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(C) In a steady-state $D.C.$ circuit, the inductor acts as a short circuit (zero resistance).First, calculate the steady-state current $i$ flowing thr

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$25 	ext{ J}$

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(C) In a steady-state $D.C.$ circuit, the inductor acts as a short circuit (zero resistance).First, calculate the steady-state current $i$ flowing through the circuit using Ohm's law:$i = \frac{V}{R} = \frac{10 	ext{ V}}{2 	ext{ }\Omega} = 5 	ext{ A}$.The magnetic energy $U$ stored in an inductor is given by the formula:$U = \frac{1}{2} L i^2$.Substituting the given values $L = 2 	ext{ H}$ and $i = 5 	ext{ A}$:$U = \frac{1}{2} 	imes 2 	imes (5)^2 = 1 	imes 25 = 25 	ext{ J}$.Therefore, the magnetic energy stored in the coil is $25 	ext{ J}$.

In the figure, the magnetic energy stored in the coil is:

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