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Capacitance, unbalanced wheatstone.

  1. May 27, 2015 #1

    Suraj M

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    1. The problem statement, all variables and given/known data
    WIN_20150527_141946.JPG
    Find capacitance across AB
    So i divided the current into q1 and q2 as shown. then i assumed that q₂>q₁.

    2. Relevant equations
    ##C=\frac{Q}{V}##
    I dont know anything after KVL KCL and loop law.

    3. The attempt at a solution
    So i split the current at T into q₃ and q₂-q₃
    by the diagram, i used Loop law for UPQTU and QRSTQ.
    this gave me a relation between q₁,q₂ and q₃.
    i got
    $$ q₁= 15q₃$$
    $$q₂=16q₃$$
    $$q₁ = \frac{15}{16} q₂$$
    using these, i used the equation..
    $$ C_{AB} = \frac{q₂ +q₁}{\frac{q₁}{5} + \frac{10}{q₁+q₃}}$$
    so i got $$C_{AB} = \frac{155}{23} µF$$
    Is this right.? because i doubt it.
     
  2. jcsd
  3. May 27, 2015 #2

    ehild

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    You work with charges instead of currents. The net charge should be zero at junctions Q and T.
     
  4. May 27, 2015 #3

    Suraj M

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    can't we do it by this method? KCL necessary? I will try.
     
  5. May 27, 2015 #4

    ehild

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    You can, but write charge instead of current. And take into account that the net charge on the connected plates is zero, and the charges on the opposite plates of a capacitor are of opposite sign.

    Show your work in detail. Otherwise I can not decide if you got the correct result or not.
     
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