# Cancelling transient state in RLC circuit

I have a problem asking the following question:

We have a LRC circuit (L,R,C all in series) with a source $E(t) = Vsin(\omega t + \phi)$. Is it possible to chose the phase $\phi$ so that we can cancel transient state and pass directly to steady state at $t = 0+$ ? The charge in the capacitor and the current in the inductor are initially zero.

Solutions book is vague and only states (translation might be bad):

Could anyone tell me why it is impossible, in a detailed enough solution?

I did a similar exercise with a RC circuit, and came easily to the conclusion that, in order to cancel steady state, I must have $\phi = arctg \left ( RC \omega \right )$. The way I solved it was to calculate for which phase I get zero in the Laplace transform of the response. The problem with a 2nd order system such as a LRC circuit is that I cannot use the same method rapidly and reliably enough because of how complicated the equations become when isolating the transient response in the s-domain.

Is there a way to know it's impossible without spending two hours in Laplace equations?

Thanks!

Here is one way. The steady state forced response of the voltage across the capacitor would be: $$v_C(t) = K\sin(wt + \theta)$$ $$i_C(t) = C \frac{dv_C}{dt} = CwK\cos(wt + \theta)$$
At time = 0, apply the two initial conditions and equate them to the pure steady state forced response: $$v_C(0) = 0 = K\sin(\theta)$$ $$i_C(0) = i_L(0) = 0 = CwK\cos(\theta)$$
It is clear that such a solution does not exist because it is not possible for: $\sin(\theta) = \cos(\theta) = 0$ for any $\theta$.
A summary of this is that the steady state response of the given forcing function is sinusoidal, so the voltage and current across the capacitor has a phase difference of 90$^\circ$. But at time zero, the initial conditions dictate the voltage and current across the capacitor to be 0. Therefore, there is no non-trivial steady state only solution at time 0.