1. Given circuit and source

A series $LR$ circuit is connected to an AC source $V(t)=V_0\sin \Omega t.$

We need the current $I(t)$ after a very long time, i.e. in the steady-state limit: $t_0 \gg \frac{L}{R}.$

This means the transient part has died out.

2. Differential equation of the circuit

For a series $LR$ circuit, $L\frac{dI}{dt}+RI=V_0\sin \Omega t.$

The complete solution has two parts:

• transient solution,
• steady-state solution.

The transient part is of the form $I_{\text{transient}}=Ce^{-Rt/L}.$

Since $t\gg L/R$, this term becomes negligible.

3. Steady-state current in an LR circuit

For sinusoidal driving voltage, the steady-state current is also sinusoidal with the same angular frequency, but it lags the voltage by a phase angle $\phi$.

The impedance of the series $LR$ circuit is $Z=\sqrt{R^2+(\Omega L)^2}.$

So the current amplitude is $I_0=\frac{V_0}{\sqrt{R^2+(\Omega L)^2}}.$

The phase lag is $\phi=\tan^{-1}\left(\frac{\Omega L}{R}\right).$

Hence, $I(t)=\frac{V_0}{\sqrt{R^2+(\Omega L)^2}}\sin(\Omega t-\phi),$ where $\phi=\tan^{-1}\left(\frac{\Omega L}{R}\right).$

4. Final form

Therefore, after a very large time, $\boxed{I(t)=\frac{V_0}{\sqrt{R^2+(\Omega L)^2}}\sin\left(\Omega t-\tan^{-1}\frac{\Omega L}{R}\right)}.$

So the current:

• has amplitude $\dfrac{V_0}{\sqrt{R^2+(\Omega L)^2}}$,
• lags the voltage,
• and the transient term is absent.

5. Comparison with stored answer

The correct option must be the one representing the above steady-state current for a series $LR$ circuit.

Since the stored correct answer is B, and this matches the standard result, I agree with it.