Animated Solution for Physics - Electromagnetic Induction: The inductance between $A$ and $D$ is

At first glance, this circuit looks like a straight line of inductors. It is incredibly tempting to just add them up as if they were in series.

But in physics, physical layout does not dictate electrical behavior. The true nature of a circuit is defined by its nodes and connections.

We must look past the drawing and trace the actual paths the current can take.

Let's start by analyzing the plain wires connecting the different parts of the circuit.

Notice the wire connecting the node before the first inductor to the node after the second inductor. Because this is an ideal wire with zero resistance, there is no voltage drop across it.

This means these two points are at the exact same electrical potential. Since the first point is connected directly to terminal $A$, both of these points are effectively node $A$.

Similarly, look at the wire connecting the node after the first inductor to the node after the third inductor.

By the same logic, these two points share the exact same potential. Since the last point is connected to terminal $D$, both of these points are effectively node $D$.

Now comes the magic of redrawing. Let's look at each inductor individually based on our new node labels.

The first inductor, $L_1$, is connected between node $A$ and node $D$.

The second inductor, $L_2$, has its left end at node $D$ and its right end at node $A$. So, it is also connected directly across $A$ and $D$.

The third inductor, $L_3$, has its left end at node $A$ and its right end at node $D$.

Suddenly, the illusion shatters! All three inductors are connected directly across the same two terminals, $A$ and $D$. This means they are perfectly in parallel.

When inductors are connected in parallel (and assuming no mutual inductance), their equivalent inductance is calculated using the reciprocal sum formula.

This is mathematically identical to how we handle resistors in parallel.

Let's substitute the given values. Each inductor has an inductance of $3\text{ H}$.

Adding these fractions is straightforward since they share a common denominator.

Taking the reciprocal of both sides, we arrive at our final answer.