Question

Two inductors of \(80 \text{ mH}\) each are joined in parallel. If the current is \(2.1 \text{ A}\), the energy stored in the combination is

• A. \(4.84 \times 10^{-2} \text{ J}\)
• B. \(7.26 \times 10^{-2} \text{ J}\)
• C. \(8.82 \times 10^{-2} \text{ J}\)
• D. \(10.85 \times 10^{-2} \text{ J}\)

Hint:

Inductors in parallel combine like resistors in parallel.

Answer 1

The Correct Option is C

Step 1: Equivalent Inductance
In parallel: $L_{eq} = \frac{L}{2} = \frac{80}{2} = 40 \text{ mH} = 0.04 \text{ H}$.

Step 2: Energy Formula
$U = \frac{1}{2} L_{eq} I^2$.

Step 3: Calculation
$U = \frac{1}{2} (0.04) (2.1)^2 = 0.02 \times 4.41$.
$U = 0.0882 \text{ J} = 8.82 \times 10^{-2} \text{ J}$.
Final Answer: (C)