# Solving an RC circuit using explicit euler

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1. Oct 7, 2015

### yugeci

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. Oct 7, 2015

### BvU

You mix up stable and good. Δt = 1 sec gives a stable solution all right !

3. Oct 7, 2015

### yugeci

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?

4. Oct 7, 2015

### BvU

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.

5. Oct 7, 2015

### yugeci

So unstable would be if the final voltage kept increasing at some exponential rate, much higher than the original value of 10 V?

6. Oct 7, 2015

### BvU

Yes. Unstable is when the error can exceed any fixed bound. Doesn't have to be exponential.

7. Oct 7, 2015

### yugeci

Alright. Thank you so much for the quick response. :)

8. Oct 7, 2015

### BvU

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.
 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... )

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