Energy stored in a coil of self-inductance 40 | vedclass

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Question

(A) The energy $U$ stored in an inductor with self-inductance $L$ carrying a current $i$ is given by the formula:$U = \frac{1}{2} L i^2$Given:$L = 40

Vedclass · Accepted Answer

$0.08$

Vedclass · Answer

(A) The energy $U$ stored in an inductor with self-inductance $L$ carrying a current $i$ is given by the formula:$U = \frac{1}{2} L i^2$Given:$L = 40 \, mH = 40 	imes 10^{-3} \, H$$i = 2 \, A$Substituting the values into the formula:$U = \frac{1}{2} 	imes (40 	imes 10^{-3}) 	imes (2)^2$$U = \frac{1}{2} 	imes 40 	imes 10^{-3} 	imes 4$$U = 20 	imes 10^{-3} 	imes 4 = 80 	imes 10^{-3} \, J$$U = 0.08 \, J$

Energy stored in a coil of self-inductance $40 \, mH$ carrying a steady current of $2 \, A$ is.....$J$

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