0
$\begingroup$

I have a series RLC circuit with an equation:

$$\frac{d^2I}{dt} + 2\alpha \frac{dI}{dt} + \omega_0^2 I = 0$$

(No outside sources affecting the circuit, only some $I_0$ was in the circuit at the beginning)

The result is, for overdamped response ($\alpha^2 - \omega_0^2 > 0$):

$$I = Ae^{(-\alpha - \sqrt{\alpha^2 - \omega_0^2})t} + Be^{(-\alpha - \sqrt{\alpha^2 - \omega_0^2})t}$$

So, if $I(0) = I_0$ then $A+B = I_0$, but that still doesn't solve A and B completely. What other initial conditions do i have to include or what other things do i have to do in order to find A and B completely?

|cite|improve this question
$\endgroup$
0
$\begingroup$

As in any other second-order ODE, the two initial conditions needed are $I(0)$ and $\frac{dI}{dt}(0)$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ And if dI/dt(0) = 0, then how do I solve it? $\endgroup$ – lkky7 Oct 27 '18 at 17:48
  • $\begingroup$ @lkky7 how about you take dI/dt at t=0 and see what happens? $\endgroup$ – ZeroTheHero Oct 27 '18 at 19:36
  • $\begingroup$ @lkky7 $dI/dt(0) = 0$ gives you an equation containing your constants $A$ and $B$, so you only have one arbitrary constant not two. $\endgroup$ – alephzero Oct 27 '18 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.