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Kirchoff's method for a weird circuit

  1. Apr 17, 2016 #1
    1. The problem statement, all variables and given/known data
    In the following circuit, the battery has emf ε = 12.7 V. The resistors are R1 = 2000 Ω . R2 = 3000 Ω, and R3 = 4000 Ω. What is the current through resistor R2 ?

    2. Relevant equations
    Kirchoff's Rules
    V=IR

    3. The attempt at a solution
    i0 (into A) - i2 (current into resistor 2) - i3 (current into resistor 3) = 0
    Then what happens at C and B?
     
  2. jcsd
  3. Apr 17, 2016 #2
    For some reason the image failed to upload. Here it is kirchoff.png
     
  4. Apr 17, 2016 #3

    berkeman

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    There may be simpler ways to solve this, but just go ahead and write the KCL equations and try it that way. Please show your work... :smile:
     
  5. Apr 17, 2016 #4

    SammyS

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    Here is a thread with a similar circuit.
     
  6. Apr 21, 2016 #5
    i0 = initial current
    i1 = current to R2
    i2 = current to R1
    i3 = current to R3
    i4 = current to lower R1
    i5 = current to lower R2

    i0 - i1 - i2 = 0
    i3 + i1 - i4 = 0
    i2 - i3 - i5 = 0
    -i2R1 - i3R3 + i1R2 = 0
    -i5R2 + i4R1 + i3R3 = 0
    12.7 - i1R2 - i4R1 = 0
     
    Last edited: Apr 21, 2016
  7. Apr 21, 2016 #6
    The currents are as follows from i0 to i5
    .0052
    .0024
    .0028
    .0004
    .0028
    .0024
    Since the ones through R1 and R2 are identical, it does seem like there is an easier way to solve this, but how?
     
  8. Apr 21, 2016 #7

    berkeman

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    Did you read through the thread that SammyS linked to in post #4? :smile:
     
  9. Apr 21, 2016 #8

    NascentOxygen

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    It is usually preferable to not involve more current variables than are needed, to minimise the amount of algebra it generates. Once you label 3 currents, the others are already defined in terms of those 3 so they can be marked on the schematic accordingly.
     
  10. Apr 21, 2016 #9
    So in the end, I only need 5 equations?

    12.7 - i1R2 - (i1 + i3)R1 = 0
    -i2R1 - i3R3 + i1R2 = 0
    i3R3 - (i1+i3)R2 + i5R1 = 0
    i0 - i1 - i2 = 0
    i2 - i3 - i5 = 0 or i1 + i3 - i2 = 0
     
    Last edited: Apr 21, 2016
  11. Apr 21, 2016 #10
    Yes, the currents through R2 are the same and so are the ones for R1 because reversing the polarities would give an identical result.

    So that means i4 = i2 and i1 = i5
    i0 - i1 - i2 = 0
    i2 - i3 - i1 = 0 or i1 + i3 - i2 = 0
    12.7 - i1R2 - i2R1 = 0
    -i2R1 - i3R3 + i1R2 = 0
    i3R3 - i1R2 + i2R1 = 0
    ?

    But I'm not all convinced about the reversing polarities proof. Who's to say that once the current passes through the bridge and all the other junctions, it's the same?
     
    Last edited: Apr 21, 2016
  12. Apr 21, 2016 #11

    gneill

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    Symmetry says so. If you can flip or rotate a circuit about one or more axes of symmetry and the result is identical to the original circuit, then the resulting circuit must behave identically to the original and the symmetric pairs of circuit elements must display the same behavior (current, potential difference).

    Take your original circuit and rotate it 180°. Doing so cannot change the circuit behavior since no connections have changed (the topology remains the same).
    upload_2016-4-21_18-15-14.png

    Note that the bridge circuit still has the same values of resistance in all the same locations. The potential supplied by the voltage source has the opposite polarity in this view (negative at top rather than positive at top), but the polarity that we assign + and - potentials and current directions are just a matter of convention. You can change from assuming positive charge carriers to negative charge carriers without affecting any equations. So the symmetry pairs of resistances must carry the same currents in both orientations.
     
  13. Apr 21, 2016 #12
    i0 - i1 - i2 = 0
    i3 + i1 - i4 = 0
    i2 - i3 - i5 = 0
    -i2R1 - i3R3 + i1R2 = 0
    -i5R2 + i4R1 + i3R3 = 0
    12.7 - i1R2 - i4R1 = 0
    Are you trying to say that the current into point A for the non-rotated circuit is the same as the current flowing into point D for the rotated circuit?
     
  14. Apr 21, 2016 #13

    gneill

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    The magnitude is certainly the same. Change the polarity of the voltage source and the signs of the currents will agree too.

    Besides, how could they be different, since the same current that leaves the battery must return to it?
     
  15. Apr 21, 2016 #14
    Right, charge is conserved. If all the resistors were different, would the current still be the same?
     
  16. Apr 21, 2016 #15

    gneill

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    Which current? Certainly the current entering (or leaving) node A via the battery is the same as the current leaving (or entering) node D via the battery.
     
  17. Apr 21, 2016 #16
    I see, so only currents through same resistors will have the same magnitude?
     
  18. Apr 21, 2016 #17

    gneill

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    Rephrase that as "The currents through the symmetrically paired resistors will be the same" and you would be right. Don't overspecify by saying "only". Some circuits may have resistors of different values that happen to have the same magnitude of current.
     
  19. Apr 21, 2016 #18
    The currents through the symmetrically paired resistors will be the same.
    What if the resistors have different resistances?
     
  20. Apr 21, 2016 #19

    SammyS

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    Will there still be the same symmetry?

    Precisely which resistances will you change?
     
  21. Apr 21, 2016 #20
    In the un-rotated figure: top left resistor is R1 top right is R2. Bottom left is R3, bottom right is R4. The bridge resistor is R5.
     
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