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Homework Help: Equvilent capacitance

  1. Jul 13, 2008 #1
    1. The problem statement, all variables and given/known data
    a potential difference V = 100 V is applied across a capacitor arrangement with capacitances C1 = 11.0 µF, C2 = 7.00 µF, and C3 = 3.00 µF. Find the following values Charge on each capacitor, potential difference on each capacitor, and stored energy in Joules.

    The circuit is set up with battery and then a mixed series and parallel. C1 and C2 are in a series on a branch and C3 is on a separate branch in parallel with both C1 and C2.

    2. Relevant equations
    In a series I find charge with (1/c1 + 1/c2) = q/v looking for q. I plugged in (1/11 + 1/7) = q/100. SOlved and got a charge of 23.4 microCoulombs each on C1 and C2.

    So now I am stuck I thought I solved the series first and then used that combine capacitance to find the charge on C3. Help

    This is the response I got:
    I would suggest finding the equivalent capacitance of all of them before applying voltage.

    As I read your problem you have C1 and C2 in series with each other but taken together in parallel with C3. With the value of the effective capacitance you can calculate total stored charge in the system. Then maybe work backwards splitting the charge between the two branches could give you some insight?

    How do I find equivalent capacitance for the system of circuits?
  2. jcsd
  3. Jul 13, 2008 #2


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    I don't believe that formula is correct.

    If I'm visualizing the circuit correctly, I don't think you need to find the total capacitance of all three resistors. As soon as you find two of the values (C,Q,V) for a capacitor you can use C=Q/V to find the other one. Do you see how that helps with C3?
  4. Jul 13, 2008 #3
    No-the only thing I can figure is that I needed to find the capacitance of 1 and 2 together because they are in a series. I thought I did this by C12=Q/V = 1/c1 + 1/c2. With this combine capacitance I can then find the charge.
  5. Jul 13, 2008 #4


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    But to find the combined capacitance in series, you add the reciprocals, and then take the reciprocal again.

    You have [itex]C_{12}= Q/V[/itex] part right, where [tex]V[/tex] is the total voltage across the combinations of [itex]C_1[/itex] and [itex]C_2[/itex], but the other equation needs to be:

    \frac{1}{C_{12}} = \frac{1}{C_1}+ \frac{1}{C_2}

    You then put the equations together and solve for the Q of C1 and C2.
  6. Jul 13, 2008 #5
    Ok- so I solved for 1/c1 + 1/c2 = 233766.2338, plugged that into Q/V = 1/(1/Ceq) = 4.277e-4 to get the charge, is this charge on both capacitors or is it split over both.
  7. Jul 13, 2008 #6


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    To answer that question, remember that the capacitors [itex]C_1[/itex] and [itex]C_2[/itex] you're talking about are in series. Does that mean that the charge or the voltage are the same for [itex]C_1[/itex], [itex]C_2[/itex], and [itex]C_{\rm eq}[/itex]?
  8. Jul 13, 2008 #7
    The charge is the same - the voltage across wach capacitor will differ
  9. Jul 13, 2008 #8


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    That sounds right to me. So you have found the charge on [itex]C_1[/itex] and [itex]C_2[/itex], which are both equal to the charge you found on [itex]C_{\rm eq}[/itex].

    What about [itex]C_3[/itex]? What did you get?
  10. Jul 13, 2008 #9
    Okay - lets go back to the begining, because I am messing up somewhere:

    C1 = 11microF
    C2 = 7microF
    C3 = 3microF

    Voltage across arrangement = 100V

    --------------------- C123 = C12 + C3 C12 = 1/(1/C1+1/C2)
    | | | C12 = 1/.233766microF
    | C1 - | C12 = 4.2778microF
    V | C3 -
    | | | C123 = 4.2778microF + 3microF
    | C2 - | C123 = 7.2778microF
    | | |

    I need to find charge on each capacitor, potential difference across each capacitor and stored energy.

    I just figured out I don't need C123....since c12 and c3 are in parallel they experience the same voltage.

    q12 = C12V = 428microC
    q3 = C3V = 300microC

    I think I got it now - I think I canfind the potential across each capacitor and the energy?
    I'll reply if I can't Thanks alph
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