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Deriving Capacitor Voltage for a Discharging Capacitor

  1. Apr 7, 2016 #1
    1. The problem statement, all variables and given/known data
    I'm trying to derive the voltage waveform across the capacitor for a discharging capacitor that has been fully charged by a DC power supply ##v_0##, i.e. ##v_c(t=0)=v_0## and then at ##t=0## the switch is flipped and the capacitor starts to discharge.
    RC.png


    2. Relevant equations

    3. The attempt at a solution

    From KVL we obtain
    $$v_c=\frac{q}{c}=iR$$
    Taking the time derivative of this and since capacitor is discharging ##-\frac{dq}{dt}=i##... $$-\frac{i}{RC}=\frac{di}{dt}\\ -\int_0^t \frac{dt}{RC}=\int_{i_0}^{i}\frac{di}{i}\\ -\frac{t}{RC}=\ln (\frac{i}{i_0})\\ i=i_0e^{-\frac{t}{RC}}$$
    Now to solve for ##i_0## at t=0 ##\frac{q}{c}=i_0Re^{-\frac{t}{RC}}## becomes $$\frac{v_0}{c}=i_0R\\ i_0=\frac{v_0}{RC}\Longrightarrow i=\frac{v_0}{RC}e^{-\frac{t}{RC}}$$
    Finally from KVL $$v_c=iR=\frac{v_0}{C}e^{-\frac{t}{RC}}$$

    which is not the correct answer.
     
  2. jcsd
  3. Apr 8, 2016 #2

    cnh1995

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    Of course it isn't. You can see it is dimensionally incorrect.(Vo/C doesn't give voltage).
    Where does this come from?
     
  4. Apr 8, 2016 #3
    I got that from the KVL equation (q/c=ir) and at t=0 so ##\frac{q(0)}{c}=i(0)R##
     
  5. Apr 8, 2016 #4

    ehild

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    Q/C=Vo, not Vo/C .
     
  6. Apr 8, 2016 #5
    Whoops, after making the proper substitution then we get ##i_0=\frac{v_0}{R}\Longrightarrow i=\frac{v_0}{R}e^{-\frac{t}{RC}}\Longrightarrow V_c=iR=v_0e^{-\frac{t}{RC}}## which is the correct expression :)
     
  7. Apr 8, 2016 #6

    ehild

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    It is useful to check the dimensions during a derivation. :smile:
     
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