Step 1: Merge the equal-potential points:
Call the node tied to A by wires "node A" and the node tied to B by wires "node B".
Node A includes: terminal A, the top wire, the right end of the 3 H coil, and the left end of the 4 H coil.
Node B includes: terminal B, the bottom wire, and the junction between the 1 H and 3 H coils.
Step 2: Place each inductor:
1 H: A to node B. 2 H: top wire (node A) to the junction (node B). 3 H: junction (node B) to the right end (node A). 4 H: node A to B.
So all four sit between node A and node B.
Step 3: Add reciprocals:
$\dfrac{1}{L} = 1 + 0.5 + 0.333 + 0.25 = 2.083 = \dfrac{25}{12}$, so $L = \dfrac{12}{25}$ H $= 0.48$ H.
Final Answer:
Option (C).
\[ \boxed{\frac{12}{25}\text{ H (C)}} \]
