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(A) The energy $E$ stored in a coil with self-inductance $L$ carrying current $I$ is given by the formula: $E = \frac{1}{2} LI^2$.Let the initial curr

Vedclass · Accepted Answer

$E_2 = \frac{E_1}{4}$

Vedclass · Answer

(A) The energy $E$ stored in a coil with self-inductance $L$ carrying current $I$ is given by the formula: $E = \frac{1}{2} LI^2$.Let the initial current be $I_1$ and the initial energy be $E_1 = \frac{1}{2} LI_1^2$.Let the new current be $I_2 = \frac{I_1}{2}$ and the new energy be $E_2 = \frac{1}{2} LI_2^2$.Taking the ratio of the two energies:$\frac{E_2}{E_1} = \frac{\frac{1}{2} LI_2^2}{\frac{1}{2} LI_1^2} = \left( \frac{I_2}{I_1} \right)^2$.Substituting $I_2 = \frac{I_1}{2}$:$\frac{E_2}{E_1} = \left( \frac{I_1 / 2}{I_1} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.Therefore,$E_2 = \frac{E_1}{4}$.

If a current flowing in a coil is reduced to half of its initial value,the relation between the new energy $(E_2)$ and the original energy $(E_1)$ stored in the coil will be

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