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Charge and Voltage on Capacitors.

  1. Mar 1, 2015 #1

    Yut

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    1. The problem statement, all variables and given/known data
    Determine the charge on each capacitor and the voltage across ech, assuming C=3yF(micro) and the battery voltage is V=4.0 Vhttps://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11006374_10152682348496806_273021698509210726_n.jpg?oh=128f2d03e7e0ad275cb982e11d827c2c&oe=54F5E264&__gda__=1425403951_252b026b8105e21db6642c5d28a9f753

    2. Relevant equations
    so,
    C23 = 2C/3 = 2microF
    Qeq = Q1 + Q23 = Ceq V <--- why is Qeq is the sum of the series capacitors, i thought they had the same charge
    Q eq = 8. micorC
    Q1 = Q23 = 4.00 microC
    V1= Q1/C = 1.33 V
    V23 = V - V1 = 2.67 V
    Q2 = Q3 = CV23 = 2.0 micro C <--- not sure how did they get this value, when I plug in the found valued I dont get 2.

    3. The attempt at a solution
     
  2. jcsd
  3. Mar 1, 2015 #2

    gneill

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    Staff: Mentor

    The problem statement is unclear to me. You state that C = 3 μF, does that mean the circuit is fundamentally made up of a several 3 μF capacitors? If so, how do you arrive at C23 = 2 μF?

    Capacitors in series, unless there are circuit changes when capacitors are already charged, should bear the same charge as you indicate.

    Is the solution that you present as Relevant equations your solution or someone else's?
     
  4. Mar 1, 2015 #3

    Yut

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    This is an example problem from one of my lectures, and I was not sure how they arrive to the solution.
    Originally the C23 capacitor, was C2 and C3 capacitors in parrallel. They combined them in to Ceq by additing them two.
     
  5. Mar 1, 2015 #4

    gneill

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    Staff: Mentor

    Can you list all the initial capacitor values and the initial circuit conditions?
     
  6. Mar 1, 2015 #5

    Yut

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    https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11042003_10152682449531806_1029796180_n.jpg?oh=fcaa90dacb0f79caf862a01ccfe11094&oe=54F5C0C6&__gda__=1425387735_717d66c414ad18a91e2dbebad58aa77e
     
  7. Mar 1, 2015 #6

    gneill

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    Staff: Mentor

    Okay, it's much more clear now that I can see the whole circuit. Their derivation of the equivalent capacitance looks fine.

    In your first post you showed and questioned the step:

    Qeq = Q1 + Q23 = Ceq V

    which I agree is incorrect. Series connected capacitors should have the same charge. So what would be your version of a solution?
     
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