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(A) The equilibrium current $I$ in the coil is given by Ohm's law: $I = \frac{V}{R} = \frac{10 \, V}{4 \, \Omega} = 2.5 \, A$.The energy $E$ stored in

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$625$

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(A) The equilibrium current $I$ in the coil is given by Ohm's law: $I = \frac{V}{R} = \frac{10 \, V}{4 \, \Omega} = 2.5 \, A$.The energy $E$ stored in the magnetic field of an inductor is given by the formula: $E = \frac{1}{2} L I^2$.Substituting the given values $L = 2 \, H$ and $I = 2.5 \, A$:$E = \frac{1}{2} 	imes 2 	imes (2.5)^2 = 6.25 \, J$.To express this in the form $.......... 	imes 10^{-2} \, J$:$E = 625 	imes 10^{-2} \, J$.Thus,the value is $625$.

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

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