In the circuit shown in the figure, inductors L₁, L₂ and L₃ all have the same inductance L. Inductor L₁ is in series with a parallel combination of L₂ and L₃. Neglect mutual inductance. If U(total) is the total magnetic energy stored in the circuit and U(L₂) is the energy stored in inductor L₂, find U(total) /U(L₂).

Question

G1 Physics · Accepted Answer

Let current I flow through L₁. Because L₂ and L₃ are identical and connected in parallel, the current divides equally: I₂ = I₃ = I ÷ 2. Energy stored in an inductor is U = (1 ÷ 2)Li². Therefore, U(total) = (1 ÷ 2)LI² + (1 ÷ 2)L(I ÷ 2)² + (1 ÷ 2)L(I ÷ 2)². Thus, U(total) = (1 ÷ 2)LI² + (1 ÷ 8)LI² + (1 ÷ 8)LI² = (3 ÷ 4)LI². For L₂, U(L₂) = (1 ÷ 2)L(I ÷ 2)² = (1 ÷ 8)LI². Hence, U(total) ÷ U(L₂) = (3 ÷ 4) ÷ (1 ÷ 8) = 6. Therefore the answer is 6.