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A coil has an inductance of 2 \mathrm{H} and resistance of 4 \Omega. A 10 \mathrm{~V} is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be_____ \times 10^{-2} \mathrm{~J}.  

Option: 1

625


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

At steady state, inductor will act as a short circuit.

\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{10}{4}=\frac{5}{2} \mathrm{~A}
\mathrm{E}=\frac{1}{2} \mathrm{LI}^{2}=\frac{1}{2} \times 2 \times\left(\frac{5}{2}\right)^{2}=\frac{25}{4}=6.25=625 \times 10^{-2} \mathrm{~J}


 

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