A wire has a resistance of \( 18 \, \Omega \) at \( 20^\circ \text{C} \) and \( 19.8 \, \Omega \) at \( 80^\circ \text{C
**Potential difference drop** across resistor when current flows because charges lose potential energy qV = I² R t as heat, field E = V/l drives drift, maintaining current. At very high E, velocity saturation or heating changes τ, causing non-ohmic behaviour. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: 19.8 = 18 [1 + α (80 - 20)] . Solve: 19.8 = 18 + 1080α ⇒ 1080α = 1.8 ⇒ α = (1.8/1080) ≈ 1.67 × 10⁻³ °C⁻¹ . 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
Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases