1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Capacitor problem

  1. Jul 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Initially, a 120V emf is connected to C1 by a switch. When the switch is flipped C1 is connected to C2 and C3 in series. What are the charge and potential difference across each capacitor?
    C1 = 15 uF
    C2 = 16 uF
    C3 = 30 uF


    2. Relevant equations

    C= q/V
    1/Ceq = 1/C1 + 1/C2 + ... + 1/Cn


    3. The attempt at a solution

    Initially with C1 is only connected to the 120V source it will charge to 120V. Once the switch is flipped C1 will become the emf source. C1 will discharge while C2 and C3 will charge.
    The initial charge on C1 I found to be Qi=CV=(15uF)(120V)= 1800 uC
    Since C1, C2 and C3 are in series, the charge on the capacitors will be the same (Q1=Q2=Q3). I assume that Qi = Q1 + Q2 + Q3, as in the charge will distribute evenly within the new circuit. So Q1=600 uC. So the charge on each capacitor is Q=600 uC. But this isn't right.
    The potential difference across each capacitor would then be V=Q/C.
    So V1=Q/C1, V2 = Q/C2, V3=Q/C3.
     
  2. jcsd
  3. Jul 30, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi tracie,

    I don't believe this is right. C2 and C3 are in series, but C1 is not in series with them. So C2 and C3 have to have the same charge, but C1 does not (necessarily) have the same charge as the others. So how are the capacitors related to each other?
     
  4. Jul 31, 2008 #3
    I know that capacitors in series q=q1+q2, V=V1+V2, and C=q/V.
    After the switch is flipped, can I strictly treat C1 as a emf source? I guess I don't understand how to relate the simple series capacitor problem to the charge on C1 as well as the potential difference on C1. I assumed that (since the problem didn't state otherwise) that C1 was connected to the 120V source until fully charged (charged to 120V). I thought after the switch is flipped that the potential difference on C1 would still be 120V, but that isn't right.
    A little help in the right direction would be appreciated. Thanks!
     
  5. Jul 31, 2008 #4

    alphysicist

    User Avatar
    Homework Helper

    In series, q=q1=q2
    I would go ahead and combine C2 and C3 into an equivalent capacitance C23. (You can undo them later.) Now think about how C1 and C23 are related. The important questions: how is the voltage of C1 and C23 related? How are the final charges on C1 and C23 related? The charges on C1 and C23 will not be equal, since they are not in series. But you've already found the initial charge on C1--how does that relate to the final charges?

    Once you answer those two questions, you can write them down as two equations. At that point you should be able to use the capacitor equation C=Q/V to help solve for C1 and C23.

    (A few hours from now I'll be going out of town for several days and won't be able to respond--so if you still have questions I'm not ignoring you.)
     
  6. Jul 31, 2008 #5
    Thank you for correcting my typo.

    I guess I can relate C23 to C1 by
    initial charge on C1= q1 + q23 (since charge is conserved)
    Since I calculated initial charge on C1 and I can calculate q23 from q23=C23*V, then I can get q1. Since q2=q3, q23/2 = q2 = q3.
    I then can say V2= q2/C2 and V3=q3/C3.
    From V=V1 + V2 + V3, I have V=120V and I just found V2 and V3, so I can find V1.

    Is that the right direction?
     
  7. Jul 31, 2008 #6

    alphysicist

    User Avatar
    Homework Helper

    Right, this is the equation you want to work with. (initial charge on C1) is just a number you know, and q1 and q23 are unknowns.

    That's the right kind of idea; but V is an unknown. (It drops from 120 as the charges spread out.) But you can use the fact that V is the same for both C1 and C23. For example, you can use the relationship you're talking about to get rid of both q1 and q23 in your equation, and then you'll have one equation with one unknown.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Capacitor problem
  1. Capacitor problem (Replies: 0)

  2. Capacitor Problem (Replies: 1)

  3. Capacitor Problem (Replies: 1)

  4. Capacitor Problem! (Replies: 4)

  5. Capacitor problem (Replies: 10)

Loading...