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(A) The magnetic energy $U$ stored in an inductor is given by the formula:$U = \frac{1}{2} LI^2$This implies that $U \propto I^2$.Let the initial curr

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(A) The magnetic energy $U$ stored in an inductor is given by the formula:$U = \frac{1}{2} LI^2$This implies that $U \propto I^2$.Let the initial current be $I_1 = 2 \ A$ and the final current be $I_2 = 3 \ A$.The ratio of the final energy $U_2$ to the initial energy $U_1$ is:$\frac{U_2}{U_1} = \left( \frac{I_2}{I_1} ight)^2 = \left( \frac{3}{2} ight)^2 = \frac{9}{4} = 2.25$Thus,$U_2 = 2.25 \ U_1$.The percentage increase in the stored energy is given by:$\Delta U \% = \left( \frac{U_2 - U_1}{U_1} 	imes 100 ight) \%$$\Delta U \% = \left( \frac{2.25 \ U_1 - U_1}{U_1} 	imes 100 ight) \% = (1.25 	imes 100) \% = 125 \%$

If the current through an inductor increases from $2 \ A$ to $3 \ A$,the magnetic energy stored in the inductor increases by: (in $\%$)

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