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Equation for dissipation of a capacitor

  1. Nov 21, 2007 #1

    I'm trying to evaluation the voltage of an RC circuit as it is dissipating a stored charge. Actually trying to do this under two different situations.

    The first is let's say the circuit has reached steady state for a given input voltage, Vo. Now I reduce the voltage instantly to some value V1 greater than zero, but less than its original value. I assume it would resemble some sort of exponential decay function, like e^(-t/tau), but how would Vo and V1 factor into the equation?

    Okay, now take the same circuit. But let's say instead of using the step function in the previous example for the voltage, it now decreases linearly over time, like V(t) =-3 Volts/min. What would the response equation look like now?

    Hopefully that makes sense, I've been wracking my brain over this for the past couple days.

  2. jcsd
  3. Nov 22, 2007 #2
    Off the top of my head...

    I think the equation is

    [tex]v(t) = (V_0-V_1)e^{-t/\tau}+V_1[/tex]

    Check this by using your 2 boundary conditions v(0) = V_0 and V(infty) = V1

    You can find the derivation of this in most basic circuits book.

    I can't imagine a cap's voltage dropping off linearly vs. time. Does anyone know how?
  4. Nov 22, 2007 #3
    i'm not trying to get the cap's voltage to drop linearly, i'm saying what if the voltage into the cap drops linearly with time, how do this change the formula you posted?

    for instance, in the formula you gave, Vo @ t=0 is 30, V1 @ t=1s is 27, V2 @ t=2s is 24, etc
  5. Nov 22, 2007 #4
    Well you would just have to go back to the basics and write the KVL equation.
  6. Nov 22, 2007 #5
    sorry, you're talking to an ME, have no idea what the KVL equation is
  7. Nov 22, 2007 #6
    Do you remember solving circuits equations by writing a loop equation to sum up all the voltage drops? Then you set this equation to zero?
  8. Nov 22, 2007 #7
    yes, i do remember doing that, basically adding up all the nodes, right?
  9. Nov 22, 2007 #8
    That was Kirchoff's current law (KCL). You could try doing that too if you want... But for a simple RC circuit. It's better to use Kirchoff's voltage law instead.

    PS: Is this a mental exercise?
  10. Nov 23, 2007 #9
    No this is something I'm having trouble with at work

    I think I have an understanding but I'm having a problem putting it all together

    KVL says the sum of the voltage in a loop is equal to zero.

    [tex]\sum V_{loop} = 0[/tex]

    so if the input voltage is a function of time

    [tex]V(t) = -3t[/tex]

    the equation for the capacitor is

    [tex]I = C \frac{dV}{dt} \Rightarrow \int\frac{dV}{dt}=\frac{I}{C} \Rightarrow V = \frac{I}{C}t +a[/tex]

    and the resistor is


    putting it all together

    [tex] -3t = \frac{I}{C}t + IR + a[/tex]

    am i on the right track so far?
  11. Nov 23, 2007 #10
    Can you just show us a simplified schematic of the circuit you've got.
  12. Nov 23, 2007 #11
  13. Nov 23, 2007 #12
    So if KVL serves me correctly...

    Vin - I R - V = 0
    [tex]V_{in} - C \frac {dv_c}{dt} R - v_c = 0[/tex]
    [tex]RC \frac {dv_c}{dt} + v_c = V_{in}[/tex]

    After this its just math...
  14. Nov 23, 2007 #13
    ok so my calc is a little rusty, but here goes: (using [tex]V_{in}(t) = At[/tex])

    [tex]RC\frac{dv_{c}}{dt} + v_{c} = At[/tex]

    integrating both sides with respect to t (not sure about this part)

    [tex]RCv_{c} + v_{c}t = \frac{At^{2}}{2}+constant[/tex]

    solving for [tex]v_{c}[/tex]

    [tex]v_{c}(RC+t) = \frac{At^{2}}{2}+constant \Rightarrow v_{c} = \frac{\frac{At^{2}}{2}+constant}{RC+t}[/tex]

    is this right up to this point? how can i solve for the constant?
  15. Nov 23, 2007 #14
    I can't seem to remember off the top of my head how to solve this diff'eqn. It's first order non homogeneous. I think you can find it in Boyce and Diprima's textbook. I think it involves integrating factors or something to that extent.
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