Two ideal inductors of inductances L₁ and L₂ are connected in parallel. A time-varying current flows into this parallel combination. If the rate of change of current in the first inductor is x and the rate of change of current in the second inductor is y , and it is given that the ratio x/y = 3 , what is the ratio of their inductances L₁/L₂ ? (Neglect mutual inductance)

Step-by-step solution

For inductors connected in parallel, the potential difference (induced emf) across each branch must be identical. The induced emf across an inductor is given by E = L dI dt . Therefore, we can write: E₁ = E₂ L₁ ( di₁ dt ) = L₂ ( di₂ dt ) Given that the rate of change of current in the first inductor is x and in the second is y , we substitute these values: L₁ x = L₂ y Rearranging to find the ratio of inductances: L₁ L₂ = y x Since we are given x/y = 3 , the inverse ratio is: y x = 1 3 Thus, L₁ L₂ = 1/3 . Answer: 1/