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Calculating equivelent resistance and equivelent capaictance

  1. May 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Below is a link to the problems, A and B.

    http://oi62.tinypic.com/28bgqao.jpg



    2. Relevant equations
    For series:
    R=∑R_i
    1/C=∑ 1/C_i

    For Parralell
    1/R=∑ 1/R_i
    C=∑ C_i



    3. The attempt at a solution
    I usually don't have any problems when it comes to calculating effective resistance, its only in capacitance, but as I always assumed as its the same principle but in reverse (i.e. the equation for capacitance in series is the same as for resistance in parallel and vice versa) I am not sure why so would greatly appreciate your input into what I am doing wrong.

    Part A:
    Top Branch:
    [itex]
    (\frac{1}{12}+\frac{1}{27})^{-1}+(\frac{1}{48}+\frac{1}{4})^{-1}+33=45Ω
    [/itex]

    Bottom Branch:
    [itex]
    18+54+(\frac{1}{11}+\frac{1}{16})^{-1}=78.52Ω
    [/itex]

    Total:
    [itex]
    (\frac{1}{45}+\frac{1}{78.52})^{-1}=28.61Ω
    [/itex]


    For Part B:
    Top Branch:
    [itex]
    2 + 20 + (\frac{1}{16}+\frac{1}{35})^{-1} = 32.98 \mu F
    [/itex]

    Bottom:
    [itex]
    37+11+43=91 \mu F
    [/itex]

    Total:
    [itex]
    32.98+91=123.98 \mu F
    [/itex]
     
    Last edited: May 12, 2014
  2. jcsd
  3. May 12, 2014 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Take another look at what is going on in this line.
     
  4. May 12, 2014 #3
    Well if I have done part A correctly, my thinking was..

    Looking at the bottom branch of part A alone, 18 and 54 were simply added because they're are in series, and then because 11 and 16 are in parallel I had to take the reciprocal of the reciprocals of 11 and 16 added together .

    So what I did then for the top branch of part B (the bit you quoted). As 2 and 20 are in parallel, simply add them, and because 16 and 35 are in series, I took the reciprocal of the reciprocals of 16 and 35 added together. The same as above but in reverse.
     
  5. May 12, 2014 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    Once you have added 2 and 20, you are faced with a string of 3 capacitances in series.
     
  6. May 12, 2014 #5

    Sorry I hope I am not being stupid/blind; just taking the top brance of part B, I can only see two, the 16 and 35 on either side of the 2 and 20 in parallel as indicated in the pic below.

    http://oi57.tinypic.com/34g4v1h.jpg
     
  7. May 12, 2014 #6

    NascentOxygen

    User Avatar

    Staff: Mentor

    What is the equivalent of 2 and 20 in parallel? Redraw the circuit with a single capacitor in place of those pair. What does it look like now?
     
  8. May 14, 2014 #7
    Ahhh right, I get it now, I think. Because once the the 20 and 2 are added together they become like as if it were just one there and then that would become the other one in series with the 16 and 35?

    So it would be like.


    Top Branch:
    [itex]
    (\frac{1}{16}+\frac{1}{2+20}+\frac{1}{35})^{-1} = 7.32 \mu F
    [/itex]

    Bottom:
    [itex]
    37+11+43=91 \mu F
    [/itex]

    Total:
    [itex]
    7.32+91=98.32 \mu F
    [/itex]
     
  9. May 14, 2014 #8

    NascentOxygen

    User Avatar

    Staff: Mentor

    Looking good.
     
  10. May 14, 2014 #9
    Thanks for your help. Do you know if my answer for the resistance one is correct?
     
  11. May 14, 2014 #10

    NascentOxygen

    User Avatar

    Staff: Mentor

    The method is right. I haven't checked your arithmetic.
     
  12. May 14, 2014 #11
    OK thanks for your help.
     
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