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Capacitor circuit

  1. Nov 6, 2013 #1
    1. The problem statement, all variables and given/known data

    A 3nF Capacitor is parallel with a 4nF Capacitor.
    And they are connected with a 5nF Capacitor in series.
    IF the external voltage supply is 20V.
    FIND (a)the p.d across each capacitor,
    and (b)the total charge in the three Capacitor.

    3. The attempt at a solution

    (a)
    p.d across the 3nF,4nF capacitor=20x(5/(7+5))
    =8.33V
    p.d across the 5nF capacitor=20-8.33
    =11.67V

    (b)I try it in two method,but they have different answers.I don't know why!

    Method 1:
    equivalent Capacitor=2.917nF
    C=Q/V
    2.917nF=Q/20
    Q=58nC

    Method 2:
    charge in 3nF Capacitor=3n x8.33
    =25nC
    charge in 4nF Capacitor=4n x8.33
    =33.3nC
    charge in 5nF Capacitor=5n x11.67
    =58nC
    tatol charge:25+33.3+58
    =116.3nC
     
  2. jcsd
  3. Nov 6, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    The difference is that you're really comparing two different sums. Consider a set of identical capacitors in series. When a current is driven through the set, charge is placed on one plate of the first capacitor from the source, but the subsequent capacitors obtain their charges from the previous capacitor in line. So internal to the set of capacitors no new charges are "created" or "lost". Their net sum will always be zero.
    attachment.php?attachmentid=63698&stc=1&d=13837432335.gif

    So while the charges on each of the capacitors in series is the same, the net charge on the set is still only one times that value. The "outside world" sees only the charge placed on the first plate of the first capacitor, and removed from the bottom plate of the last capacitor. The Q = VCnet gives you that 1x value.

    In your problem you've essentially charged the series string of capacitors then disassembled them and summed the individual charges.
     

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