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Finding resistance

  1. Mar 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A number of resistors of values R1 = 5.0 Ω, R2 = 98.0 Ω, and R3 = 16.0 Ω are connected as shown in the figure. What is the resistance between points A and B?


    2. Relevant equations



    3. The attempt at a solution

    Ok so the 4 R_1's and R_3 are in series on the end which gives a resistance of 36.0 ohm. This resistance is in parallel with R_2 so it's 1/ (1/first resistance + 1/R2 ) =26.3ohm. The 4 R1's and R2 are in series so its R1(4)+R2 =118. Now this resistance is in parallel with the first R2 so its 1/ (1/118 +1/98)= 53.5ohm + the 2 R1's at the start = 63.5

    So 63.5 ohm + 26.3 ohm = 89.8 ohm but this isnt right. Can someone help me out please.
     

    Attached Files:

  2. jcsd
  3. Mar 29, 2008 #2

    Doc Al

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    Staff: Mentor

    Redo that last step. The 4 R1s are in series with the effective resistance calculated in the previous step, not with R2 alone.
     
  4. Mar 29, 2008 #3
    so R1(4) +R2 + 26.3 ohm= 144.3 ohm, so then do I just follow the steps I have already used?
     
  5. Mar 29, 2008 #4

    Doc Al

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    No, you're still making an error. As you work from right to left, you replace resistors with their equivalent resistance.

    (1) You replaced 4R1 + R3 with their equivalent = 36 ohms
    (2) Now R2 & 36 ohms are in parallel, replace them with their equivalent = 26.3 ohms
    (3) Now you have 4R1 in series with 26.3 ohms (not R2; R2 is gone!)
    ... and so on
     
  6. Mar 29, 2008 #5
    Oh ok ,I understand it good now, thank- you very much!
    26.3ohm+4R1= 46.3
    Parallel with R2 so 1/ (1/98+1/46.3)= 31.44ohm + 2R1 on the very left = 41.4 ohm :)
     
  7. Mar 29, 2008 #6

    Doc Al

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    You got it. :wink:
     
  8. Mar 29, 2008 #7
    Ok so theres another part of the quesiton I'm having trouble with I worked through it but its the first time Ive seen anything like it.

    A-C was I1, C-D was I3 D-B was I1

    I1-I3=0
    I3-I1=0
    I1-I3=I3-I1
    2 I1-2 I3=0
    I3=I1
    E-I3R-I1R=0
    E=I1R-I1R=0
    E-2I1R=0
    E= (2I1R)/(2R) I1=E/2R= 110.0V/2(98.0ohm) = 0.561A
     

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  9. Mar 29, 2008 #8

    Doc Al

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    At junction C (or any other junction) it's the total current flow that must add to zero. You counted the current from the A-C (I1) and D-C (I3), but what about the branch going to the right out of C?

    What are you given for this part of the problem? The total voltage across A-B? If so, do the opposite of what you did before. This time work backwards from left to right, using the results of your previous work to find the various currents.
     
  10. Mar 29, 2008 #9
    So are my working relevant at all? Or do I need to scrap them completely and start over?
     
  11. Mar 29, 2008 #10

    Doc Al

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    Start over. Your first equation, I1-I3=0, is incorrect.
     
  12. Mar 29, 2008 #11
    kk Ive got work so I cant solve now, Ill be back later, you may not be here but thankyou very much for the help today!
     
  13. Mar 30, 2008 #12
    So will my first equation have to consist of I1,I2,I3,I4,I5,I6 because of the complexity of the circuit?
     
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