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Question

(D) The current $i$ flowing through the coil when connected to a $100 \, V$ battery is given by Ohm's law: $i = \frac{V}{R} = \frac{100 \, V}{10 \, \O

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$250 \, J$

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(D) The current $i$ flowing through the coil when connected to a $100 \, V$ battery is given by Ohm's law: $i = \frac{V}{R} = \frac{100 \, V}{10 \, \Omega} = 10 \, A$.The energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} L i^2$.Substituting the given values $L = 5 \, H$ and $i = 10 \, A$:$U = \frac{1}{2} 	imes 5 	imes (10)^2 = \frac{1}{2} 	imes 5 	imes 100 = 250 \, J$.Therefore, the energy stored in the coil is $250 \, J$.

$A$ coil of resistance $10 \, \Omega$ and an inductance $5 \, H$ is connected to a $100 \, V$ battery. The energy stored in the coil is:

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