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Capacitors in parallel question

  1. May 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A 4μF capacitor is charged to a charge of 200 μC,and a 3μF capacitor is charged to a charge of 300μC. Both the capacitors are connected so that the positive plates are connected together. What is the total loss of energy in the process?
    2. Relevant equations


    3. The attempt at a solution
    Loss of energy= E initial -E final
    By using Energy =1/2 (Q^2)/c
    The energy I get is the final energy or initial energy?
    I'm kind of confused.

    Thanks.
     
  2. jcsd
  3. May 16, 2016 #2
    Think of the final situation as one ‘big’ capacitor C=c1*c2/(c1+c2). Q=q1+q2. Now you can calculate energies…
     
  4. May 16, 2016 #3

    gneill

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    Staff: Mentor

    Start by finding the initial energy stored in the system. That is, the sum of the energies stored in the two separate capacitors.

    I note that the problem statement specifies that the positive plates of the two capacitors are connected together, but doesn't mention the negative plates. Is the problem statement quoted exactly or have you presented it in your own words? The reason I ask is that it is possible that it is meant to be a bit of a trick question...
     
  5. May 16, 2016 #4
    By using that equation what we get is the capacity in series, but when positive plate is connected together, isn't it suppose to be parallel? Capacity =C1+C2
     
  6. May 16, 2016 #5
    Thanks.Yes it is quoted exactly.I assume that the statement is trying to show that the capacitors are arranged in parallel?
     
  7. May 16, 2016 #6

    gneill

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    Staff: Mentor

    Okay, if that's the case then assuming the capacitors are connected in parallel, what's the total capacitance of the combination?
     
  8. May 16, 2016 #7
    Total Capacitance = 3μ+4μ =7μ
    So the final energy =1/2 (200μ+300μ)^2/7μ
    Then do the initial energy separately = (1/2 (200μ)^2/4μ )+(1/2(300μ)^2/3μ)
     
  9. May 16, 2016 #8

    gneill

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    Staff: Mentor

    Okay, what are your results for initial and final energy? BY the way, "μ" alone is not a unit. You should get into the habit of attaching the actual units so that you can trace issues with unit consistency should they arise. Thus μF and μC are valid units for capacitance and charge.
     
  10. May 16, 2016 #9
    Ok take note of that, Initial energy =0.02Joule
    Final energy =0.01786Joule

    Total loss of energy=0.02-0.01786 = 2.14×10^-3 Joule

    Is that correct?
     
  11. May 16, 2016 #10

    gneill

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    Staff: Mentor

    Looks good.
     
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