Step 1: Understanding the Concept:
The figure has four inductors, 1 H, 2 H, 3 H and 4 H, joined by wires. We need to find which points are the same electrical node, because wire segments of zero resistance join the same potential.
Step 2: Key Formula or Approach:
For inductors in parallel (no mutual inductance): \(\dfrac1{L_{eq}} = \dfrac1{L_1} + \dfrac1{L_2} + \dfrac1{L_3} + \dfrac1{L_4}\).
Step 3: Detailed Explanation:
Point A is joined by a wire, over the top of the figure, to the right-hand end of the 3 H inductor and to the left end of the 4 H inductor, and also to the top of the 2 H inductor.
The bottom wire joins point B to the junction between the 1 H and 3 H inductors, which is also the lower end of the 2 H inductor.
So the 1 H inductor has one end at A and the other end at the node connected to B. The 2 H inductor has one end at the node connected to A and the other at the node connected to B. The 3 H inductor has one end at the node connected to B and the other at the node connected to A. The 4 H inductor has one end at the node connected to A and its other end at B.
All four inductors are connected between A and B in parallel:
\[ \frac{1}{L_{eq}} = \frac11 + \frac12 + \frac13 + \frac14 = \frac{12 + 6 + 4 + 3}{12} = \frac{25}{12} \]
\[ L_{eq} = \frac{12}{25} \text{ H} \]
Final Answer:
The equivalent inductance is \(\dfrac{12}{25}\) H, option (C).
\[ \boxed{\frac{12}{25}\text{ H (C)}} \]
