#wire resistance

Potentiometer, Conductivity and Special Cases

A wire of length \( 8 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) has a resistance of \

**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) . Rearrange: rho = (R A/l) . Substitute: rho = (16 × 2 × 10⁻⁶/8) = 4 × 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 - I r and P = I²R, evaluation yields 4.0 × 10⁻⁶ Ω m, consistent

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

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Wheatstone Bridge and Meter Bridge

A wire has a resistance of \( 10 \, \Omega \) at \( 20^\circ \text{C} \) and \( 12 \, \Omega \) at \( 100^\circ \text{C}

**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. Use: R_t = R₀ [1 + α (T - T₀)] . Given: R₀ = 10 Ω , R_t = 12 Ω , T = 100° C , T₀ = 20° C . Substitute: 12 = 10 [1 + α (100 - 20)] . Solve: 12 = 10 + 80α ⇒ 80α = 2 ⇒ α = (2/80) = 0.025 × 10⁻² = 2.5 × 10⁻⁴ °C⁻¹ . Applying I = n e

Ref: NCERT > Physics Book > Current Electricity > Wheatstone Bridge and Meter Bridge

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Temperature Dependence of Resistance and Resistivity

A wire has a resistance of \( 35 \, \Omega \) at \( 20^\circ \text{C} \) and \( 36.4 \, \Omega \) at \( 60^\circ \text{C

**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. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: 36.4 = 35 [1 + α (60 - 20)] . Solve: 36.4 = 35 + 1400α ⇒ 1400α = 1.4 ⇒ α = (1.4/1400) = 1.0 × 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 - I r and P = I²R,

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

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Temperature Dependence of Resistance and Resistivity

A wire of length \( 10 \, \text{m} \) and cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) has a resistance of

**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Substitute: rho = (20 × 1 × 10⁻⁶/10) = 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 - I r and P = I²R, evaluation yields 2.0 × 10⁻⁶ Ω m,

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

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Electric Current, Drift Velocity and Mobility

A wire of length \( 7 \, \text{m} \) and resistance \( 14 \, \Omega \) is stretched to \( 14 \, \text{m} \). What is the

**Drift velocity** v_d = I/(n e A), I current (A), n number density of conduction electrons (m⁻³) ≈8.5×10²⁸ m⁻³ for copper, e =1.6×10⁻¹⁹ C, A cross-sectional area (m²). Typical v_d ≈10⁻⁴ m/s for 1 A in mm² wire, slow despite fast signal propagation due to electric field establishment. Volume constant: l A = l' A' ⇒ A' = (A/2) . New resistance: R' = (rho l'/A') = (rho (2l)/(A/2)) = 4 (rho l/A) = 4R = 4 × 14 = 56 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

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 \( 12 \, \text{m} \) and cross-sectional area \( 3 \times 10^{-6} \, \text{m}^2 \) has a resistance of

**Resistance** R = ρ l/A, ρ resistivity (Ω·m), l length (m), A area (m²), ρ = m/(n e² τ) from Drude model, τ average collision time. Ohm's law V = I R holds when ρ constant, independent of V. Volume constant stretching l→2l implies A→A/2, so R' = ρ·2l/(A/2)=4R, resistance quadruples when length doubles at constant volume. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Substitute: rho = (24 × 3 × 10⁻⁶/12) = 6 × 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 =