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Solving an RC circuit using explicit euler

  1. Oct 7, 2015 #1
    1. The problem statement, all variables and given/known data

    Hi there. I have a simple RC circuit with a battery voltage of 10 V, R = 1 Ω, C = 1 F and a switch. I want to use the Explicit Euler (forward divided difference) to solve the equation and check for stability, rather than using a ODE. I am finding the equation for when the capacitor is fully charged and then the battery removed, standard situation etc.

    2. Relevant equations

    C. dv/dt + V(t)/R = 0 (loop equation for the circuit)

    3. The attempt at a solution

    The first thing I attempted to do was rewrite / rearrange the circuit in the form v(tnew) = v(told)(1 + a.Δt)n then check what value of t the equation would be unstable for.

    dv/dt + V(t)/RC = 0

    Rewriting dv/dt as (Vf - Vi)/Δt,

    (Vf - Vi)/t + V(t)/RC = 0
    ... leads to

    Vf = Vi - Δt/RC
    Vf = Vi(1 - Δt/RC) ----- equation 1

    So a = -1/RC = -1, giving

    v(t) = Vi(1 - Δt)
    v(t) = 10(1 - Δt)

    For v(t) to be unstable, |1 - Δt| > 1 ∴ t should be 2sec or more.

    However, logically speaking shouldn't the equation be unstable at values of t > 1 sec. Because if you plug Δt = 1 sec, the final voltage is always 0. Where have I made a mistake?
     
  2. jcsd
  3. Oct 7, 2015 #2

    BvU

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    You mix up stable and good. Δt = 1 sec gives a stable solution all right !
     
  4. Oct 7, 2015 #3
    OK for t = 1 sec, V is constantly zero which is still OK I guess, but for t = 1.1 sec, equation 1 gives -1V. This shouldn't be possible?
     
  5. Oct 7, 2015 #4

    BvU

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    What you mean is: that isn't good. But it is still stable: V does not exceed any bound for growing t.
    [Iedit] I should correct that: the error does not exceed any bound.
     
  6. Oct 7, 2015 #5
    So unstable would be if the final voltage kept increasing at some exponential rate, much higher than the original value of 10 V?
     
  7. Oct 7, 2015 #6

    BvU

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    Yes. Unstable is when the error can exceed any fixed bound. Doesn't have to be exponential.
     
  8. Oct 7, 2015 #7
    Alright. Thank you so much for the quick response. :)
     
  9. Oct 7, 2015 #8

    BvU

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    Note that your differential equation has a very decently behaving analytical solution.


    Perhaps that's why the gap between a 'good' Δt (*) and the unstability criterion is rather wide.
    [edit] I think that I have to withdraw that. No basis for such a statement, I suppose

    (*) Just play around with excel and check that a 1 mV upper bound on the error requires Δt < 0.5 msec or thereabouts

    PS I suppose with "a ODE" you mean some wrapped integrator that comes with your program ? (Not shrink wrapped, because then you might see that it could well be an explicit Euler too... :smile: )
     
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