Physics - EMI/AC Question with Solution | TestHub

The phase difference between the source voltage and the steady-state current in an L-R series circuit is $\alpha ={\tan }^{-1}(\omega L/R)$, so A is correct.

An ideal inductor prevents instantaneous current change. Since there was no current before $t=0$, the current through the inductor immediately after closing the switch must be zero. So B is correct.

For no transient current, the steady-state current $I_{steady}(t)=I_0\sin (\omega t+\varphi -\alpha )$ must be zero at $t=0$. This requires $\varphi -\alpha =n\pi$, which means $V(0)=V_0\sin (\varphi )$ is generally not zero (unless $\alpha$ is also $0$ or $\pi$, i.e., $L=0$). Thus C is incorrect. The average power dissipated in steady state is $P_{avg}=V_{rms}{I}_{rms}\cos (\alpha )=\frac{V_0}{\sqrt{2}}\frac{V_0}{\sqrt{2}Z}\frac{R}{Z}=\frac{V_0^{2}R}{2Z^2}=\frac{V_0^{2}R}{2(R^2+(\omega L{)}^2)}$. So D is correct.