A

Step 1: Analyzing the circuit when the switch is closed
After the switch is closed at time t=0, the circuit contains an inductor (L = 400 mH = 0.4 H) and two resistors (R1 = 2 Ω and R2 = 2 Ω) in series.

Step 2: Finding the total resistance
The total resistance in the circuit is:
$$R_{total} = R_1 + R_2 = 2 \, \Omega + 2 \, \Omega = 4 \, \Omega$$

Step 3: Using the formula for current growth in an RL circuit
The current through the circuit as a function of time is given by:
$$I(t) = \frac{E}{R_{total}}(1 - e^{-\frac{R_{total}}{L} t})$$
where E = 12 V, R_total = 4 Ω, and L = 0.4 H.

Substituting the values:
$$I(t) = \frac{12}{4}(1 - e^{-\frac{4}{0.4} t}) = 3(1 - e^{-10t})$$

Step 4: Finding the potential drop across the inductor
The potential drop across an inductor is given by:
$$V_L(t) = L \frac{dI(t)}{dt}$$

Calculating the derivative of I(t):
$$\frac{dI(t)}{dt} = 3 \cdot 10 e^{-10t} = 30 e^{-10t}$$

Thus, the potential drop across the inductor is:
$$V_L(t) = 0.4 \, H \cdot 30 \, e^{-10t} = 12e^{-10t}$$

Step 5: After steady state
After the steady state is reached (t → ∞), I(t) approaches 3 A and the switch is then opened. Immediately after opening, the inductor will act to maintain the current.

The current through R1 just before opening the switch is 3 A, and since the inductor will generate the same current in the opposite direction, the direction through R1 will be the same as before but of decreasing magnitude due to the discharge through the resistors.

Final Answer
The potential drop across L as a function of time is:
$$V_L(t) = 12e^{-10t} \text{ V}$$ and the direction of current through R1 after opening the switch remains the same initially (3 A) as it decays exponentially over time.