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: Understanding the Question:
Two identical inductors are in parallel. We need to find the total energy stored for a given total current.
Step 2: Key Formula or Approach:
1) Equivalent inductance for parallel inductors: \(\frac{1}{L_p} = \frac{1}{L_1} + \frac{1}{L_2}\).
2) Energy stored in an inductor: \(U = \frac{1}{2} L I^2\).
Step 3: Detailed Explanation:
Given: \(L_1 = L_2 = 80 \text{ mH} = 80 \times 10^{-3} \text{ H}\).
Current \(I = 2.1 \text{ A}\).
Equivalent inductance \(L_p\):
\[ L_p = \frac{L_1 \cdot L_2}{L_1 + L_2} = \frac{80 \times 80}{80 + 80} = \frac{6400}{160} = 40 \text{ mH} \]
\[ L_p = 40 \times 10^{-3} \text{ H} \]
Total energy stored \(U\):
\[ U = \frac{1}{2} L_p I^2 = \frac{1}{2} (40 \times 10^{-3}) \times (2.1)^2 \]
\[ U = 20 \times 10^{-3} \times 4.41 \]
\[ U = 88.2 \times 10^{-3} \text{ J} \]
\[ U = 8.82 \times 10^{-2} \text{ J} \]
Step 4: Final Answer:
The energy stored is \(8.82 \times 10^{-2} \text{ J}\).