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RC capacitor discharge

  1. Mar 1, 2012 #1
    1. The problem statement, all variables and given/known data

    The capacitor in an RC circuit with a time constant of 25 ms is charged to 10 V. The capacitor begins to discharge at t = 0 s

    2. Relevant equations
    τ=RC
    ΔVC=(ΔVC)e^-t/τ


    3. The attempt at a solution
    part a) At what time will the charge on the capacitor be reduced to half its initial value?

    10 ln 5 = (10)e^-τ/.025s
    ln 5 = -τ/.025s
    ln 5(.025s)= τ
    τ=0.04 s

    part b) At what time will the energy stored in the capacitor be reduced to half its initial value?

    Not sure, but any hints as to my next step would be appreciated...
     
  2. jcsd
  3. Mar 1, 2012 #2

    gneill

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    Staff: Mentor

    Where did the 10 ln 5 in the first line come from?
    The result for your time value looks a bit high.
    What's an expression for the energy stored on a capacitor?
     
  4. Mar 1, 2012 #3

    ehild

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    Homework Helper
    Gold Member

    Is the same ΔVC on both sides? That would mean e^-t/τ=1 all time.

    Where does that ln5 come?

    ehild
     
  5. Mar 2, 2012 #4
    Where did the 10 ln 5 in the first line come from?

    I used 10 V * ln 5 to signify that ln 5 was half of the initial value.

    Instead I should have used 5
    ΔV final=(ΔV initial)e^-t/τ

    5 = 10e^-τ/.025s
    0.5 = e^-τ/.025s
    ln 0.5 = -τ/.025
    ln 0.5(.025) = -τ
    τ= 0.013 s

    Uc=0.5 C (ΔVc)^2 or Q = C ΔVc

    Easy to solve if I have C or Q
     
  6. Mar 2, 2012 #5

    gneill

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    Staff: Mentor

    Why not write v(t) = vi e-t/τ ?
    Check the calculation on that last step.
    C is just a proportionality constant; it won't affect the relative change in magnitude of the function.

    Write the expression for the energy stored on a capacitor versus voltage. Plug in your expression for the voltage versus time.
     
  7. Mar 2, 2012 #6
     
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