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Asked in GSEB Board March 2019 · Coil geometry, cores and combinations
Given: five identical inductors, L = 3 H, forming a bridge: A–C and C–B on top, A–D and D–B below, and the fifth between C and D.
Idea: the bridge is balanced — the arms on the two sides are in the same ratio (3:3 = 3:3) — so C and D are always at the same potential.
With no emf across it, the C–D inductor carries no current and can be removed.
Upper route A–C–B: 3 + 3 = 6 H in series. Lower route A–D–B: also 6 H.
Two 6 H routes in parallel: 1/(L_eq) = 1/6 + 1/6, so L_eq = 3 H.
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