In the given circuit below,the inductance val | vedclass

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(D) Let $L_1 = L_2 = L_3 = L$.
The inductors $L_2$ and $L_3$ are connected in parallel,so their equivalent inductance $L_{23}$ is given by $\frac{1}{

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$6$

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(D) Let $L_1 = L_2 = L_3 = L$.
The inductors $L_2$ and $L_3$ are connected in parallel,so their equivalent inductance $L_{23}$ is given by $\frac{1}{L_{23}} = \frac{1}{L_2} + \frac{1}{L_3} = \frac{1}{L} + \frac{1}{L} = \frac{2}{L}$,which means $L_{23} = \frac{L}{2}$.
The total inductance of the circuit is $L_t = L_1 + L_{23} = L + \frac{L}{2} = \frac{3L}{2}$.
The total magnetic energy stored in the circuit is $U_t = \frac{1}{2} L_t I^2 = \frac{1}{2} (\frac{3L}{2}) I^2 = \frac{3}{4} LI^2$,where $I$ is the total current flowing through the circuit.
Since $L_2$ and $L_3$ are in parallel and have equal inductance,the current $I$ splits equally between them. Thus,the current through $L_2$ is $I_2 = I/2$.
The magnetic energy stored in the $L_2$ inductor is $U_l = \frac{1}{2} L_2 I_2^2 = \frac{1}{2} L (I/2)^2 = \frac{1}{2} L (I^2/4) = \frac{1}{8} LI^2$.
Therefore,the ratio $U_t / U_l = (\frac{3}{4} LI^2) / (\frac{1}{8} LI^2) = \frac{3}{4} 	imes 8 = 6$.

In the given circuit below,the inductance values of $L_1$,$L_2$,and $L_3$ are the same. The magnetic energy stored in the entire circuit is $U_t$ and that stored in the $L_2$ inductor is $U_l$. The ratio $U_t/U_l$ is . . . . . . . (Ignore the mutual inductance,if any.)

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