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Asked in NEET 2025 · Wheatstone bridge
The battery, the 1/3 Ω, 5.5 Ω and 1.5 Ω resistors are in series and feed the network between B and E.
Between B and E: B to A is 5 Ω; B to D is 2.5 Ω (C to D is a plain wire); A to E is 3 Ω (A to F is a wire); D to E is 1.5 Ω; and 6 Ω joins A to D.
5/3=(2.5)/(1.5), so this is a balanced Wheatstone bridge and the 6 Ω carries no current.
R_BE=((5+3)(2.5+1.5))/((5+3)+(2.5+1.5))=(8×4)/(12)=8/3 Ω.
Total: 8/3+1/3+1.5+5.5=10 Ω.
I=5/(10)=0.5 A.
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