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

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{E_1}{E_2}=\frac{\frac{1}{2} L\left(\frac{I}{2}ight)^2}{\frac{1}{2} L^2}=\frac{1}{4} \ & E_2=\frac{1}{4} \mathrm{E} 1\end{aligned}$ $	herefore \mathrm{E}_1-\mathrm{E}_2=\frac{3}{4} \mathrm{E}_1$.

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

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