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 \(L\), so \(L_1 = L_2 = L_3 = L\).

Let the total current flowing through the circuit be \(I\). This current flows entirely through \(L_1\).

Since \(L_2\) and \(L_3\) are connected in parallel and have equal inductances, the current \(I\) divides equally between them. Therefore, the current through \(L_2\) is \(\dfrac{I}{2}\) and the current through \(L_3\) is \(\dfrac{I}{2}\).

The magnetic energy stored in an inductor is given by \(U = \dfrac{1}{2} L i^2\).

The energy stored in the inductor \(L_2\) is:
\(U_l = \dfrac{1}{2} L \left(\dfrac{I}{2}\right)^2 = \dfrac{1}{8} L I^2\)

The equivalent inductance of the entire circuit is:
\(L_{eq} = L_1 + \dfrac{L_2 L_3}{L_2 + L_3} = L + \dfrac{L \cdot L}{L + L} = L + \dfrac{L}{2} = \dfrac{3L}{2}\)

The total magnetic energy stored in the entire circuit is:
\(U_t = \dfrac{1}{2} L_{eq} I^2 = \dfrac{1}{2} \left(\dfrac{3L}{2}\right) I^2 = \dfrac{3}{4} L I^2\)

The ratio of the total energy to the energy stored in \(L_2\) is:
\(\dfrac{U_t}{U_l} = \dfrac{\dfrac{3}{4} L I^2}{\dfrac{1}{8} L I^2} = \dfrac{3}{4} \times 8 = 6\)

Answer: \(6\)