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Find the individual charge (and potential difference?) on each capacitor

  1. May 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Four capacitors are connected as shown in the figure below. (Let C = 18.0 µF.)
    26-p-023-alt.gif

    Find the equivalent capacitance between points a and b

    Calculate the charge on each capacitor, taking ΔVab = 17.0 V

    2. Relevant equations
    Q = CΔV

    C = Q / ΔV


    ΔV1 = ΔV2 = ΔVtot (for parallel capacitors)

    Q tot = Q1 + Q2 (charge on parallel capacitors, it says)


    ΔVtot = ΔV1 + ΔV2 (for series capacitors)

    Q = Q1 = Q2 (charge on series capacitors, it says)



    3. The attempt at a solution
    The only thing I could solve was the equivalent capacitance across, which is 6 x 10-6

    And I broke it up like this (and it's correct):
    http://desmond.imageshack.us/Himg213/scaled.php?server=213&filename=prob1m.jpg&res=landing [Broken]


    I have NO IDEA at all how to determine the individual charge on each of them, even with the formulas...

    I can see ΔV across series is supposed to add up to 17V but no idea how.

    Where should I start on that?


    Edit: ok used Q = CV for the (20 x 10-6 f) capacitor and got (102 x 10-6 C) because charge is supposed to be the same across series
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 5, 2012 #2
    What you have done so far is correct...now find the total capacitance of the 2 in series.... (You were able to find the equivalent of C and the 3uF in series !!!) then you should be able to get the total charge.
     
    Last edited: May 5, 2012
  4. May 5, 2012 #3
    ok i got them all

    you need to just mainly worry about using total charge Q = Q1 = Q2 for series capacitors and V tot = V1 = V2 for parallel capacitors, just to find the individual potentials / charges.

    Then you need to work your way down with these and the other two formulas.
     
  5. May 5, 2012 #4
    thats correct !!
     
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