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. Balance condition: (R₁/R₂) = (R₃/R₄) . Substitute: (14/28) = (10/R₄) . Solve: 0.5 = (10/R₄) ⇒ R₄ = (10/0.5) = 20 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 20 Ω,