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Bridge voltage divider

  1. Jan 9, 2016 #1
    • Thread moved from the technical physics forums, so no Homework Help Template is shown.
    Good morning,I'm french and I need help for this exercice.
    It's a exercice it is an exercise on the voltage gain,and on the first scheme there are the correction of my teatcher,but I but I was wondering if we can't calculate the voltage gain with a bridge voltage divider like I did.(It's in french but only the formula are important).
    The second scheme,it's the same thing(the method of my teacher ,and bridge voltage divider),but is it good?
    160109064247913760.jpg
    160109064312572686.jpg
     
  2. jcsd
  3. Jan 9, 2016 #2

    BvU

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    First diagram: yes. One more step and you can see that the ##V_s\over V_e## ratios are identical !

    Second diagram: I don't see what your professeur did ?

    But you want to check your ##1\over Z_{eq}## !
     
  4. Jan 9, 2016 #3
    Hello,thank you for your help :),so my teacher did not do the second exercise,but I tried to use his methode.
    His method is in red on this diagram(or scheme ):
    Is-it a question?
    No I don't want to check my 1/Zeq,this calculation is right I think but I have not always trusted me.
    I must speak better in english to better understand you and write better.
    But yes professeur=professor in english :).
    16011002031976569.jpg
     
  5. Jan 10, 2016 #4

    BvU

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    Method of your professeur does not give 1: numerator and denominator are different.

    Time to point out that method of professeur and your method are not different: $$ {V_s\over V_e} = {Z' I_e \over \left ( Z + Z' \right ) I_e} = {Z' \over Z + Z' } $$
     
  6. Jan 10, 2016 #5

    BvU

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    Of course not$${1\over 2} = {1\over 3} + 0.16667 \Rightarrow 2 = 3 + 6 \ \ \ \ {\rm ?} $$
     
  7. Jan 10, 2016 #6
    Yes yes,I have forgot one thing,and to answer your question no that does not mean it,but yes effectively the methods are similar.
    Thank you very much !
    I wish you a good day :).
     
  8. Jan 10, 2016 #7

    BvU

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    Avec plaisir. You're welcome and I'll be glad to look at your result for the (R//C) / (R+C + R//C) case...
     
  9. Jan 10, 2016 #8
    Normally equal to Zeq :
    160110041719803814.jpg
     
  10. Jan 10, 2016 #9

    BvU

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    So far, so good !
     
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