Two different coils have self-inductance L_1 | vedclass

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(D) Given: $L_1 = 8 \, mH$,$L_2 = 2 \, mH$,and $\frac{di_1}{dt} = \frac{di_2}{dt} = k$ (constant).$1$. Induced voltage: $V = L \frac{di}{dt}$. Since $

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(D) Given: $L_1 = 8 \, mH$,$L_2 = 2 \, mH$,and $\frac{di_1}{dt} = \frac{di_2}{dt} = k$ (constant).$1$. Induced voltage: $V = L \frac{di}{dt}$. Since $\frac{di}{dt}$ is the same for both,$\frac{V_2}{V_1} = \frac{L_2}{L_1} = \frac{2}{8} = \frac{1}{4}$. Thus,option $(b)$ is correct.$2$. Power supplied: $P = V \cdot i$. Given $P_1 = P_2$,so $V_1 i_1 = V_2 i_2$. Therefore,$\frac{i_1}{i_2} = \frac{V_2}{V_1} = \frac{1}{4}$. Thus,option $(a)$ is correct.$3$. Energy stored: $W = \frac{1}{2} L i^2$. The ratio is $\frac{W_2}{W_1} = \left( \frac{L_2}{L_1} ight) \left( \frac{i_2}{i_1} ight)^2 = \left( \frac{1}{4} ight) \left( 4 ight)^2 = \frac{1}{4} 	imes 16 = 4$. Thus,option $(c)$ is correct.Since $(a)$,$(b)$,and $(c)$ are all correct,the answer is $(d)$.

List-$I$	List-$II$
$A$. Magnetic induction	$I$. $MLT^{-2}A^{-2}$
$B$. Magnetic flux	$II$. $ML^2T^{-2}A^{-2}$
$C$. Magnetic permeability	$III$. $ML^0T^{-2}A^{-1}$
$D$. Self inductance	$IV$. $ML^2T^{-2}A^{-1}$

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Match the List-$I$ with List-$II$:
List-$I$ List-$II$
$A$. Magnetic induction $I$. $MLT^{-2}A^{-2}$
$B$. Magnetic flux $II$. $ML^2T^{-2}A^{-2}$
$C$. Magnetic permeability $III$. $ML^0T^{-2}A^{-1}$
$D$. Self inductance $IV$. $ML^2T^{-2}A^{-1}$

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