Practice question

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Resistance is given by R = (rho l/A) . Rearrange: rho = (R A/l) . Given: R = 2 Ω , A = 1 × 10⁻⁶ m² , l = 1 m . Substitute: rho = (2 × 1 × 10⁻⁶/1) = 2 × 10⁻⁶ Ω m . 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