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Asked in CBSE AIPMT 2006 · Wheatstone bridge
Battery: long plate on the left, so the left corner is at V and the right corner at 0.
Upper branch 4 Ω–4 Ω: A sits halfway, V_A=V/2.
Lower branch 1 Ω–3 Ω: the 1 Ω drops a quarter of V, so V_B=V-V/4=(3V)/4.
V_B>V_A, and 4/4 eq1/3 so the bridge is not balanced.
Current in the wire flows from B to A.
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