A series $$LR$$ circuit is connected to a voltage source with $$V(t) = V_0\sin(\omega t)$$. After a very large time, current $$I(t)$$ behaves as $$(t_0 \gg \frac{L}{R})$$:

In an $$LR$$ circuit, the ratio $$\tau = \frac{L}{R}$$ is known as the inductive time constant.

The total current in such a circuit consists of two parts: the transient response (which decays exponentially) and the steady-state response.

The condition $$t_0 \gg \tau$$ means that several time constants have passed, causing the transient component ($$e^{-t/\tau}$$) to die out completely.

Once the transients have vanished, the circuit reaches a steady state where the current oscillates at the same frequency as the source voltage $$V(t) = V_0 \sin(\omega t)$$.

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

The steady-state current $$I(t)$$ is given by $$I(t) = I_0 \sin(\omega t - \phi)$$

Where Amplitude ($$I_0$$): $$I_0 = \frac{V_0}{Z}$$ (a constant value), and Phase Lag ($$\phi$$): $$\phi = \tan^{-1}\left(\frac{\omega L}{R}\right)$$.

After a very large time ($$t_0 \gg L/R$$), the current becomes a stable sinusoidal oscillation. Therefore, Option D is the correct graphical representation.