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Find the total resistance

  1. May 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the total resistance of this circuit:
    resisters.jpg

    each resistance in this circuit is 20ohms.
    (r1=r2=r3=r4=r5=r6=r7=r8=20ohms)

    2. Relevant equations
    Rt=r1+r2+r3... for series
    1/Rt=1/r1+1/2r+1/r3... for parallel


    3. The attempt at a solution

    since r4 and r5 is parallel to r6
    so r4 to r6
    = 1/(1/(20+20)+1/20)
    =13.333333333333
    r7 is in series with r4 to r6
    r4 to r7
    =13.333333333333+20
    =33.333333333333
    r3 is parallel with r4 to r7
    r3 to r7
    =1/(1/33.333333+1/20)
    =12.5
    r8 is in series with r3 to r7
    r3 to r8
    =12.5+20
    =32.5
    r2 is parallel with r3 to r8
    r2 to r8
    =1/(1/32.5+1/20)
    =12.38
    r1 is in series with r2 to r8
    rt
    =20+12.38
    =32.38

    yea i got it and i think it's right except my friend said it's wrong but i can't figure out where
     
  2. jcsd
  3. May 3, 2009 #2
    Replace r4 to r6 by a single resistor since you solved for its resistance. Then check if that single resistor is in series with r7. Keep redrawing the pic with the simplified resistors and work your way to the left.
     
  4. May 3, 2009 #3
    isn't that what i did on my solution?
     
  5. May 3, 2009 #4
    I will give you a starting point. The new resistor you found in your first step is not in series with r7. If you drew it properly in the schematic it would in fact be in parallel with r3 and r7.
     
  6. May 3, 2009 #5

    LowlyPion

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    Homework Helper

    Looks to me for the configuration shown like:

    (((((R4 + R5) || R6) + R7) || R3) + R8) || R2) + R1

    That comes out to 20 + 260/21 = 32.381 Ω
     
  7. May 3, 2009 #6
    Edit: Nevermind, I looked at it again and see how R7 can be in series with the new resistor. It is like an optical illusion :)

    So what you did looks correct.
     
  8. May 3, 2009 #7

    LowlyPion

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    Homework Helper

    Your equation looks to depend on R8 being connected to the far node, the junction of R5, R6, R7, and not the node between R3 and R7
     
  9. May 3, 2009 #8
    Yeah, I was wrong. I have always been bad with mazes as well.
     
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