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Energy dissipated in circuit with two capacitors

  1. Oct 30, 2009 #1
    1. The problem statement, all variables and given/known data
    A 2.0-microfarad capacitor is charged to 250 V. It is then connected to an uncharged 1.0 microfarad capacitor through a 2.2 ohm resistor, by closing switch S in the figure .

    Find the total energy dissipated in the resistor as the circuit comes to equilibrium. Hint: Think about charge conservation.
    Express your answer using two significant figures.

    The answer to the question is 21 millijoule.

    2. Relevant equations
    C=Q/V
    Capacitators in series carry the same charge.
    P=I*V
    I=V/R
    P=I^2*R
    U=1/2 * Q * V^2 (energy stored in a capacitor)

    3. The attempt at a solution

    I don't really know what to do. I guess charge moves from the charged capacitor to the uncharged one, until the both charges on the capacitors are equal. Moving charge is a current, and this current should give an energy dissipation in the resistor.

    I tried some calculations on paper, but there's not much use writing them here, since I seem to need help with the basic approach of this problem.

    Thanks in advance,
    Merlione.
     
  2. jcsd
  3. Oct 30, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    Hi Merlione, welcome to PF.
    After closing the circuit, at equilibrium, what is the common voltage across each capacitor?
    Find the initial energy in 2 μF capacitor. Find the final energies in the two capacitors after equilibrium. Find the difference in the energies to find energy dissipated in the resistance.
     
  4. Nov 1, 2009 #3
    Thanks for your reply! And for the welcome. :)

    I still haven't figured it out.

    The initial energy is given by
    [tex]

    u=\frac{1}{2}*C*V^2

    [/tex]

    Which gives me 62.5 millijoule.

    Then, I thought... Once the circuit has reached equilibrium:
    Since capacitors in series carry the same charge, the charge present on the first capacitor will distribute evenly among the two capacitors when the circuit is closed.

    [tex]

    V1=\frac{Q}{C1}
    V2=\frac{Q}{C2}
    [/tex]

    And this should give me the voltages across each capacitor? Which are 125V and 250V.
    However when I calculate the energy stored in the capacitors then, and add them up, I get an energy of 46.8mJ.

    The difference is 15.6 mj, and not 21 mj, which is the right answer. So, obviously I'm erring somewhere.

    I have to say I do not quite understand this. Shouldn't I include the R parameter somewhere in a formula?

    [edit]

    Through a sort of backward-calculating-with-the-answer, using the energy formula
    [tex]

    u=\frac{1}{2}*C*V^2

    [/tex]
    Filled in the ' replacement' capacitance of the the capicitances, added reciprocally, and U of 21 millijoule, I derive a voltage of 250 volts.

    Which is the voltage across the capacitor in the stage before eq. What I wonder is, would it be correct to say that, once this switch is closed, the voltage across the total circuit is 250 volts? And you can simply use this along with the replacement capacitance to find a new stored energy...?

    A bit confused and in doubt, waiting for the helper who'll make it a little more clear....
    ~Merlione.
     
    Last edited: Nov 1, 2009
  5. Nov 1, 2009 #4

    rl.bhat

    User Avatar
    Homework Helper

    The R parameter decides the time taken by the system to reach the equilibrium.
    Find the equivalent capacity when the two capacitors are connected in series.
    Then find the charge Q drawn from the source. This charge is common for both the capacitors. Then find the individual energies in the capacitor.
     
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