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Homework Help: Simple C-R circuit Problem

  1. Jun 2, 2012 #1
    Hi friends,
    I have a big issue in a simple C-R circuit. Thank you in advance if anyone would help me in this case.
    https://fbcdn-sphotos-a.akamaihd.net/hphotos-ak-ash4/301868_2238324694917_1752023214_1085083_1481697696_n.jpg [Broken]

    In this problem we have to find out after closing the switch 's' how much charge will flow from Y to X.

    Well I have got the answer also but I have confusion in that only.
    The doubt is as follows.

    Initially I found the potential difference across 3μF, 6μF capacitors and the 3Ω
    and 6Ω resistor. The readings are as follows respectively - 6 V, 3 V, 3 V and 6 V. According to these the charge on each capacitor is 18μC cause both are in series. But after closing the switch 6μF and 6Ω comes in parallel. The P.D. across and 6Ω resistor is 6 V and hence the P.D across 6μF becomes 6 V. And hence the P.D. across 3μF capacitor becomes 3 V. Now according to the relation "Q = CV" the new charges on 6μF and 3μF are 36μC & 9μC. Hence the amount of charge from Y to X is (36μC - 9μC = 27μC).

    Friends my confusion is after being in parallel with 6Ω resistor why P.D. across 6μF becomes 6 V why not some other value. Someone said it is due to 6Ω is at higher potential. If that so why this theory is not applied with 3μF and 3Ω resistor, here 3μF is at higher potential 6 V.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 6, 2017
  2. jcsd
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