-240deg (-ve). As the circuit is not filtered by an inductor, the distortion factor factor will be high and DC current will contain high order harmonics. This will make the waveform look like a short impulse as shown in the answer and not the square wave you are used to seeing. So the only answer that makes sense is D.

I don't see how blown Fuse F3 has any impact on the current waveform. The phase voltages will be unbalanced but the current Ia waveform essentially remains the same ( magnitude will change) as far as I can see. Maybe someone can comment on that.


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## K_Nova (Feb 20, 2013)

knight1fox3 said:


> For this particular problem, I don't think that graph is helpful.
> 
> To quote another poster in that thread, the key here is understanding what is actually being asked.
> 
> ...


Can someone please clarify the latter portion of the second statement in the above quote: "and (2) one negative peak, as it will carry the return current from phase B".

I can't seem to grasp my mind around this, on how conceptually this is happening. Thanks!


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## knight1fox3 (Feb 20, 2013)

Did you check out the link that someone else posted and read through all the information? There is a pretty neat simulation toward the end which shows a graphical representation of what is being explained. Take a look at that and see if that helps



afacemire said:


> check out this applet. it helps:
> 
> http://services.eng.uts.edu.au/~venkat/pe_html/ch05s1/ch05s1p1.htm


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## K_Nova (Feb 20, 2013)

Thanks Knight, that was a very helpful link!

So just to reinforce the point I took (2) screen shots, let me know if my below explanation conforms to the solution's explanation.

Lets assume V_R is V_A from problem #519, and V_Y is V_B from problem #519. Now, the first image displays the positive peak phase A will see and the second image displays the negative peak phase A will see (the return current from Phase B). Is that correct?


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## EE_PE_AP (Feb 23, 2013)

K_Nova said:


> Thanks Knight, that was a very helpful link!
> 
> So just to reinforce the point I took (2) screen shots, let me know if my below explanation conforms to the solution's explanation.
> 
> Lets assume V_R is V_A from problem #519, and V_Y is V_B from problem #519. Now, the first image displays the positive peak phase A will see and the second image displays the negative peak phase A will see (the return current from Phase B). Is that correct?


Hi K-Nova where did you take these shots from? I mean source? I would want to read that too. Please tell me the source book or a web link if you dont mind. Thanks in advance


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## knight1fox3 (Feb 24, 2013)

knight1fox3 said:


> Did you check out the link that someone else posted and read through all the information? There is a pretty neat simulation toward the end which shows a graphical representation of what is being explained. Take a look at that and see if that helps
> 
> 
> 
> ...


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