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Different EMFs joined in parallel

  1. Jun 17, 2017 #1
    1. The problem statement, all variables and given/known data

    We need to find I1,I2,I3 (current through all the three branches) 20170617_113748-1.jpg
    2. Relevant equations
    I know that if emfs *of same value* are connected in parallel,then they can be replaced with a single emf of that same value,and that's where the problem comes in.There are 3 emfs,all of different values connected in parallel.
    As far as the rest of the curcuit goes,
    ▪KVL (Kirchoffs Voltage Law):Sum of potential drops across a closed loop is zero
    ▪Voltage difference across conducting wire=0
    3. The attempt at a solution
    You'll forgive me if the diagram's a little crowded.Basically,I assumed potential to be 0 at a point for reference purposes, and then labelled all other parts of the circuit corresponding to that.

    Now,as I said in the last line of the previous section,

    ▪Voltage difference across conducting wire=0
    That means that V(at A)=V(at B)=V(at C)
    (Assumin I1 as x,I2 as y,I3 as z)
    That would give us our first set of equations:

    Now,applying KVL in loop ABEF,
    However,on solving these equations,I ain't getting the right answer.

    Help appreciated..
  2. jcsd
  3. Jun 17, 2017 #2


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    You can use nodal analysis. Write the KCL equation for the left node (junction of three resistors) in terms of the node voltage. Assume the negative terminal node of the batteries at 0V. The only unknown in the equation will be the node voltage.
    Try Millman's theorem.
  4. Jun 17, 2017 #3


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    It is easier for me to work out the problem than to see you you where you went wrong. Also it would be easier for me to see whether I went wrong if you had given the answer which apparently you know.

    Yes, essentially as cnh says, you seem to realise that the voltage, we'll call it V, at A, B and C has to be the same despite the differing battery EMFs. From that I would work out the currents I1, I2, I3 in terms of V: their sum equals the current through the 6 Ω resistor, which gives you an equation with only V as unknown . I get for V, 95/13 ≈ 7.3 V is that what it's supposed to be?
  5. Jun 17, 2017 #4
    95/12,and yes,it's giving the corect answer for I1,I2 and I3.
  6. Jun 17, 2017 #5
    Thanks for the link to this theorem.It's awesome,but the answer coming from it is SLIGHTLY different (95/11 whereas it should be 95/12).
    Is this expected or did I do something wrong?
  7. Jun 20, 2017 #6


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    Yes /12 not /13, I can't add up. :redface:

    But was important to realise it was in the right ballpark - there is a lot of conductance than those three parallel branches.
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