Let the self-inductances of the coupled inductors be $L_a = 3$ H and $L_b = 2$ H, and the mutual inductance be $M = 4$ H. Since both currents enter the dotted terminals, the mutual inductance term in the voltage equations will be $+M$.
The voltage equations for the coupled inductors in Figure (a) are:
$$v_1 = L_a \frac{di_1}{dt} + M \frac{di_2}{dt} = 3 \frac{di_1}{dt} + 4 \frac{di_2}{dt}$$
$$v_2 = M \frac{di_1}{dt} + L_b \frac{di_2}{dt} = 4 \frac{di_1}{dt} + 2 \frac{di_2}{dt}$$
The voltage equations for the T-equivalent circuit in Figure (b) are:
$$v_1 = (L_1 + L_3) \frac{di_1}{dt} + L_3 \frac{di_2}{dt}$$
$$v_2 = L_3 \frac{di_1}{dt} + (L_2 + L_3) \frac{di_2}{dt}$$
Comparing the coefficients, we get:
$$L_1 + L_3 = 3$$
$$L_3 = 4$$
$$L_3 = 4$$
$$L_2 + L_3 = 2$$
From these equations, we obtain $L_3 = 4$ H, $L_1 = 3 - 4 = -1$ H, and $L_2 = 2 - 4 = -2$ H. This corresponds to option (B).
However, given that option (A) is provided as the correct answer, let's explore an alternative T-equivalent representation for coupled inductors, although it's less standard. If we assume a T-model where:
$L_1 = L_a + M = 3 + 4 = 7$ H
$L_2 = L_b + M = 2 + 4 = 6$ H
$L_3 = -M = -4$ H
Then the voltage equations for this T-model would be:
$v_1 = L_1 \frac{di_1}{dt} + L_3 \frac{di_2}{dt} = 7 \frac{di_1}{dt} - 4 \frac{di_2}{dt}$ (Does not match the original equations)
There seems to be an inconsistency between the standard T-equivalent model and the provided correct answer. Based on the standard derivation, option (B) is the correct representation. Assuming the provided answer key is accurate, there might be a specific, non-standard convention being used for the T-equivalent in this context. Without further information on this specific convention, we proceed with the standard derivation.
Final Answer: (B)