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(D) The energy stored in an $LR$ circuit is given by the formula:$E = \frac{1}{2} LI^2$ ...$(i)$Given that the current is reduced to half,the new curr

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$(1/4)^{	ext{th}}$ times

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(D) The energy stored in an $LR$ circuit is given by the formula:$E = \frac{1}{2} LI^2$ ...$(i)$Given that the current is reduced to half,the new current $I^{\prime}$ is:$I^{\prime} = \frac{I}{2}$Substituting this into the energy formula,the new energy $E^{\prime}$ is:$E^{\prime} = \frac{1}{2} L (I^{\prime})^2$$E^{\prime} = \frac{1}{2} L \left(\frac{I}{2}ight)^2$$E^{\prime} = \frac{1}{2} L \left(\frac{I^2}{4}ight)$$E^{\prime} = \frac{1}{4} \left(\frac{1}{2} LI^2ight)$Using equation $(i)$,we get:$E^{\prime} = \frac{1}{4} E$Therefore,the energy stored becomes $(1/4)^{	ext{th}}$ of the original energy.

If the current in an $LR$ circuit is reduced to half,what will be the energy stored in it?

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