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Help on resistance

  1. Aug 19, 2004 #1
    hi guys,

    i need to calculate the substitution resistance between the points A and B of the chain in the attached picture. Each rectangle is a resistance R , except the resistance on the right-top of the figure. That has value 2R ; and the solution has to be (13/11)R. I know that we have to use Kirchoff's laws, yet I do not seem to be able to solve it, i Get (9/5)R. Maybe you can.

    Last edited: Feb 9, 2006
  2. jcsd
  3. Aug 19, 2004 #2


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    How are you sure it's (13/11) R? It looks like just R to me -- the middle resistance is inconsequential, since it's between two nodes that are always at the same potential.

    - Warren
  4. Aug 19, 2004 #3
    yes, you are right. But i made a little mistake. All resistances are R except the resistance on the right -top. It is 2R not R

    Solution is certainly (13/11) R

  5. Aug 20, 2004 #4
    Well, the only thing I was able to do is to apply the two Kirchoff's laws to the chain. In the first knod I have for the currents : I_0=I_1+I_2 Then the two knods in the middel (say I_3 is the current in the middle) = I_1=I_3+I_4 and below I_2+I_3=I_4

    Then the second law , I follow the two closed subchains clockwisely

    first chain : -I_1R-I_3R = I_2R
    second subchain : -I_4R+I_5R+I_3R = 0

  6. Aug 20, 2004 #5
    google it. I always hated those resistance exercise, they can be especially difficult. This one is not too difficult though. You could use Millman, Norton, Thevenin... whatever. Here is an elementary solution :
  7. Aug 20, 2004 #6
    [tex] R_{eq}[/tex]=[tex]\frac{v}{i}[/tex] ........(1)

    node 1 :
    i + [tex]\frac{v_{1}-v}{R}[/tex] + [tex]\frac{v_{2}-v}{R}[/tex] = 0

    i = [tex]\frac{-v_{1}-v_{2}+2R}{R}[/tex] ....... (2)

    node 2 :
    [tex]\frac{v-v_{1}}{R}[/tex] + [tex]\frac{0-v_{1}}{2R}[/tex] + [tex]\frac{v_{2}-v_{1}}{R}[/tex] = 0

    [tex]v_{2}[/tex] = [tex]\frac{-2v + 5v_{1}}{2}[/tex] ..... (3)

    node 3:

    [tex]\frac{v_{1}-v_{2}}{R}[/tex] + [tex]\frac{0-v_{2}}{R}[/tex] + [tex]\frac{v-v_{2}}{R}[/tex] = 0

    [tex]v_{1}[/tex] = -v + [tex]3v_{2}[/tex] ....... (4)

    Use (3) & (4) and substitute the answers into (1) get :

    [tex]v_{2}[/tex] = [tex]\frac{7v}{13}[/tex]
    [tex]v_{1}[/tex] =[tex]\frac{8v}{13}[/tex]
    [tex]R_{eq}[/tex] = [tex]\frac{13R}{11}[/tex]

    Node 1, 2 and 3 use Kirchoff's current law.
    Last edited: Aug 20, 2004
  8. Aug 20, 2004 #7
    thanks to all of you for helping me out...

  9. Aug 20, 2004 #8
    you're welcome Marlon :wink: I love electronics :yuck:
    :biggrin: :tongue2:
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