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Question

A Wheatstone bridge with \( R_1 = 8 \, \Omega \), \( R_2 = 16 \, \Omega \), \( R_3 = 12 \, \Omega \),
\( R_4 = 24 \, \Omega \) has a \( 10 \, \text{V} \) battery across AC and a galvanometer (\( 5 \, \Omega
\)) across BD. What is the current through the galvanometer?

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Explanation

**Meter bridge** uses uniform wire of length 1 m, balance length l gives R_unknown = R_known·l/(100-l). Principle same as Wheatstone, with wire resistances proportional to lengths, allowing unknown resistance determination from length ratio. Check balance: (R₁/R₂) = (8/16) = 0.5 , (R₃/R₄) = (12/24) = 0.5 . Bridge is balanced. Since balanced, I_g = 0 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation yields 0 A,

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