RLC specific current-time value

Question

I previously asked this question and got the answer I was looking for

Series RLC Circuit Step Response

I have a solution to a second order ODE that is

$ i(t) = A{\rm e}^{{S_1}t} + B{\rm e}^{{S_2}t} $

And again what I thought would be simple has me beaten, I want to calculate the t for specific values of I

So at what value of t do I get say 10A (for arguments sake)

Clearly there are going to be two solutions, rising edge and falling edge if you like

I have tried a lot of things but because A is negative then I cant take the Ln, so I treid swapping sides and it just got a bit hairy, I ended up with something that gave me an answer almost correct but only for the rising edge I think I need a quadratic with two solutions but I have no idea how I get that from the above!

Answer 1

The problem can be analytically solved for some values of $S_1$ and $S_2$, for instance if $S_2 = 2S_1$ then, the time $t$ at which you get the current $i$ is

$$ t(i) = \frac{1}{S_1}\ln \left(\frac{A + \sqrt{A^2 -4Ai}}{2A} \right) $$

You can also solve it for $S_2 = 3S_1$, but it gets complicated very quickly. Instead I suggest you to use Newton's method to calculate $t$:

• Start with a reasonable assumption for $t_0$
• Iterate $$ t_{k+1} = t_k - \frac{A(e^{S_1 t} - e^{S_2 t}) - i_{\rm target}}{A(S_1e^{S_1 t} -S_2 e^{S_2 t})} $$

until $i(t_k) \approx i_{\rm target}$

EDIT

In the situation $S_2 = 2 S_1$, call $x = e^{S_1 t}$ such that $x^2 = e^{2S_1 t} = e^{S_2 t}$, then the equation can be written as

$$ i = Ax-Ax^2 ~~~\Rightarrow~~~ Ax^2 -Ax + i = 0 $$

whose solution is

$$ x = \frac{A + \sqrt{A^2 - 4A i}}{2A} = e^{S_1 t} $$

Taking $\ln$ at both sides we get to

$$ S_1 t = \ln \left( \frac{A + \sqrt{A^2 - 4A i}}{2A} \right) $$