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Kirchhoff's Loop Rule Problem.

  1. Dec 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Actually, the main question is to find out the potential difference between the points 'A' and 'D'.
    The cell with emf 'E' has not been shown in the main question, but I have assumed it to be connected between 'A' and 'D'. Each resistor is of 1 Ohm.

    2. The attempt at a solution
    While solving the question, I applied Kirchhoff's Laws
    Let 'I' be the current and let 'I1' be the current through the branch QR.
    Since the total resistance offered by the branches (QR, PS, TU) is same (One ohm) , the currents through them will also be the same.
    Assuming E > 3 V ( where 'E' is the emf of the cell)

    Applying Kirchhoff's Current Law,
    I = 3 I1
    Applying KVL to the loop BAPQRDCB,
    Δ V = 0
    So, I(1) + I1(1) + 3 = E ....(1)
    Similarly,
    When KVL is applied to the loop BAPSDCB,
    I(1) + I1 (1) + 2 = E ....(2)
    (1) = (2), as E = E,
    Which, on solving, gives 2 = 3!
    Where did I go wrong?


    uyy.png
     
    Last edited: Dec 30, 2014
  2. jcsd
  3. Dec 30, 2014 #2

    ehild

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    That is not true. The potential difference is the same across the three branches: VP-VS. As there are different cells in every branch, the voltages across the resistors are also different.

    You can denote the currents flowing in the branches I1, (upper branch) I2 (middle branch) and I3 (bottom branch). Then I=I1+I2+I3. Write the loop equation for the loops which contain E and the series 1 ohm and one of the branches.

    kircloop.JPG
     
  4. Dec 30, 2014 #3

    BvU

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    1) What was shown in the main question ? Nothing there? Or a short-circuit ? Was anything mentioned about E at all ?

    No. There are voltage sources too !
    2) so that gets rid of I = 3 I1. You will need to split up I in I1, I2, I3.

    [edit] ah, ehild was faster !
     
  5. Dec 30, 2014 #4
    No, only the portion of the figure between the points A and D has been given.

    But, I've assumed E > 3V.. So the current will pass according to the cell with emf 'E' only? Am I right?
     
  6. Dec 30, 2014 #5

    SammyS

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    No. You are not correct about that.

    Use Kirchhoff's Rules, as suggested by the title of your thread.

    Don't introduce anything extra into the circuit.
     
  7. Dec 30, 2014 #6

    ehild

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    Was the current I given? Or were you asked to give the potential difference in terms of I?
     
  8. Dec 31, 2014 #7
    No. I added it myself..
     
  9. Dec 31, 2014 #8
    What would have happened if there was a question with the same circuit diagram and a cell with emf (E) 12 V was present there?
     
  10. Dec 31, 2014 #9

    ehild

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    If it is no indication that current flows through the leftmost resistor, you have to assume it zero. There is current in the three branches only, and find the potential difference between points A and D.
     
  11. Dec 31, 2014 #10

    SammyS

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    The question in your Original Post asks the potential difference between A and D. If you introduce an ideal voltage source, E, as you have the answer becomes trivial. The potential difference between A and D is just that of the voltage source.

    It would help if you would state the problem with the wording as it was given to you.
     
  12. Jan 1, 2015 #11
    So, the mistake lies in that "assumption" itself. It messed everything up.
    Thanks a lot!
     
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