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Finding Potential and Charge across a Capacitor in a Multi-Loop Circuit

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data

    A 20.0 V battery is connected across capacitors of capacitances C1 = C6 = 2.9 μF and C3 = C5 = 2.00C2 = 2.00C4 = 4.7 μF.

    http://imgur.com/ssxTC

    What are
    (c) V1 and (d) q1 of capacitor 1,


    (e) V2 and (f) q2 of capacitor 2,


    (g) V3 and (h) q3 of capacitor 3,



    2. Relevant equations

    Ceq = c+c+c+c in parallel, 1/Ceq = 1/c+1/c+1/c in series

    C=q/V


    3. The attempt at a solution

    I solved A and B for this problem, namely, the equivalent capacitance, and the charge.

    C4 = 2.35, C2 = 2.35, C3 = 4.7, C5 = 4.7, C1 = 2.9, C6 = 2.9

    1/C5 + 1/C3 = 2.35 + C2 + C4 = 7.05

    C1 + C6 = 5.8

    1/7.05 + 1/5.8 = 3.18e-6, which is the correct value for the C equivalence

    Thus, 3.18e-6 = q/V(20) = q = 63.6e-6F

    After this, I formed a small loop with the battery, C4 and C6.

    Their C = 2.35 + 2.9 = 5.25e-6 = q/V(20), such that q = 1.05e-4

    V = q/C4 = 44.68, but I know this can't be right because the potential can't be greater than the original potential, 20

    I'm not certain where to go from here, in terms of finding the potential across or charge across these 3 resistors.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 23, 2011 #2
    The 1/7.05 is wrong and should be closer to 1/7.7 ? See,
     

    Attached Files:

    • C043.jpg
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  4. Sep 24, 2011 #3
    Yeah, I got that at first too, but if you notice from the 'trick question', it's 2.00C4 = 4.7, and with algebra is 2.35. Then 2.00C2 = 2.00C4 = 2.35 as well.

    I got the equivalent capacitance of the circuits correct, and the charge is also correct. It's just the potential over the resistors that I don't understand how to get.

    Thanks for replying though.
     
  5. Sep 24, 2011 #4
  6. Sep 24, 2011 #5
    I read it wrong, thanks for pointing that out. Will try again.
     
  7. Sep 24, 2011 #6
    I got your numbers, minus some rounding errors. See,

    The thing to remember is the charge on two capacitors in series is the same and the voltage on capacitors in parallel is the same.

    Good luck!
     

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