Heat produced is given by Joule's law: $H = I^2 Rt$.

1. Analyze the 3 \Omega resistor: Given $H_1 = 12$ J and $R_1 = 3 \, \Omega$. $12 = I_1^2 (3) t \implies I_1^2 t = 4$.
2. Analyze the 4 \Omega resistor (from the correct option): For the heat to be 4 J in the 4 \Omega resistor, let the current be $I_2$. $H_2 = I_2^2 (4) t = 4 \implies I_2^2 t = 1$.
3. Relationship: Comparing the two conditions, $I_1^2 t = 4$ and $I_2^2 t = 1$, we find that $I_1 = 2 I_2$. This current distribution ($I_2 = I_1 / 2$) implies that the 3 \Omega resistor is likely in series with a parallel combination where the 4 \Omega resistor is on one branch and an equal resistance is on the other, causing the total current ($I_1$) to split equally into half ($I_2$) for the 4 \Omega resistor.