Circuit Analysis. Capacitor in AC and DC circuit.

Discussion in 'Homework Help' started by VinceClortho, May 1, 2016.

  1. VinceClortho

    Thread Starter New Member

    Mar 30, 2016
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    (sorry for the link but I couldn't get a picture to upload) http://imgur.com/h8NbFJs I really don't know where to begin on this one. I'm familiar with capacitors to an extent but this type of question was never really covered in our class. I tried an equation Vmin=Vdc -Vp but I know that isn't right.

    Capacitor_question.jpg

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    Last edited by a moderator: May 2, 2016
  2. JoeJester

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    Apr 26, 2005
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    You can begin by describing the AC signal ... Draw it out and lable the amplitudes
     
  3. wayneh

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    Sep 9, 2010
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    Your formula might be helpful if you define the terms better. What's Vmin? How do you get Vp? See Joe's comment.
     
  4. VinceClortho

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    Mar 30, 2016
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    Something like this? http://imgur.com/T7N83Dl

    Capacitor_question_solution_try1_reduced.jpg

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    Last edited by a moderator: May 2, 2016
  5. VinceClortho

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    Mar 30, 2016
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    As I understand it Vmin is the minimum voltage of the sine wave. Vp is the peak value of the sine wave and is found by multiplying Vrms by 1.414.
     
  6. Jony130

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    Yes, but notice that -20V RMS ( -28.2V peak) will kill the electrolytic capacitor so you must "shift" this voltage up.
     
  7. JoeJester

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    @VinceClortho

    Your drawing is incorrect. Review your question and correct the drawing.

    Review the terms RMS, peak, average, and peak to peak with respect to a sine wave signal.
     
    Last edited: May 2, 2016
  8. VinceClortho

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    Mar 30, 2016
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    So does "shifting" it up mean introducing a DC voltage greater than peak? As I understand it that means the voltage will never reach negative.
     
  9. VinceClortho

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    I'm afraid I don't understand.
     
  10. Jony130

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    Yes, something like this.
     
  11. VinceClortho

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    http://imgur.com/KYRiFa1 Here are some new values. I still don't quite understand where I went wrong, would you mind showing me?

    Capacitor_question_solution_try2_reduced.jpg

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    Last edited by a moderator: May 2, 2016
  12. MrAl

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    Jun 17, 2014
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    Hi,

    You want to make sure the capacitor voltage never goes negative.
    The sine voltage goes negative without the DC voltage source, but with the DC source having the correct value the cap voltage will never go negative and thus the cap will be ok. Note the polarity of the cap. That means it cant go negative or the cap blows out.

    Since you know the sine voltage, figure out what DC voltage it would take, when added to the sine voltage, to keep the cap voltage always positive.
     
  13. VinceClortho

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    Mar 30, 2016
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    Assuming I use the general rule of one standard unit larger I would think that 29.3 V would be correct. Vp=28.3; Vmin=Vdc-Vp= 29.3V-28.3V =1V. Is this correct?
     
  14. JoeJester

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    I wanted you to realize the relationships illustrated in this graphic

    [​IMG]


    Here is a 1 kHz signal. what can you sat about them?
    AC.png
     
    Last edited: May 2, 2016
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  15. VinceClortho

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    Thank you very much for this and your help. Thanks everyone for the help.
     
  16. WBahn

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    Mar 31, 2012
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    @JoeJester Note that the Average Value for the green waveform shown is 0V. I think it is trying to get at the average value of the full-wave rectified version of the green waveform.
     
  17. JoeJester

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  18. MrAl

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    Hi,

    The mathematical average of a sine wave is zero, but the average voltage of a sine wave voltage source is taken to be 2/pi times the peak, which is about 0.637 times the peak, and this is correct even though it goes plus and minus and would normally average out to zero over any integer number of full cycles. Mathematically that is the average of abs(sin(wt)) which is the average of the absolute value of the sine wave which of course is the average of the ideal full wave rectified sine voltage.
     
  19. WBahn

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    Taken to be that by who (whom?)? I'm not being snide, I've just never seen (that I recall) the claim that the average voltage of a pure sine wave is anything other than zero (before the graphic that JoeJester posted, anyway). If I pass a signal through a low pass filter (of appropriately low cutoff frequency) I will see the average value of the waveform. If I pass a pure sine wave through such a filter I will see zero (or very nearly zero) at the output. I will NOT see 2/pi times the peak. A sine wave and a full-wave rectified sine wave are two fundamentally different waveforms.
     
  20. JoeJester

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    @MrAl , @WBahn

    That graph illustrates the average position on that half cycle, not the whole wave. If we did the same for the negative half, it would end with a zero net sum.

    I don't recall if my instructor from 1973 specified it was for just that half cycle. It was my mistake for not specifying the values were only for the half cycle specified on the graph.

    Of course, if we connected an analog multi-meter to that signal, the signal is full wave rectified prior to the input of the meter circuit.

    WBahn is correct because I failed to clarify that graph with text.
     
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