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Asked in GUJCET 2026 · Wheatstone bridge
Given: from the figure AB=2 Ω, BC=2 Ω, AD=2 Ω, and between D and C the unknown X with 6 Ω in parallel across it; the galvanometer lies on BD and the cell of emf 6 V across A and C.
Idea: at balance the galvanometer carries no current, so (AB)/(BC)=(AD)/(DC). The emf never enters; balance depends on the four arms alone.
2/2=2/(DC), so the whole DC arm must be 2 Ω.
That arm is X in parallel with 6 Ω: 1/X+1/6=1/2, so 1/X=1/3.
X=3 Ω.
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