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: (22/44) = (15/R₄) . Solve: 0.5 = (15/R₄) ⇒ R₄ = (15/0.5) = 30 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 30 Ω,