Practice question

**Wheatstone bridge balance** condition R₁/R₂ = R₃/R₄, R₄ = R₂ R₃/R₁, when galvanometer current zero, potentials at midpoints equal. At balance, no current through galvanometer, enabling precise resistance measurement independent of source voltage. No current flows when the bridge is balanced, i.e., R₁ / R₂ = R₃ / R₄ , making the potentials at the galvanometer’s nodes equal, so the potential difference across it is zero. 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 Ratio of resistances in opposite arms is equal,