Ratio of Total Magnetic Energy to Energy in Single Inductor

To find the ratio of the total magnetic energy stored in the circuit ($U_t$) to the magnetic energy stored in the $L_2$ inductor ($U_l$), we analyze the current distribution through each inductor.

Let the equal inductance values be: $L_1 = L_2 = L_3 = L$

Let $I$ be the main current entering the circuit.

1. Current through each inductor:

   • The main current $I$ flows directly through inductor $L_1$. Therefore: $I_1 = I$
   • Since $L_2$ and $L_3$ are connected in parallel and have identical inductances ($L_2 = L_3 = L$), the current $I$ splits equally between them: $I_2 = I_3 = \frac{I}{2}$

2. Energy stored in inductor $L_2$ ($U_l$): $U_l = U_2 = \frac{1}{2} L_2 I_2^2 = \frac{1}{2} L \left(\frac{I}{2}\right)^2 = \frac{1}{8} L I^2$

3. Total energy stored in the entire circuit ($U_t$): The energy stored in each individual inductor is: $U_1 = \frac{1}{2} L_1 I_1^2 = \frac{1}{2} L I^2$ $U_2 = \frac{1}{8} L I^2$ $U_3 = \frac{1}{2} L_3 I_3^2 = \frac{1}{2} L \left(\frac{I}{2}\right)^2 = \frac{1}{8} L I^2$

   Summing these gives the total magnetic energy: $U_t = U_1 + U_2 + U_3 = \frac{1}{2} L I^2 + \frac{1}{8} L I^2 + \frac{1}{8} L I^2 = \frac{3}{4} L I^2$

4. Ratio of total energy to energy in $L_2$ ($U_t / U_l$): $\frac{U_t}{U_l} = \frac{\frac{3}{4} L I^2}{\frac{1}{8} L I^2} = \frac{3}{4} \times 8 = 6$