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Circuits - Calculating Effectice Resistance

  1. Feb 25, 2014 #1
    Sorry for my crude drawings but I thought it was easiest and quickest this way to post the question.

    1. The problem statement, all variables and given/known data

    Calculate the effective resistance for the circuits below.

    A. circuit_A.jpg

    B: circuit_B.jpg


    2. Relevant equations
    [tex] R_{series}=\sum\limits_{i=1}^n R_i \\
    \frac{1}{R_{parallel}}=\sum\limits_{i=1}^n \frac{1}{R_i}
    [/tex]


    3. The attempt at a solution

    We have never been told how to calculate the effective resistance where both series and parallel occur, only how to do it if it is series and how to do it if it is parallel. But using intuition for part A i just did the bit in series and then the bit in parallel and added them like below.

    [tex]
    \sum\limits_{i=1}^n R_i = 2+2=4 \\
    \sum\limits_{i=1}^n \frac{1}{R_i} = \frac{1}{2}+\frac{1}{2} = 1 ∴ R=1 \\
    ∴R_{total}=4+1=5Ω
    [/tex]

    For part B I am scratching my head as to where to start because the whole thing is in parallel and then there are parrellel bits inside of that iyswim. Any hints on how to approach part B would be much appreciated, thanks.
     
    Last edited: Feb 25, 2014
  2. jcsd
  3. Feb 25, 2014 #2
    Forget about the rest of the circuit .First consider only 12,4,16 Ohm resistors in the top branch .What do you think ? Are they in series or parallel ? Can you calculate their effective resistance ?
     
    Last edited: Feb 25, 2014
  4. Feb 25, 2014 #3
    Right ok, I think I see what your getting at. So as the 12,4 and 16 are in parrallel I can calculate the effective resistance of just that, ignoring the rest (for now), and replace the 12,4,16 branches with the effective resistance. Ok I will have a go now.
     
  5. Feb 25, 2014 #4
    Yes...that is correct .
     
  6. Feb 25, 2014 #5
    OK so for part B then I broke it down to the individual branches to make it easier on the eye.

    Top Branch:
    [tex]
    3+9+(\frac{1}{12}+\frac{1}{4}+\frac{1}{16})^{-1}=14.53Ω[/tex]

    Middle Branch[tex]
    20+10=30Ω[/tex]

    Bottom Branch[tex]
    5+(\frac{1}{8}+\frac{1}{33})^{-1}=6.44Ω[/tex]

    Putting it all together[tex]
    (\frac{1}{14.53}+\frac{1}{30}+\frac{1}{6.44})^{-1}=3.88Ω
    [/tex]
     
  7. Feb 25, 2014 #6
    Nice work :thumbs:
     
  8. Feb 25, 2014 #7
    Thanks for your help :)
     
  9. Feb 25, 2014 #8
    You are welcome :smile:
     
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