To analyze the given circuit, we can use the equation for induced electromotive force (EMF) in an inductor. The formula is:

$ \varepsilon - \frac{L \, dI}{dt} - I \cdot R = 0 $

For this circuit, we have:

$ \varepsilon = 12 \, \text{V} $ (the applied EMF)

$ L = 3 \, \text{H} $ (the inductance of the inductor)

$ \frac{dI}{dt} = 8 \, \text{A/s} $ (the rate of change of current)

Assuming the direction of current is such that the change in current is negative, substitute these values into the equation:

$ 12 - 3 \times (-8) - I \times 12 = 0 $

Solving for $ I $ (the current in the circuit):

$ 12 + 24 - 12I = 0 $

$ 36 = 12I $

$ I = \frac{36}{12} = 3 \, \text{A} $

Thus, the current at the moment when the resistance $ R $ is $ 12 \, \Omega $ is $ 3 \, \text{A} $.