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(B) The energy stored in an inductor is given by the formula $U = \frac{1}{2} L i^2$.When a coil is connected to a battery of $EMF$ $E$ and resistance

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$8$

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(B) The energy stored in an inductor is given by the formula $U = \frac{1}{2} L i^2$.When a coil is connected to a battery of $EMF$ $E$ and resistance $R$,the steady-state current $i$ is given by $i = \frac{E}{R}$.Substituting the values: $E = 10 \ V$,$L = 4 \ H$,and $R = 5 \ \Omega$.First,calculate the steady-state current: $i = \frac{10 \ V}{5 \ \Omega} = 2 \ A$.Now,calculate the stored energy: $U = \frac{1}{2} 	imes 4 \ H 	imes (2 \ A)^2$.$U = 2 	imes 4 \ J = 8 \ J$.

$A$ coil of resistance $5 \Omega$ and inductance $4 H$ is connected to a $10 \ V$ battery. The energy stored in the coil is: (in $J$)

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