Question:
If an inductor coil of self-inductance 2 H stores 25 J of magnetic energy, then the current I passing through it is:
If an inductor coil of self-inductance 2 H stores 25 J of magnetic energy, then the current I passing through it is:
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This is the magnetic equivalent of capacitive energy $1/2 CV^2$.
Updated On: Apr 28, 2026
- 25 A
- 10 A
- 15 A
- 2 A
- 5 A
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The Correct Option is
Solution and Explanation
Step 1: Concept
Magnetic energy $U = \frac{1}{2} L I^2$.
Step 2: Analysis
Given $U = 25\ J$ and $L = 2\ H$.
Step 3: Calculation
$25 = \frac{1}{2} (2) I^2 \implies 25 = I^2 \implies I = 5\ A$. Final Answer: (E)
Magnetic energy $U = \frac{1}{2} L I^2$.
Step 2: Analysis
Given $U = 25\ J$ and $L = 2\ H$.
Step 3: Calculation
$25 = \frac{1}{2} (2) I^2 \implies 25 = I^2 \implies I = 5\ A$. Final Answer: (E)
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