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. Balance condition: (R₁/R₂) = (R₃/R₄) . Substitute: (16/32) = (20/R₄) . Solve: 0.5 = (20/R₄) ⇒ R₄ = (20/0.5) = 40 Ω . 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 40 Ω,