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I have to calculate \$i(t)\$ at règime, by knowing that \$v_s(t)=2\cos(\omega t + \pi/2)\$, \$R=1\Omega\$, and that, at the generator frequency, \$X_c=-1\Omega\$ and \$X_L=1\Omega\$.

schematic

simulate this circuit – Schematic created using CircuitLab

I know how to solve this by using the phasor domain. That is: using the current divider formula (after I substituted the series \$v_s\ -\ R\$) with the Norton equivalent (current generator \$v(s)/R\$ in parallel with \$R\$).

But something just came across in my mind: at règime, doesn't the inductor and the capacitor behave, respectively, as a short circuit and an open circuit? By reasoning that way $$i(t)=\frac{v(s)}{2R}=\frac{\cos(\omega t + \pi/2)}{R}$$

But how can I proceed? Is it correct to reason that way?

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    \$\begingroup\$ What does "at regime" mean? \$\endgroup\$ – The Photon Jun 15 '15 at 20:52
  • \$\begingroup\$ After a sufficient time such that the temporary effects have gone away. \$\endgroup\$ – sl34x Jun 15 '15 at 20:54
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    \$\begingroup\$ I would call it "steady state" \$\endgroup\$ – Eugene Sh. Jun 15 '15 at 20:54
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    \$\begingroup\$ The usual word for that in English is "in steady state". \$\endgroup\$ – The Photon Jun 15 '15 at 20:54
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    \$\begingroup\$ Anyway. The inductor and capacitor are "short" and "open circuit" in steady state, only if the steady state is for DC. In your case it's AC, so the components response is not that simple (phasors, as said). \$\endgroup\$ – Eugene Sh. Jun 15 '15 at 20:56

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