NCEES #519

[robertplant22]

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

 

[knight1fox3]

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.

 

[robertplant22]

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.

 

[knight1fox3]

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.

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.

 

[ecbahr]

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

 

[Silkworm]

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.

 

[K_Nova]

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.

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.

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!

 

[knight1fox3]

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

check out this applet. it helps:

http://services.eng.uts.edu.au/~venkat/pe_html/ch05s1/ch05s1p1.htm

 

[K_Nova]

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?

 

[EE_PE_AP]

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

 

[knight1fox3]

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

check out this applet. it helps:

http://services.eng.uts.edu.au/~venkat/pe_html/ch05s1/ch05s1p1.htm