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Calculating equivalent capacitance, a question

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    Capacitance for series and parallel connected capacitors; C=Q/delta_V, derivations of last formula

    3. The attempt at a solution
    Part A

    C1 and C2 appear to be connected in series. Therefore, use appropriate formulas.

    (1/Cs)=(1/3C)+(1/C) = 3C/4 (answer I got is "flipped" here due to the way the formula is stated)

    Alright, so we have the equivalent capacitance for C1 and C2. Now move on to C3.

    (C1+C2) + C3 appear to be parallel-connected. So,

    Cp=(3C/4) + 5C = 23C/4

    I also tried 23/4C in case it couldn't read my entry. A preview confirms it is what I intend, and yet it is wrong. Tried different combos, all wrong. I'm certain I did this correct... what did I do wrong here?

    Part B
    I'm not sure how to even solve this part. I actually guessed on this. At first I thought higher capacitance = more charge storage, but that was wrong.

    Anyway, Capacitance = Q/delta_v. So they want charge, so I derive Q=C(delta_V).

    delta_V is not provided, so I don't understand what they want. I'm not even sure how I'm supposed to input it... obviously, integer values are what they want. I am clueless on this problem :(

    Part C
    I understand V2=V3 because they are connected in parallel. Why is V1 bigger than the other two, though?

    As far as I can derive, delta_v=Q/C. So, shouldn't a bigger C mean a smaller delta_V? Am I supposed to use the answers I supposedly got in Part B?

    Part D

    I don't understand. C3 is the last to be charged (am I thinking of this the wrong way?), and so I am thinking it shouldn't have be affecting the other caps at all. Plus, shouldn't the charges on C1 and C2 be equal, because they are connected in series?

    EDIT: I thought about it a little more. C2 and C3, connected in parallel, so therefore they "share" a charge. Because of this, If C3 increases, C2 has no choice but to decrease in charge. Since one could argue that C1 is connected to C3 via a series connection, then that would mean charge is the same - thus meaning that if C3's Q increases, C1's charge MUST increase as well. Is that right?


    I would really appreciate any assistance! thanks :smile: This is also my first post here.
    Last edited: Feb 4, 2007
  2. jcsd
  3. Feb 5, 2007 #2
    Before I begin, let me say that I'm studying for this exam so my answers may be completely wrong...but here's what I think:

    Well for part 1, you're on the right track. I think you need to figure out the equivalent capacitance for C2 and C3 before you figure out the equivalent capacitance between C1 and C2...

    I'm not sure about part 2, but for part 3, if you follow the circuit, from + to -, there is a potential drop across the first capacitor so thats why V1>V2 which equals V3

    Not sure about part4...havent gotten a chance to review that part of the material yet.
  4. Feb 5, 2007 #3
    Part A:
    This is a wrong assumption! C1 is in series with the effective capacitance due to the parallel combination of C3 and C2.

    C' due to C2 and C3 in parallel = C2 + C3 = C + 5C = 6C

    This C' is in series with C1. Therefore C1 and C' are in series

    C eff = C1*C'/(C1+C') = 3C*6C/(3C+6C)= 18C^2/9C = 2C

    Part B:
    In series combination, the charge on each capacitor will be same (This follows from conservation of charge). Here C1 and C' are in series. The charge given to C' is actually distributed among C2 and C3 in the ratio C2:C3 (because V2=V3 and Q=CV. Therefore Q2:Q3 = C2:C3).

    Suppose a charge Q flows through the circuit when it closed. Q1 = Q'= Q.
    Q2:Q3 = 1:5. Thus Q3>Q2.

    Conclusively Q1>Q3>Q2

    Part C:
    Since C2 and C3 are in parallel the potential difference across them will be equal. This V2=V3. This narrows us to option 1 and 3. From Part A, we can see that C1:C' = 1 : 2

    In series combination, the charge on each capacitor will be same (This follows from conservation of charge).

    We know that V= Q/C

    Hence V1:V' = (1/C1):(1/C') = C':C1 = 2:1

    This V1=2V'
    and V'=V2=V3

    This answer is V1>V2=V3.

    Part D:
    Yes u're in the wrong direction. Time's not a factor here coz we're not asked about the transient currents.

    If C3 is increased Q2:Q3 = C2:(C3+x). Thus still more of the charge given to C' as a whole goes to C3.

    Meanwhile.... C'= C2+C3+x
    and thus C eff also has to increase. Therefore more charge get accumulated on C1 and C'. This Q1 and Q3 increases while Q2 decreases.

    Note: Q2 need not decrease fro its original value (I hope this doesn't confuse you!) . It depends upon the increase in C3! If it don't click that well please feel free to ask.
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