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Stored Energy Difference Between Capacitors

  1. Feb 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Three capacitors of C1 = 6.7 micro-farads, C2 = 19.4 microfarads, and C3 = 9.3 microfarads are all connected in series to a 9 volt battery. The stored energy in C1 is found (U1s). The same three capacitors are connected in parallel to the same battery. The stored energy in C1 is found again in this configuration (U1p). What is the difference in stored energy in the parallel configuration compared to the series configuration in microjoules? That is determine U1p - U1s=

    2. Relevant equations
    U1p = 1/2C1V^2
    U1s = 1/2 Q^2/2C1
    VCeq = Q

    3. The attempt at a solution

    Using the formulas I got U1p = 271.35 and U1s = 64.24. Thus U1p - U1s should equal 207.15 but the correct answer is 176.42. Can anyone see what I'm doing wrong?
     
    Last edited by a moderator: Feb 19, 2015
  2. jcsd
  3. Feb 19, 2015 #2

    berkeman

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    Staff: Mentor

    I'm not understanding how you are handling the series case. What are the voltages across the 3 caps in the series case?
     
  4. Feb 19, 2015 #3
    I wasn't given the voltage across each, just the voltage of the battery. For a series I found the equivalent capacitance using (1/C1 + 1/Cn)^-1. And then I plugged that into the Q = CeqV equation to find Q. Then I used Q in the stored energy equation with C1. I hope that made sense.
     
  5. Feb 19, 2015 #4

    berkeman

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    Staff: Mentor

    I don't think that will work. You need to find the voltage division among the 3 series capacitors, so you can calculate how much energy is stored on C1 in the series configuration. The total series capacitance doesn't help, I don't believe.

    What equation(s) would you use to figure out how the 9V source divides among the 3 series capacitors? Since they are in series, what can you say about charge flow as they are charged up to 9V total across all 3 caps?
     
  6. Feb 19, 2015 #5
    I'm not really sure what equation I could use. But I know in series the charge in each would be the same which is why I used the equivalent capacitance to find the charge using Q=CV.
     
  7. Feb 19, 2015 #6

    berkeman

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    Right, since the series current is the same for all 3 caps, and current I = dQ/dt, then each cap will get the same amount of charge on it during the charging process. And since Q=CV, you know what the voltage is across each cap. If you use that voltage for C1, do you get the right answer for the ratio of the stored energies?
     
  8. Feb 19, 2015 #7
    Well when I do the math I get Q = 29.34. Then I plug that into 1/2 (Q^2/C1) and I get 64.24 for the stored energy of the series capacitor. Then the difference between the two is 207.15 but the correct answer is 176.42.
     
  9. Feb 19, 2015 #8

    gneill

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    You can use the voltage or the charge to find the energy stored on a capacitor:

    $$E_c = \frac{1}{2} C V_c^2 = \frac{1}{2} \frac{Q_c^2}{C}$$

    I'm seeing a result that's very close to yours (rounding of intermediate results could explain the difference). It could be that values in the problem have been "updated" without changing the answer key.
     
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