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Series-Parallel Circuit problem, afraid I'm tripping over my work

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data
    physicshelp.jpg


    I want to solve for the Req, the total I, and total Power first. Then, I want to solve for the V, I, and Power for ALL of the resisters.

    2. Relevant equations
    I'm using V = IxR and P = IV for this. Also the formulas for finding Req for series (R1+R2) and parallel(1/R1 + 1/R2).



    3. The attempt at a solution
    I have solved just about everything except for the 60 ohm resister. I am not sure if my numbers are adding up correctly and if possible I would like some help. By help I mean could someone try solving this as well so I may check my answers and hopefully help me understand this stuff more cleary.

    Thank you.
     
  2. jcsd
  3. May 27, 2013 #2

    CWatters

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    You need to show us some working. What did you get for the total resistance?
     
  4. May 27, 2013 #3
    Total resistance was .1217, Req was 41.09, total power is .61

    For the 10ohm:
    V = 1.22V
    I = .1217A
    P = .15W

    For 20ohm:
    V = 1.51V
    I = .0756A
    P = .11W

    For 30ohm:
    V = 2.27V
    I = .0756A
    P = .17W

    I attempted the 50 and 40ohm resisters and im almost certain they're wrong. If someone could show me what they have for all of this and/or check my current answers, that'd be great.

    The 40,50, and 60 ohm ones are giving me trouble.
     
  5. May 27, 2013 #4

    CWatters

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    Agreed.
    I get Req = 31.09 + 10 = 41.09 Ohms
    Power = V2/R = 0.61 W

    Agreed.

    Then what I would do is calculate the voltage drop across the top group of resistors. Can be done several ways but using the potential divider rule...

    V = (5 * 31.09)/(41.09) = 3.783V

    Then apply the potential divider rule four times to that voltage to calculate the voltages on the following "four" resistors..

    30Ohm
    20Ohm
    60Ohm
    and the 50//40 = 22.222.Ohm

    Knowing the voltage on all the resistors you can calculate I and P for each including the 40 and 50Ohm individually.

    For example the voltage drop across the 60 Ohm is...

    = (3.783V * 60) / (60 + 50//40)
    = (3.783V * 60) / (60 + 22.222)
    = 2.83V
     
  6. May 27, 2013 #5
    This is very helpful! We haven't learned the divider rule yet, but I'm sure we will. Thank you!
     
  7. May 28, 2013 #6

    CWatters

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    Ah ok.

    You can also do it by using Ohms law. for example the voltage drop across the top set is..

    V = I * R
    = .1217A * 22.222.

    The voltage divider rule typically allows you to avoid working ou the current explicitly. There is a primer here although perhaps skip straight to the section on the Resistive divider...

    If you have ever covered "ratios" in other subjects it will be quite easy.

    http://en.wikipedia.org/wiki/Voltage_divider

    There is also a current divider rule but perhaps get to grips with the voltage divider first.

    http://en.wikipedia.org/wiki/Current_divider
     
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