#electrical resistance

Electric Current, Drift Velocity and Mobility

What is the primary source of energy dissipation in a resistor carrying current?

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Energy dissipation in a resistor occurs as electrons collide with lattice ions, transferring kinetic energy gained from the electric field into thermal energy (heat) via lattice vibrations. 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 Collisions with

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

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Temperature Dependence of Resistance and Resistivity

A wire has a resistance of \( 25 \, \Omega \) at \( 20^\circ \text{C} \) and \( 27.5 \, \Omega \) at \( 90^\circ \text{C

**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. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: 27.5 = 25 [1 + α (90 - 20)] . Solve: 27.5 = 25 + 1750α ⇒ 1750α = 2.5 ⇒ α = (2.5/1750) ≈ 1.43 × 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

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

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Resistance, Resistivity and Ohm's Law

A wire of length \( 4 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) 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 = (8 × 2 × 10⁻⁶/4) = 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 =

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

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Potentiometer, Conductivity and Special Cases

What causes the potential difference across a resistor to drop when current flows through it?

**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. Current through a resistor ( I = V / R ) causes a potential drop ( V = I R ) as electrical energy is converted to thermal energy due to collisions with lattice ions. 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 > Potentiometer, Conductivity and Special Cases

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PHYSICS

A wire of length 6 m and cross-sectional area 5 × 10⁻⁶ m² has a resistance of 12 Ω . What is the resistivity of t

Given: A wire of length 6 m and cross-sectional area 5 × 10⁻⁶ m² has a resistance of 12 Ω . What is the resistivity of the material? These values define the system as per NCERT data. Formula: Resistance: R = rho l/A. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Rearrange: rho = R A/l . Substitute: rho = frac12 × 5 × 10⁻⁶⁶= 10 × 10⁻⁶= 1.0 × 10⁻⁵Ω m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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