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Deriving expressions for total capacitance

  1. Feb 13, 2014 #1
    Hello,

    my timetabled experiments for lab work are a week or two ahead of my E&M course which means i'm doing the experiments before being introduced to the theory.

    As part of my prep work I need to come up with an expression for the total capacitance of a number of different circuits involving a combination of parallel and in-series circuits.

    I would really appreciate it if someone could look at what I have done so far and offer some help with the last one that I am stuck on.

    Thanks!

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

    Derive expressions for the total capacitance expected in part 2.3.3

    (Circuit diagrams attached)

    2. Relevant equations

    Ctotal = C1+C2 + C3... ( for capacitors in parallel)

    1/Ctotal = 1/C1 + 1/C2 + 1/C3 ... (for capacitors in series)
    3. The attempt at a solution

    Circuit (i)

    Capacitor 1, 2 and 3 are in parallel with one another and therefore add to give Ctotal = C1+C2 + C3

    Circuit (ii)

    Capacitor 3 and 4 are in series with one another, so add to give 1/C3,4 = 1/C3 + 1/C4

    Capacitor 2 and 'capacitor 3,4' are in parallel so add to give C2,3,4 = (C3,4) + C2

    Capacitor 1 and Capacitor 2,3,4 are in series with one another so 1/C1,2,3,4 = 1/C1 + 1/C2,3,4

    Circuit (iii)

    Capacitors 1, 2 and 3 are all in parallel so add as C1,2,3 = C1 + C2 + C3

    Similarly Capacitors 4, 5 and 6 add to give C4,5,6 = C4 + C5 + C6

    These two grouping add to give 1/C1,2,3,4,5,6 = 1/(C1,2,3) + 1/(C4,5,6)

    Circuit (iv)

    I'm really struggling to figure out how to break apart the capacitors into groups and which ones to consider etc.

    I know it's asking a lot to pick through the above but i'd really appreciate some help here.

    Thanks :)
     

    Attached Files:

    Last edited: Feb 13, 2014
  2. jcsd
  3. Feb 13, 2014 #2

    BvU

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    You are doing just fine. In iv you make life easier by drawing A higher and B lower. Then you see 2,3,4 in series and 1 in parallel with that.

    Four in series caused a problem in an earlier thread today. I don't think you will stumble, though.
     
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