A

Step 1: Write down the expression for the voltage across the capacitor, which is given by v(t) = V_M \cos(\omega t).

Step 2: The current I(t) through the capacitor can be described using the relationship between current, capacitance, and voltage.
The current through a capacitor is given by the equation \( I(t) = C \frac{dV(t)}{dt} \).

Step 3: Calculate the derivative of the voltage:
\( \frac{dV(t)}{dt} = \frac{d}{dt}(V_M \cos(\omega t)) = -V_M \omega \sin(\omega t) \).

Step 4: Substitute the derivative back into the current equation:
\( I(t) = C\left(-V_M \omega \sin(\omega t)\right) = -C V_M \omega \sin(\omega t).

Step 5: In a poorly conducting medium, we also need to consider the active resistance (R). The steady-state current in the presence of resistance is governed by Ohm's law. Therefore, we can write:\
\( I(t) = \frac{V(t)}{R} = \frac{V_M \cos(\omega t)}{R}.

Step 6: To summarize the total current with respect to the alternating voltage and time dependence, we have two components: the displacement current is related to the capacitance, and the conduction current is given by Ohm's law. The steady-state current would then focus on the sinusoidal terms produced by both capacitance and resistance. Thus the expression for I(t) would be summed correctly.

Step 7: Taking these into account and simplifying leads to: \(I(t) = \frac{V_M \omega C}{R} \sin(\omega t + \phi) \) where \(\phi\) is the phase shift due to the reactance of the capacitor which can be derived if needed.

Therefore, the steady-state current flowing in lead wires includes the effects of both the capacitor behavior and resistance, resulting in a sinusoidal changing current dependent on the applied voltage frequency.