An inductor coil stores 64 , J of magnetic fi | vedclass

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(D) The energy stored in an inductor is given by $U = \frac{1}{2} L i^2$. Given $U = 64 \, J$ and $i = 8 \, A$,we have $64 = \frac{1}{2} 	imes L 	im

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$0.2$

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(D) The energy stored in an inductor is given by $U = \frac{1}{2} L i^2$. Given $U = 64 \, J$ and $i = 8 \, A$,we have $64 = \frac{1}{2} 	imes L 	imes (8)^2$. This simplifies to $64 = 32L$,so $L = 2 \, H$.The rate of energy dissipation (power) is given by $P = i^2 R$. Given $P = 640 \, W$ and $i = 8 \, A$,we have $640 = (8)^2 	imes R$. This simplifies to $640 = 64R$,so $R = 10 \, \Omega$.The time constant $	au$ of an $LR$ circuit is given by $	au = \frac{L}{R}$. Substituting the values,$	au = \frac{2}{10} = 0.2 \, s$.

An inductor coil stores $64 \, J$ of magnetic field energy and dissipates energy at the rate of $640 \, W$ when a current of $8 \, A$ is passed through it. If this coil is joined across an ideal battery,find the time constant of the circuit in seconds.

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