To solve this problem, we need to understand the relationship between voltage and current in a capacitor.$\\$Given:$\\ V(t) = V_0 \sin(\omega t) \\$The current $I(t)$ through a capacitor is given by the equation:$\\ I(t) = C \frac{dV(t)}{dt} \\$Substitute $V(t) = V_0 \sin(\omega t)$ into the equation:$\\ I(t) = C \frac{d}{dt}(V_0 \sin(\omega t)) \\$Differentiate $V_0 \sin(\omega t): \\ \frac{d}{dt}(V_0 \sin(\omega t)) = V_0 \omega \cos(\omega t) \\$Thus, the current is:$\\ I(t) = C V_0 \omega \cos(\omega t) \\$We know that $\cos(\omega t) = \sin\left(\omega t + \frac{\pi}{2}\right) \\$Therefore, $I(t)$ can be rewritten as:$\\ I(t) = C V_0 \omega \sin\left(\omega t + \frac{\pi}{2}\right) \\$This shows that the current $I(t)$ leads the voltage $V(t)$ by $90^\circ. \\$However, the correct option states that the current leads by $180^\circ, \\$which is incorrect based on our analysis.$\\$Let's analyze the energy consumption:$\\$The energy consumed by a capacitor over a full cycle is given by:$\\ E = \int_0^{T} V(t) I(t) \, dt \\$Since $I(t) = C \frac{dV(t)}{dt},$ we have:$\\ E = \int_0^{T} V(t) \cdot C \frac{dV(t)}{dt} \, dt \\$This integral represents the energy stored and released by the capacitor,$\\$which over a full cycle results in zero net energy consumption.$\\$Therefore, the correct behavior is described by Option 4:$\\$Over a full cycle, the capacitor C does not consume any energy from the voltage source.$\\$Thus, the correct option is Option 4.