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Question

In a Wheatstone bridge, if the bridge is unbalanced, what determines the direction of 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. When unbalanced ( R₁ / R₂ neq R₃ / R₄ ), the potentials at the galvanometer’s nodes differ. The current flows from the higher-potential node to the lower one, determined by the relative voltage drops across the bridge arms. 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 Potential difference

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