Two identical inductors, each of inductance L, are connected in configurations P and Q. In P, they are in series and points a and b span only the upper inductor; in Q they are in parallel and points a and b span the parallel pair. The same time-varying total current I(t) flows into each configuration. Neglect mutual inductance. Find E(P)/E(Q). Image
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Question

Two identical inductors, each of inductance L, are connected in configurations P and Q. In P, they are in series and points a and b span only the upper inductor; in Q they are in parallel and points a and b span the parallel pair. The same time-varying total current I(t) flows into each configuration. Neglect mutual inductance. Find E(P)/E(Q). Image

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G1 Physics · Accepted Answer

Step 1: In P, the full input current I(t) passes through the upper inductor. Thus the emf magnitude between a and b is E(P) = L|dI/dt|. Step 2: In Q, equal inductors in parallel carry I/2 each and have equivalent inductance L/2. Step 3: Their terminal emf magnitude is E(Q) = (L/2)|dI/dt|. Step 4: Ratio E(P)/E(Q) = L/(L/2) = 2. Final answer: 2 (Option D).