An induction coil stores 32 J of magnetic en | vedclass

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(A) Let the inductance of the coil be $L$ and its resistance be $R$.The magnetic energy stored in the inductor is given by $U = \frac{1}{2} L I^2$.Giv

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$0.20$

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(A) Let the inductance of the coil be $L$ and its resistance be $R$.The magnetic energy stored in the inductor is given by $U = \frac{1}{2} L I^2$.Given $U = 32 \ J$ and $I = 4 \ A$,we have:$32 = \frac{1}{2} 	imes L 	imes (4)^2$$32 = \frac{1}{2} 	imes L 	imes 16$$32 = 8L \Rightarrow L = 4 \ H$.The power dissipated as heat is given by $P = I^2 R$.Given $P = 320 \ W$ and $I = 4 \ A$,we have:$320 = (4)^2 	imes R$$320 = 16 	imes R$$R = \frac{320}{16} = 20 \ \Omega$.The time constant $	au$ of an $RL$ circuit is defined as $	au = \frac{L}{R}$.Substituting the values:$	au = \frac{4}{20} = 0.2 \ s$.

An induction coil stores $32 \ J$ of magnetic energy and dissipates energy as heat at the rate of $320 \ W$ when a current of $4 \ A$ is passed through it. Find the time constant of the circuit when the coil is joined across a battery.

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