A voltage $E(t)$ is applied to a RCL Circuit connected in series. The charge on the capacitor at any time, $t$ in seconds is given by :

$$q’’ + 9 q’ + 14 q =E(t) = \frac{1}{2} \sin t $$

Given that there is no initial charge, but an initial current of $1$ amperem what is the charge in the steady state ?

First, I solved this differential equation using Laplace transform and got

$$q(t) = \frac{1}{500} (110 e^{-7t} - 101e^{-7t} + 13 \sin t - 9 \cos t)$$

What does steady state mean ?

Why is the answer for the charge in steady charge $ q(t) = \frac{1}{500} (13 \sin t - 9 \cos t )$ (Without the part that I’ve taken out.)

  • $\begingroup$ Your solution doesn't include $E(t)$ in any way- it should, shouldn't it? $\endgroup$ – Brian Borchers Aug 20 '18 at 12:12
  • $\begingroup$ Yes, I’ve updated it. The E(t) was originally there and I was told that e(t) is $ 1/2 \sin t $ so that I can find $q(t)$ $\endgroup$ – user185692 Aug 20 '18 at 12:15
  • $\begingroup$ I would assume "steady state" means the state that the system eventually settles in, i.e. as $t\gg 0$. $\endgroup$ – Arthur Aug 20 '18 at 12:16

This is a somewhat loose use of the term “steady state”. It's used here to mean that the transient exponentials have decayed and only the permanent oscillation remains. I'd rather call it a “quasi-steady state”. It's not steady if you focus on the time domain, where the quantities are still changing, but when dealing with electrical circuits, one often focuses on the frequency domain, and with that in mind, a single sine wave with a fixed frequency and amplitude is something like a steady state.


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.