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Homework Help: Power Electronics: Choppers RLE Load

  1. Jan 3, 2018 #1
    1. The problem statement, all variables and given/known data
    upload_2018-1-3_13-45-55.png

    2. Relevant equations
    I rms for triangular waveform = I(max)/ (1.732)
    R = V/I
    3. The attempt at a solution
    Since there is RLE load, current waveform is triangular. It is
    upload_2018-1-3_13-47-48.png

    Source current is
    upload_2018-1-3_14-6-51.png
    Average of I source. I used integration. Assumed Time period = 10 seconds. So T on = 5 seconds. (question says duty = 0.5)

    So integrating I got average source current = 14.7 A
    Average source voltage = 200 V
    So input resistance = 200 / 14.7 = 13.6A

    But I'm not able to get the first one. I get answer as 26.33 A But that is not even in the option.
    I used formula square I(load) = square dc component (26.05) + square (rms of triangular waveform)
     
    Last edited: Jan 3, 2018
  2. jcsd
  3. Jan 3, 2018 #2

    NascentOxygen

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    Staff: Mentor

    It says minimum is 32.75A and maximum is 26.05A. Something seems wrong, since we expect max > min.

    So I'd say there's a printing error, maybe two, and in either the data or the answer options. Would (A) 26.17A satisfy you? If not, you could play around with various changes in the answers to see whether you can stumble across consistent values.
     
  4. Jan 3, 2018 #3
    Yes the printing error occured to me as well. I guess only the max n min words should be exchanged. Cause I'm getting right answer for second part of question.
    How do you get 30.17A as answer?
     
  5. Jan 3, 2018 #4

    NascentOxygen

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    Staff: Mentor

    I looked for a single digit to change that would turn what's printed into something close to what you're looking for.
     
  6. Jan 3, 2018 #5
    Book says answer is A.
    Is this correct method?
    upload_2018-1-3_19-34-30.png
     
  7. Jan 3, 2018 #6

    NascentOxygen

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    Staff: Mentor

    Your calculations are probably right. You answered while I was still adjusting my comment. I settled on 26.17 as a likely option.

    Maybe if you worked it out exactly, using the exponentials instead of approximating to straight lines, it might match the 26.17A answer exactly?
     
  8. Jan 3, 2018 #7
    So instead of triangular ur saying exponential waveform? e-Rt/L?
     
  9. Jan 3, 2018 #8

    NascentOxygen

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    Staff: Mentor

    Well, only if you want to test whether my hunch is right. :oldsmile: The homework would not expect you to, the triangular approximation is fine. Though in the real world you should show that the triangular approximation is justified and not simply assume it is, i.e., not just hope it is.
     
  10. Jan 3, 2018 #9
    I got something like this:
    upload_2018-1-3_20-4-23.png
     

    Attached Files:

  11. Jan 3, 2018 #10
    I guess triangular waveform is justified when the inductance is very big. If resistance is very big then waveform of current = waveform of voltage = square.
    I'm getting integrating value as 27.6A
    How did you integrate so fast?
     
  12. Jan 3, 2018 #11

    NascentOxygen

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    Staff: Mentor

    I haven't worked out an answer. I see insufficient information given to do so, in this instance.

    The rising exponential, A(1-e-t/T) has magnitude A = (200-E)/R – 26.05
    and T = L/R = unspecified
    and passes through the point 6.7 at t=?? (5ms your nomination)

    Too many unknowns to evaluate to a number, I think.
     
  13. Jan 3, 2018 #12
    I assumed T to be 10 seconds and since duty cycle is 0.5 we get inductor storing energy for 5 seconds and releasing energy for 5 seconds.
     
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