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Proving Kirchoff's Laws algebraically

  1. Feb 1, 2012 #1
    1. The problem statement, all variables and given/known data
    From the attached file using Kirchoff's laws prove that V AB = V AD and that i1 = i/2

    2. Relevant equations
    V1 + V2 + V3 +... Vn = 0 (The sum of the voltage applied and dropped across the components is zero.
    i1 + i2 = i3 (The sum of the branch currents is equal to the total current)

    3. The attempt at a solution
    I thought I was ok with algebraic manipulation but this has stumped me :-( any pointers would be appreciated as always!
     

    Attached Files:

  2. jcsd
  3. Feb 1, 2012 #2

    vela

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    What did you get when you tried to apply Kirchoff's laws to the circuit?
     
  4. Feb 1, 2012 #3
    Hi, as there are no figures given it looks like I'm expected to do prove everything using algebra...

    I've calculated hypothetic values - say V = 12v, R total = 16Ω and i total = 0.75A and everything seems to add up but I'm afraid my algebra manipulation skills are somewhat limited.

    For example I've tried that if: V ab = V ad and i1 = i - i2 and i2 = i -i1 (Kirchoffs 2nd Law) then i1 R0 = i2 R0 but don't know where to go now
     
  5. Feb 1, 2012 #4

    vela

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    You can't assume VAB=VAD because that's what you're trying to prove.

    Applying KCL to node A gives you ##i = i_1 + i_2 ##. I'm assuming all currents are flowing left to right. What do you get if you apply it to nodes B, C, and D?
     
  6. Feb 1, 2012 #5
    B: i3 = i1
    C: i = i3 + i4
    D: i4 = i2
     
  7. Feb 1, 2012 #6

    vela

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    Great! So we can eliminate i3 and i4 because we know they're equal to i1 and i2 respectively, and the equation for node C tells us the same thing as the equation for node A.

    Now apply KVL to the loop ABCDA. What do you get?
     
  8. Feb 1, 2012 #7
    VT - i1R0 - i1R0 - i2R0 - i2R0 = 0
    VT = VAB + VBC + VDC + VAD
    VT = R0 (i1+i1+i2+i2)
    VT = R0 (2i1 + 2i2)

    But I don't understand how V AB can be described as V AD

    or even

    VT = VAB + VBC + VDC + VAD
     
    Last edited: Feb 1, 2012
  9. Feb 1, 2012 #8

    vela

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    This isn't quite right. Across the four resistors, Ohm's Law tells us the potential differences are

    VAB=i1R0
    VBC=i1R0
    VAD=i2R0
    VDC=i2R0

    where the higher potential is at the left end of each resistor (where the current enters). So starting at A and going clockwise around the loop, we first encounter a voltage drop VAB and then another drop VBC, but going from C to D, the potential increases by VDC because we are going from right to left, and again from D to A by VAD. So KVL gives us

    - VAB - VBC + VDC + VAD = 0

    Note that VT doesn't enter into the picture because the battery is not part of the loop ABCDA. So plugging in the values above, you get

    -i1R0-i1R0+i2R0-i2R0=0
     
  10. Feb 1, 2012 #9
    Oh, I see! So the direction of flow has a direct effect on the sign of the variable! Cool, there are a few more questions in this assignment but I think I understand.

    Thank you for all your help!
     
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