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Circuit theory -- Conditions to balance a resistive bridge circuit

  1. Jan 21, 2015 #1
    image.jpg 1. The problem statement, all variables and given/known data
    Establish the condition required to make the current through Ze in Figure 1 zero (i.e. For the voltage Vx to equal Vy).

    2. Relevant equations


    3. The attempt at a solution
    I'm not sure if I'm on the right line here, hopefully someone can point me in the right direction if it's wrong...

    If Zb and Zc were the same value then the voltage drop across them would be equal, likewise Za and Zd would also have to be the same value to have voltage Vx equal to Vy. In this condition there will be no current flowing through Ze.

    Is my attempt correct or at least on the right path?

    Thanks in advance
     
  2. jcsd
  3. Jan 21, 2015 #2

    NascentOxygen

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    Hi CG139. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    You've indicated here a special case of the more general right answer. :)
     
    Last edited by a moderator: May 7, 2017
  4. Jan 21, 2015 #3

    phinds

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    To add to NO's statement, you have selected not only a special case but a VERY special case ... that is, it is NOT enough that Zb=Zc or that Za=Zd it would be necessary that BOTH of those conditions occur at the same time. This is a very special case indeed.

    Think about the relationships between Za to Zb and Zc to Zd

    That is, basically your statement

    If Zb and Zc were the same value then the voltage drop across them would be equal, likewise Za and Zd would also have to be the same value to have voltage Vx equal to Vy. In this condition there will be no current flowing through Ze.

    Is very limited. Specifically, "If Zb and Zc were the same value then the voltage drop across them would be equal" is a very limited case because it then REQUIRES that Za=Zd
     
  5. Jan 21, 2015 #4

    NascentOxygen

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    It is better to allow the poster some opportunity to think about the response, rather than leaping in and doing the thinking for him. In the long run, depriving a poster of the opportunity to review his own work is performing a disservice.

    A little restraint can provide a poster the necessary time to process what his attention has been drawn to.

    I know it's tempting to scramble to fill in what a previous poster has left unsaid. It was alluded to but left unstated for good reason.
     
  6. Jan 21, 2015 #5

    phinds

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    Yeah, you are right. I normally do that and in fact I sometimes chide others for not doing it. Thanks.
     
  7. Jan 21, 2015 #6
    Thanks NascentOxygen and Phinds, this subject has had me scratching my head a lot and I'm not sure I've fully understood the replies. Do I just need to expand my statement to include that both conditions need to occur at the same time? When I wrote my original statement I was thinking that, i just didn't make it clear- will try harder.
     
  8. Jan 21, 2015 #7

    BvU

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    Dear CG,

    Can you write down an expression for Vx using the variables in this figure ?
     

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  9. Jan 21, 2015 #8
    Hi BvU,

    Vx = (Za / (Zb+Za)) *Vin

    I struggled to get the format how I wanted it. I've tried to use Latex but don't think I can use it as I'm using a tablet rather than a laptop.
     
    Last edited: Jan 21, 2015
  10. Jan 21, 2015 #9

    BvU

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    Very well. Now Vy in this picture:
     

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  11. Jan 21, 2015 #10
    Vy = (Zd / (Zc+Zd)) *Vin
     
  12. Jan 21, 2015 #11

    NascentOxygen

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    Going back to your original diagram. Given these values, calculate ZC for the balance condition (i.e., no current through Ze)

    ZA = 5k,
    ZB = 3k
    ZD = 12k
    ZC = ?
     
  13. Jan 21, 2015 #12

    BvU

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    Dear CG,

    As O2 isn says more or less, we now go towards vx = vy.
    Simply grab together vx = vy, #8 and #10. You see that Vin can be divided out.
    That already gives you an adequate answer for the exercise. But it can be made more compact and more elegant by inverting on both sides and subtracting 1.
     
  14. Jan 22, 2015 #13
    NascentOxygen,

    Zc = (Zb * Zd) / Za

    Zc = (3k * 12k) / 5k

    Zc = 7.2k

    These values would balance the circuit (Vx=Vy).


    BvU,

    Sorry, but I don't fully understand #12. Apologies for my poor knowledge, I'm a long way from my comfort zone with this and the material from my Uni doesn't seem to help (not helped by the fact my maths is also poor).

    By 'grab together' do you mean simplify to create one expression?


    Thanks to you both for your help so far, it is very much appreciated!
     
  15. Jan 22, 2015 #14

    BvU

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    No need to apologize. Yes, grab together is an undiplomatic way to say: if x=u and y=v and x=y then u = v. Mathematicians might say "use symmetry and transitivity", physicists just grab :)

    Vx = (Za / (Zb+Za)) *Vin (your post),
    Vy = (Zd / (Zc+Zd)) *Vin (your post), and from the exercise wording:
    Vx = Vy (we're not interested in either of these two, they just have to be equal)

    "Grab together" is then luring you to write (Za / (Zb+Za)) *Vin = (Zd / (Zc+Zd)) *Vin
    Vin in itself isn't all that interesting either, bringing us to (Za / (Zb+Za)) = (Zd / (Zc+Zd))

    This time you have a general answer, not just something for a special case.
    (a few pathological cases have to be excluded, though: Vin = 0, Zb+Za = 0, Zc+Zd=0; no big deal)

    That (Za / (Zb+Za)) = (Zd / (Zc+Zd)) already is an adequate answer for the exercise. But it can be made more compact and more elegant by inverting on both sides and subtracting 1. That way you get an expression the exercise is really asking for.
     
  16. Jan 22, 2015 #15

    NascentOxygen

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    Can you express the result in words? I'm still not sure that you recognize the general solution ....:nb)
     
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