A

Step 1: When the switch is closed at position (1), the capacitor begins to charge through the resistor R. The current I in the circuit during this charging phase can be described by the equation:
$$ I(t) = \frac{V}{R} e^{-\frac{t}{RC}} $$
where V is the voltage across the capacitor. As the capacitor charges, the current decreases exponentially, which aligns with the option [A].
Step 2: After a long time, the capacitor is fully charged, and at t = t', when the switch is moved to position (2), the capacitor starts to discharge through the second resistor. The initial current during discharge is determined by the charge stored in the capacitor at that moment and also decreases exponentially.
Step 3: Maximum current depends on the capacitor charge and can be influenced by the resistor values and capacitance (R, C) as indicated in option [C].
Step 4: The time to achieve 50% charging of the capacitor is also influenced by the resistance (R) according to the time constant, which shows its significance in option [D].
However, focusing purely on the current behavior during charging, the correct option when considering the current falls significantly is option [A]. Therefore, A.