Question

Two inductors of 88 mH each are joined in parallel. The current passing through this combination is 4 A . The energy stored in this combination of inductors in joule is

• A. 0.176
• B. 0.352
• C. 0.704
• D. 0.880

Hint:

Find the equivalent inductance of the parallel pair and use U = half L I squared.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Two equal inductors in parallel give \(L_{eq}=\dfrac{L}{2}\).

Step 2: Equivalent inductance
\(L_{eq}=\dfrac{88}{2}=44\ \text{mH}=0.044\ \text{H}\).

Step 3: Energy
\(U=\dfrac{1}{2}L_{eq}I^{2}=\dfrac{1}{2}\times0.044\times4^{2}\).

Step 4: Calculation
\(U=0.022\times16=0.352\ \text{J}\).

Final Answer:
The energy stored is 0.352 J. \[ \boxed{\text{(B) }0.352} \]