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

Let the inductance of each inductor be $$L$$ and the total current entering the circuit be $$I$$.

Since $$L_2$$ and $$L_3$$ are identical and connected in parallel, the current divides equally between them.

$$I_2=I_3=\frac{I}{2}$$

The magnetic energy stored in an inductor carrying current $$I$$ is  $$U=\frac{1}{2}LI^2$$

For $$L_1$$,  $$U_1=\frac{1}{2}LI^2$$

For $$L_2$$,  $$U_{L_2}=\frac{1}{2}L\left(\frac{I}{2}\right)^2$$

$$U_{L_2}=\frac{1}{8}LI^2$$

Similarly, for $$L_3$$,  $$U_{L_3}=\frac{1}{8}LI^2$$

Therefore, the total magnetic energy stored in the circuit is  $$U_t=U_1+U_{L_2}+U_{L_3}$$

$$U_t=\frac{1}{2}LI^2+\frac{1}{8}LI^2+\frac{1}{8}LI^2$$

$$U_t=\frac{4+1+1}{8}LI^2$$

$$U_t=\frac{3}{4}LI^2$$

Hence,  $$\frac{U_t}{U_{L_2}}=\frac{\frac{3}{4}LI^2}{\frac{1}{8}LI^2}$$

$$\frac{U_t}{U_{L_2}}=6$$

Hence, the correct answer is 6.