Question
Download Solution PDFA current of 4.0 A is maintained in a coil of self-inductance 8.0 mH. The energy stored in the coil is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Self-inductance:
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Self-inductance is the property of the current-carrying coil that resists or opposes the change of current flowing through it.
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This occurs mainly due to the self-induced emf produced in the coil itself.
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The emf induced in the coil is given by, \(e=-L\frac{dI}{dt}\) where L = self-inductance
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The energy stored in the coil is, \(E=\frac{1}{2}LI^2\) where I = current
Calculation:
Given,
The current, I = 4.0 A, self-inductance of a coil, L = 8.0 mH
The energy stored in the coil is calculated by the given formula,
\(E=\frac{1}{2}LI^2\)
\(E=\frac{1}{2}\times 8\times 10^{-3}\times 4^2 \)
E = 64 mJ
Last updated on May 26, 2025
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