# Complex Imaginary Test 1 Problem 1

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#### cos90

##### cos(90)=0
I'm really struggling with these power electronics questions.

Why is the answer not 125 * sqrt(3) * sqrt(2)? The voltage provided is phase voltage and the voltage over the diodes would be line voltage?

I'm really struggling with these power electronics questions.

Why is the answer not 125 * sqrt(3) * sqrt(2)? The voltage provided is phase voltage and the voltage over the diodes would be line voltage?
I too feel the same. The output in this case across the load is L-L voltage that means  Voutput (rms)=125*sqrt3 and peak= sqrt Vrms=125*sqrt3*sqrt2. Does anyone else has other opinion

I'm really struggling with these power electronics questions.

Why is the answer not 125 * sqrt(3) * sqrt(2)? The voltage provided is phase voltage and the voltage over the diodes would be line voltage?
It was a 3-Phase AC circuit before it was rectified, but now it's a pulsating DC circuit.

The temptation to treat this like a run-of-the-mill 3-Phase AC circuit is why this problem would be a good one to include on a real PE exam, except one of the wrong solutions would be the result of the 3-Phase calculation to determine if the examinee really understands the underlying principles of rectification.

I thought the same thing when I ran across this problem during my studies, but I did some reading at the following sites:

http://tinyurl.com/y9p2vkbt

It was a 3-Phase AC circuit before it was rectified, but now it's a pulsating DC circuit.

The temptation to treat this like a run-of-the-mill 3-Phase AC circuit is why this problem would be a good one to include on a real PE exam, except one of the wrong solutions would be the result of the 3-Phase calculation to determine if the examinee really understands the underlying principles of rectification.

I thought the same thing when I ran across this problem during my studies, but I did some reading at the following sites:

http://tinyurl.com/y9p2vkbt

I still have a feeling that Voltage across load is Line line voltage of AC and so it is sqrt3*sqrt2*Vp-rms . I am attaching a part of the explanation given at this link. Can someone drwa the o/p wave shape which proves otherwise.  https://www.pantechsolutions.net/powerelectronics-tutorials/introduction-of-three-phase-half-full-wave-converter
View attachment 9808

I guess what I'm trying to say is that the load isn't seeing the line-to-line rectified voltage...it's seeing the sum of three single-phase voltages that have been fully rectified, so each rectified phase voltage is simply sqrt 2*Vrms.

There could be 6 or 1,000 phases that were fully rectified and it wouldn't make a difference to what the load sees.

I guess what I'm trying to say is that the load isn't seeing the line-to-line rectified voltage...it's seeing the sum of three single-phase voltages that have been fully rectified, so each rectified phase voltage is simply sqrt 2*Vrms.

There could be 6 or 1,000 phases that were fully rectified and it wouldn't make a difference to what the load sees.
I am sure you are not kidding. I know I will not be able to explain the concept by messages but the voltage output is same as read by the voltmeter in red in attached figure and that too appears in the explanation I attached earlier and that too suits my logic. I think the link I shared earlier explain it quite nicely. The peak voltage  here is peak of line to line and not the phase to neutral and the answer should be 1.72*1.41*125=303V

View attachment 9817

I am sure you are not kidding. I know I will not be able to explain the concept by messages but the voltage output is same as read by the voltmeter in red in attached figure and that too appears in the explanation I attached earlier and that too suits my logic. I think the link I shared earlier explain it quite nicely. The peak voltage  here is peak of line to line and not the phase to neutral and the answer should be 1.72*1.41*125=303V

View attachment 9817
Where you have your voltmeter in your sketch will read sqrt3 * 125V, or 216.5V...that is not where Vab is measured from.

Vab is measured from the outputs of the rectifier, as shown in my attachment.

@BigWheel again I understand the same thing. In the given link it is I/P 400V Line to line (230V Phase to nuetral) ac and load voltage peak is  sqrt 3*Vline= 400*sqrt3= 565 this is same as sqrt2*sqrt3*230 as 230 is the phase voltage there. View attachment 9825

@BigWheel again I understand the same thing. In the given link it is I/P 400V Line to line (230V Phase to nuetral) ac and load voltage peak is  sqrt 3*Vline= 400*sqrt3= 565 this is same as sqrt2*sqrt3*230 as 230 is the phase voltage there. View attachment 9825
And the voltmeter Vab you have shown will read the same as voltmeter I showed v; as the voltage drop in diodes is taken as zero.

This answer on this source agrees with us and says the V,phase,rms is 230V. Times sqrt(2), 325, for peak amplitude and times sqrt(3) , 565 for line voltage.

The SPICE simulation below the answer agrees with these values.

It seems this is another mistake in CI. I don't really care about the mistake because I was able to figure out what he wanted based on the multiple choice answers but it is frustrating my inner (not very good) student.

@BigWheel again I understand the same thing. In the given link it is I/P 400V Line to line (230V Phase to nuetral) ac and load voltage peak is  sqrt 3*Vline= 400*sqrt3= 565 this is same as sqrt2*sqrt3*230 as 230 is the phase voltage there. View attachment 9825
You're right. I missed the connection. I interpreted the question as what would be the peak voltage of the given phase voltage at Vab (perhaps that was the intended question), but the question simply asks what what is the peak voltage at Vab, so your answer is right.

I think you meant 400 * sqrt2 = 565V, though.

Thanks.

You're right. I missed the connection. I interpreted the question as what would be the peak voltage of the given phase voltage at Vab (perhaps that was the intended question), but the question simply asks what what is the peak voltage at Vab, so your answer is right.

I think you meant 400 * sqrt2 = 565V, though.

Thanks.
@cos90 and @BigWheel thanks for finally getting it right. Yes it is the inner student which remains agitated till you get an logic aligned answer. It remains agitated till either the logic is proved right it is otherwise proved. It gives you a relief when it is decided else the your CPU (brain) will show busy sign (that rotating circle) specially if you are at discussion with a person who you know is knowing the subject.I had done all CI and found quite a few wrongs there including this question..