The equivalent inductance between A and B is equal to

Equivalent inductance calculation between points A and B

The circuit simplifies when recognizing that all four inductors connect between common nodes N (A and D) and M (C and B). This places the 2H, 3H, 4H, and 6H inductors in parallel.

The equivalent inductance $L_{eq}$ for parallel inductors satisfies:

$\frac{1}{L_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6}$

Combining fractions with common denominator 12:

$\frac{1}{L_{eq}} = \frac{6 + 4 + 3 + 2}{12} = \frac{15}{12} = \frac{5}{4}$

Thus, $L_{eq} = \frac{4}{5}$ H, corresponding to option A.