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DC circuit analysis using Reciprocity Theorem

  1. Jul 12, 2015 #1
    1. The problem statement, all variables and given/known data
    In the following circuit I need to find current I1 if the given parameters are E1=0V,E2=15V,E3=60V
    In the second case, given parameters are I2=100mA,I3=40mA,E1=30V,E2=E3=0V
    In both cases, R2=R3=R4=R5=300 ohm
    2. Relevant equations
    Using reciprocity theorem for DC circuits.

    3. The attempt at a solution
    I used reciprocity theorem on the second case, moving the source E1 between the nodes where I3 flows. From there I1=-I3=-40mA. Is this correct? This is the second case. Could someone help how to find current I1 in the first case?

    Thanks for replies.
     

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  2. jcsd
  3. Jul 13, 2015 #2

    Hesch

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    Draw 4 loop currents in the diagram: L1 . . L4.
    Make 4 equations, using Kirchhoffs voltage law ( KVL ).
    Solve the 4 equation, and you have found I1.
     
  4. Jul 13, 2015 #3
    Resistors R1 and R6 are not given. Is it possible to find them, and then the current?
     
  5. Jul 13, 2015 #4

    Hesch

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    Something is wrong here.
    In the first case you know: E1=0V,E2=15V,E3=60V and R2=R3=R4=R5=300 ohm. That's all.

    You could find a possible solution by:

    Assume R1 = infinit → R1 and E1 can be removed. It doesn't change anything.
    Assume R6 = infinit → R6 can be removed.
    Now R3 ends blind and can be removed.

    Find I2 and I3. I1 = 0.

    That's a possible ( and easy ) solution.

    Or do you have to find I1 as a function of R1 and R6? : I1(R1,R6) ?
     
    Last edited: Jul 13, 2015
  6. Jul 13, 2015 #5

    NascentOxygen

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    In case (1) you are given the values of three voltage sources and asked to find the current through a resistor of unknown value? With the information provided, I'm sure that can't be done.

    In case (2) where there is only one voltage source, then the current from that source cannot be less than 100mA. You don't need any special theorem to show this; if you have a resistance network and a single voltage source, you can't have greater current through any resistor than is being supplied by the source.

    It would have been far better had you presented just a single problem in this thread; mixing two like you have is thoroughly confusing. :mad:
     
  7. Jul 13, 2015 #6

    SammyS

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    I'm pretty sure that the second case can be solved, the value of E1 not being needed.
     
  8. Jul 14, 2015 #7

    Hesch

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    ? ?
    E1 = 30V in the second case.
     
  9. Jul 14, 2015 #8

    SammyS

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    I was able to solve for i1 without using the fact that E1 = 30V .
     
  10. Jul 14, 2015 #9

    Hesch

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    Show me.
     
  11. Jul 14, 2015 #10

    SammyS

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    @gruba

    I didn't use reciprocity. I got a different answer for the second problem.

    (I should have mentioned this earlier.)
     
  12. Jul 14, 2015 #11

    SammyS

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    @Hesch

    Since OP has not solved this, I won't give the full solution.

    The current though R2 is ##\ i_1+i_2\ ## downward.

    The current though R4 is ##\ i_1+i_2+i_3\ ## upward.

    Furthermore, E2 = 0 so R2 and R4 are in parallel .
     
  13. Jul 14, 2015 #12

    Hesch

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    Ok, thank you.
    I'll have a look at it ( maybe using my 4 equations in #2 ).
     
  14. Jul 14, 2015 #13
    How do you know that current through R3 flows upward?
    If you use Kirchhoff's law for nodes, current through R4 (I3 - current though R2), (again how to determine direction).
    How did you get I1+I2+I3
    These currents are not in relation by any node.
    I think R2 and R4 not in parallel.

    Could you show your solution, I can't figure it out.
     
  15. Jul 14, 2015 #14

    Hesch

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    I just wondered if some currents would be altered due to altering E1, but they won't. Only the value of R1 will be changed, everything else is constant.

    And well, I had to use 5 equations to solve the problem, ( are you reading this gruba? 5 equations! ). But then I also know the values of R1 and R6 :smile:.
    Hint: R6 = 2 * R1 at E1 = 30V.

    If the value of E1 becomes too low, the value of R1 becomes negative. I don't know if that makes sense?
     
  16. Jul 14, 2015 #15
    Can you show how you have done it?
     
  17. Jul 14, 2015 #16

    SammyS

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    image-png.85867.png
    I stated that current through R4 is upward. (That's using the direction of the currents as given in the figure.)

    This is for the second problem so that I2=100mA, I3=40mA, E1=30V, E2=E3=0V .

    E2 = 0 , therefore, R2 and R4 have the same voltage drop across them and they are in parallel.

    Furthermore, E3 = 0 which leads to R4 and R5 also being in parallel. That's not needed for finding ##\ i_1\ ##.

    The junction law tells you that the current through R4 is upward.

    Let me state again, that these directions are for the assumed current directions given in the figure. It may well be that the value found for ##\ i_1\ ## is a negative value of sufficient size so that the resulting current through R4 is downward.

    As a further note: I am not using reciprocity for this solution, so I have not switch location of sources as you did.
     
  18. Jul 14, 2015 #17

    SammyS

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    That's not how we operate here at PF .

    We try to help you solve problems, but it's you that solves.
     
  19. Jul 14, 2015 #18

    Hesch

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    I've attached the diagram with 4 loops drawn. From the 4 loops you can setup 4 equations, but the loop-current L3 is known ( 0.04A ), so you need only equations as for L1, L2, L4. Then you need two equations:

    L3 = 0.04A
    L2 - L1 = 0.1A

    Now, the voltage drop across R6 = L4 * R6 and across R1 = L1 * R1. Just call these voltage drops V6 and V1. Then you have 5 equations with 5 unknown:
    L1, L2, L4, V1, V6.

    Example, L1:

    -( 300Ω +R1 )L1 - 300Ω*L4 - V1 = 30V

    Setup the last 4 equations and solve them.

    To find the current through R3, you just sketch an arrow for example upward. Now the current will be L4+L1. Maybe the result is negative, but it's because the current actually goes downward ( positive ).
     

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  20. Jul 14, 2015 #19

    Hesch

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    Correction:
    Should have been:

    - 300Ω * L1 - 300Ω * L4 - V1 = 30V
     
  21. Jul 31, 2015 #20

    NascentOxygen

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    It seems that no one has been able to see how to analyze by applying the Reciprocity Theorem, specifically, to this problem.

    gruba, has your course provided you with the Reciprocity Theorem solution?
     
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