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(C) The energy stored in an inductor (coil) is given by the formula $E = \frac{1}{2} L I^2$,where $L$ is the self-inductance and $I$ is the current.Le

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Decrease by $75 \%$

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(C) The energy stored in an inductor (coil) is given by the formula $E = \frac{1}{2} L I^2$,where $L$ is the self-inductance and $I$ is the current.Let the initial current be $I_1 = I$ and the initial energy be $E_1 = \frac{1}{2} L I^2$.If the current is reduced by $50 \%$,the new current becomes $I_2 = I - 0.5 I = 0.5 I = \frac{I}{2}$.The new energy stored in the coil is $E_2 = \frac{1}{2} L (I_2)^2 = \frac{1}{2} L (\frac{I}{2})^2 = \frac{1}{2} L (\frac{I^2}{4}) = \frac{1}{4} E_1$.The change in energy is $\Delta E = E_1 - E_2 = E_1 - \frac{1}{4} E_1 = \frac{3}{4} E_1$.Since $\frac{3}{4} = 0.75$,the energy decreases by $75 \%$.

If the current flowing through a coil is reduced by $50 \%$,then the energy stored in the coil will

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