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Capacitors in Parallel and Series

  1. Oct 8, 2005 #1
    Hello. I have a brain fart and this combination circuit has me stumped. Here is a crude representation of the circuit. I had no problem finding the total capacitance (1.34uF), but the second part asks you to find the charge on each of the capacitors, assuming that the potential difference between b and a is 300 V.

    2uF 5uF
    a.---| |----- ----.b
    3uF ------||----------
    Ok, I know that the Q2=Q5 and Q3=Q7. The total charge is 402 uC, so Q3=Q7=402uF. Where do I go from here. I have tried all combinations...
  2. jcsd
  3. Oct 8, 2005 #2


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    Homework Helper

    Okay, I don't see that circuit of yours. Could you go maybe by circuit flow :P
  4. Oct 8, 2005 #3

    The circuit from point a goes to a capacitor of 3uF, then leads to 2 capacitors in series (2uF and 5uF) that are together parallel to a capacitor of 1uF. From the parallel circuit section you get to point b. Does that make sense? I tried to draw it out...but I guess it didn't work:) Thanks!
  5. Oct 8, 2005 #4


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    So 3uF+( (2uF + 5uF)//(1uF ) ?
    Your diagram just suggests otherwise. Anyway.
    To calculate the charge on a capacitor in steady state, you use:

    Q = CV

    The only problem then would be finding V across each capacitor.
    The voltage division rule applies to capacitors as V1 = C2/(C1+C2)
  6. May 8, 2009 #5
    1. With equivalent capacitance, first calculate charge with Q=CV.
    2. Then calculate voltage across 3uF using V=Q(as obtained above)/3uF
    3. Voltage across the parallel combination of 1uF with series combination of 2uF and 5uF will Vab (300v)-V(across 3uF, as obtained above, in step 2). This will be voltage across 1UF.
    4. Using voltage division formula calculate voltages across 2uF and 5uF; sum of voltages across 2uF and 5Uf will be same as voltage across 1uF
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