Question

Two identical inductors are connected in parallel. If the parallel combination stores \(40\ \text{J}\) of magnetic field energy when a total current of \(2\ \text{A}\) is passed through it, the value of the self-inductance of each coil is

• A. \(20\ \text{H}\)
• B. \(60\ \text{H}\)
• C. \(40\ \text{H}\)
• D. \(10\ \text{H}\)

Hint:

Two identical inductors in parallel give \(L/2\); then use \(U=\tfrac{1}{2}L_{eq}I^2\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For two identical inductors \(L\) in parallel (no mutual coupling), \(L_{eq}=\dfrac{L}{2}\).

Step 2: Use the energy
Stored energy \(U=\dfrac{1}{2}L_{eq}I^{2}=\dfrac{1}{2}\cdot\dfrac{L}{2}\cdot4=L\).
Given \(U=40\ \text{J}\), so \(L=40\ \text{H}\).

Step 3: Check
Each coil carries \(1\ \text{A}\), so the total energy is \(2\times\dfrac{1}{2}\times40\times1^{2}=40\ \text{J}\), which matches.

Final Answer:
Each coil has a self-inductance of 40 H. \[ \boxed{\text{(C) }40\ \text{H}} \]