Three pure inductors each of inductance $6\text{ H}$ are connected as shown in the figure. Their equivalent inductance between the points 'P' and 'Q' is
Choose the correct answer from the options given below
Show Hint
For $n$ identical components connected in parallel, the equivalent value is simply the individual component value divided by the total number of branches ($R/n$ or $L/n$). Here, we can quickly compute $L_{eq} = \frac{6\text{ H}}{3} = 2\text{ H}$ in one step.
Step 1: Understanding the Question:
The problem shows a network of three identical inductors, each with an inductance $L_1 = L_2 = L_3 = 6\text{ H}$, connected by cross-over wiring paths. We need to find the total equivalent inductance between terminals P and Q.
Step 2: Key Formula or Approach:
We can analyze the circuit by labeling the nodes to find the electrical potential connections. If all three inductors share the same pair of node potentials across their terminals, they are configured in a parallel arrangement.
The equivalent inductance $L_{eq}$ for three inductors connected in parallel is calculated using the reciprocal formula:
$$\frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}$$
Step 3: Detailed Explanation:
Let's trace the node paths through the connecting wires:
Terminal P connects directly to the left side of the first inductor. The top cross-over wire extends this node P directly to the junction between the second and third inductors. Therefore, one terminal of all three inductors is connected to node P.
Terminal Q connects directly to the right side of the third inductor. The bottom cross-over wire extends this node Q directly to the junction between the first and second inductors. Therefore, the other terminal of all three inductors is connected to node Q.
Since each of the three inductors has one side connected to node P and the opposite side connected to node Q, all three inductors are connected in
parallel across terminals P and Q.
Substitute $6\text{ H}$ for each inductor into the parallel combination formula:
$$\frac{1}{L_{eq}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6}$$
Invert the fraction to solve for $L_{eq}$:
$$\frac{1}{L_{eq}} = \frac{1}{2} \implies L_{eq} = 2\text{ H}$$
Step 4: Final Answer:
The total equivalent inductance of the network is $2\text{ H}$, which matches option (D).
