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What is the attenuation factor?

  1. Sep 5, 2016 #1
    1. The problem statement, all variables and given/known data
    the question has a transformer, the hv side of which is connector to a pot that is said to be an attenuator in the question with attenuation factor = .8?
    the turns ratio of transformer is 1:1.125.
    I have to find HV side voltage after the attenuation...i know attenuator reduces voltage but how does the factor come into picture?


    2. Relevant equations
    transformer HV Voltage = n1/n2*LV Voltage = 1.125/1*230 = 258.75

    3. The attempt at a solution
    I considered attenuator as resistor with .8 value of total value. like a pot and took output voltage as V=V(HV)*.8/(.8+.2)=V *.8 = 258.75*.8 = 207V

    I dont think i used the attenuatior factor rightly. Google suggested a log formula. Like 10 raised to A/20. where A is factor.
    I'm not sure how to proceed.

    Any help?
     
  2. jcsd
  3. Sep 6, 2016 #2

    BvU

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    Hi,

    Not an expert, but I agree with what you did - provided you sketched the whole problem (in particular: no load)

    The attenuation you quote is for purposes like audio where they express attenuation in dB (1 Bel = factor 10 in power so 20 dB is a factor 10 in voltage)
     
  4. Sep 6, 2016 #3
    Thanks for the reply. Here is the print screen of the homework question paper...
    Homework%20Question_zpskkd05azf.png
     
  5. Sep 6, 2016 #4

    BvU

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    Ok, so that is a different problem than the one you calculated. First of all your turns ratio is not 1:1.125 but 1:1.25 .
    On what you call (als a bit confusing) the LV side the voltage isn't 230 but 1000 Volt. (Although that isn't really important because the exercise asks for ratios :smile:)
    Re-doing your calculation with the 1.25 should give you ##\displaystyle {V_{YZ1}\over V_{WX1}}## which already narrows it down.
    Now the other way around for [edit: wx/yz, not yz/wx] ##\displaystyle {V_{WX2}\over V_{YZ2}} ## . What do you propose ?

    And yes, the attenuation factor 0.8 is simply as you assumed.
     
  6. Sep 6, 2016 #5
    How do you calculate VYZ1? What formula for using attenuation do you suggest?
     
  7. Sep 6, 2016 #6
    Sorry about the errors posted in the problem...I thought I had mugged up the question, but I was wrong...sorry again.
     
  8. Sep 6, 2016 #7

    BvU

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    No need to apologize. You learn and you have the right to make mistakes.
     
  9. Sep 6, 2016 #8
    The answer in the book is C. I don;t know how that comes. When Vwx1 is 100 V. V of HV side is 100*1.25 = 125V.
    So input to attenuator is 125V. How do i calcuate Vyz1 based on this? Attenuation factor is .8
     
  10. Sep 6, 2016 #9

    BvU

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    See, I make mistakes too: 100 V instead of 1000 V :smile:

    If you attenuate 125 V by a factor of 0.8 you get 125 V x 0.8 = 100 V !

    ( :rolleyes: the 100/100 is a likely result for this first part -- any of the other answers makes it unnecessary to solve the second part. I sort of gave that away by saying "narrows it down" instead of "gives the answer").
     
  11. Sep 6, 2016 #10
    Hmm...I had to read your posts a few times and go back at the question. I guess you're right. Attenuation simply works as o/p = i/p * attenuation factor.
    The whole log and decibels stuff doesn't apply here.
     
  12. Sep 6, 2016 #11

    BvU

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    Right. Now, how do we choose between (B) and (C) ?
     
  13. Sep 6, 2016 #12
    I did that. The Vyz2 is 100 V. So o/p of attenuator is 100V. So i/p of attenuator is 100V divided by attenuation factor = 100/0.8 = 125V.
    So HV of transformer is 125V.
    LV of transformer is 125 * N1/N2 = 125*1/1.25 = 100V. So Vwx2 is 100V.
    So ratio is 100/100 V.
    So correct answer is C.
     
  14. Sep 6, 2016 #13

    BvU

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    Ha, you fell for it. An attenuator attenuates. It does not amplify.
     
  15. Sep 6, 2016 #14
    Darn...this is so confusing. So how is C the answer? What is the formula for attenuator? the formula for transformer is Voltage ratio = turns ratio. is there a simple way to solve this problem?:frown:
     
  16. Sep 6, 2016 #15

    BvU

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    You're almost there. Your transformer is ideal, meaning V x I is the same on both sides. So what is the input current on the righthand side ?
     
  17. Sep 6, 2016 #16
    Well as per KCl, current input at right side of attenuator = current input to transformer + current through attenuator.
    so I (input to attenuator) = 100V /(attenuator resistance) + I transformer input current...
    Sorry i know this is the wrong track. How can you find current without load or B-H curve or resistance?
     
  18. Sep 6, 2016 #17

    BvU

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    Look at it from the other side (the lefthand side): it is open, so there is no current. V x I = 0 on that side. Ideal transformers have V x I the same on both sides. So there is no current going into the transformer on the righthand side. Therefore there is no current going "upward" from Y to the transformer.
    That means the voltage drop over that top part of the "pot" is ... ?
    So the righthand side of the transformer sees how many Volts ?
     
  19. Sep 6, 2016 #18
    But if there is no current in the transformer HV side, then transformer action wouldn't take place. No current, so no flux, so no linkage and no emf?
    This is then an open circuit test, i guess.
    So the right side of transformer would see input voltage from right side that is 100V, as the attenuator isn't really doing anything if no current is flowing...
     
  20. Sep 6, 2016 #19
    then the answer comes to 80V/100V that is B.
     
  21. Sep 6, 2016 #20

    BvU

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    I agree. I think it was a nice exercise (looks innocent but has some substance), but you might have a different opinion :smile: .
    All clear now ? Then on to the next challenge :wink:
     
    Last edited: Sep 6, 2016
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