#cross‑sectional area

Potentiometer, Conductivity and Special Cases

What happens to the current in a conductor if its cross-sectional area is doubled while keeping the potential difference

**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. Resistance R = rho l / A . If A doubles, R becomes R/2 . Current I = V / R , so if R halves and V is constant, I doubles. 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 It

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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Temperature Dependence of Resistance and Resistivity

A copper wire carries \( 3.4 \, \text{A} \) with a drift speed of \( 1.0 \times 10^{-4} \, \text{m/s} \). If \( n = 8.5

**Resistivity temperature variation** ρ_t = ρ₀[1+α(T-T₀)], α ≈4×10⁻³ /°C for copper, 1.7×10⁻⁴ /°C for nichrome. Given R=60 Ω at 30°C, α=1.7×10⁻⁴ /°C, T=330°C, ΔT=300°C, R_t=60[1+1.7×10⁻⁴×300]=60×1.051=63.06 Ω, modest increase for nichrome due to small α. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (3.4/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁴) . Calculate: A = (3.4/1.36 × 10⁵) ≈ 2.5 × 10⁻⁵ m² . 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,

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

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Electric Current, Drift Velocity and Mobility

A copper wire carries \( 3 \, \text{A} \) with a drift speed of \( 9 \times 10^{-5} \, \text{m/s} \). If \( n = 8.5 \tim

**Current and drift relation** I = n e A v_d shows current proportional to drift velocity and area. For A=6×10⁻⁷ m², I=1.8 A, n=8.5×10²⁸ m⁻³, v_d =1.8/(8.5×10²⁸×1.6×10⁻¹⁹×6×10⁻⁷)=2.2×10⁻⁴ m/s, illustrating small drift speed even for ampere currents. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (3/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 9 × 10⁻⁵) . Calculate: A = (3/1.224 × 10⁵) ≈ 2.45 × 10⁻⁵ m² . 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

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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Resistance, Resistivity and Ohm's Law

A wire of length \( 3 \, \text{m} \) and cross-sectional area \( 3 \times 10^{-6} \, \text{m}^2 \) has a resistance of \

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Given: R = 5 Ω , A = 3 × 10⁻⁶ m² , l = 3 m . Substitute: rho = (5 × 3 × 10⁻⁶/3) = 5 × 10⁻⁶ Ω m . Applying I = n e A v_d, R = ρ l/A,

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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