current electricity electrical resistivity is an important Class 12 Physics topic where students learn how resistance depends on material, length, area, temperature and conductivity.

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Current Electricity | Temperature Dependence | Resistance

Current Electricity - Temperature Dependence of Resistance

current electricity temperature dependence of resistance is explained with formula sheets, derivations, temperature coefficient of resistance, resistivity, conductivity, error-based questions, SVG graphs and exam-level practice for CBSE Class 12, NEET, JEE Main, JEE Advanced, AP, IB, IGCSE, A-Level and Olympiad Physics.

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If you are facing difficulty understanding Temperature Dependence of Resistance, Temperature Coefficient of Resistance, Resistivity, Conductivity, Error-Based Questions or IIT-JEE level Current Electricity concepts, contact Kumar Sir for one-to-one Physics guidance.
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1. Formula Sheet First

Rt = R0(1 + αΔT)Resistance at temperature t.
Rt = R0(1 + αT)When initial temperature is 0°C.
ΔR = R0αΔTChange in resistance.
ΔR/R0 = αΔTFractional change.
α = ΔR/(R0ΔT)Temperature coefficient.
ρt = ρ0(1 + αΔT)Temperature dependence of resistivity.
σ = 1/ρConductivity is reciprocal of resistivity.
% change = (ΔR/R0) × 100Percentage resistance change.
ΔR/R = Δρ/ρ + ΔL/L - ΔA/AError-type relation.
ΔR/R = αΔTTemperature-only change.
[α] = K-1SI unit of α is per kelvin.
dim(α) = Θ-1Dimension of temperature coefficient.
Symbols: Rt = final resistance, R0 = initial resistance, α = temperature coefficient, ΔT = temperature change, ρ = resistivity, σ = conductivity, ΔR = resistance change.
Unit note: Since ΔR/R0 is dimensionless, α has unit °C-1 or K-1.

2. What Is Temperature Dependence of Resistance?

Temperature changes the motion of atoms and charge carriers. In metals, lattice ions vibrate more strongly at higher temperature, electron collisions increase, relaxation time decreases, drift velocity becomes harder to maintain and conductivity decreases.

Atomic explanation: Heating increases lattice vibrations, so electrons collide more frequently.
Relaxation time: More collisions reduce relaxation time and mobility.
Conductivity: Since σ = ne²τ/m, lower τ makes metallic conductivity decrease.

3. Temperature Coefficient of Resistance

Temperature coefficient α is the fractional change in resistance per degree change in temperature.

1
Fractional change in resistance = ΔR/R0.
2
Per degree temperature change means divide by ΔT.
3
Therefore α = ΔR/(R0ΔT).

4. Derivation of Rt = R0(1 + αΔT)

1
Start with definition: α = ΔR/(R0ΔT).
2
Multiply both sides by R0ΔT: ΔR = R0αΔT.
3
Since ΔR = Rt - R0, write Rt - R0 = R0αΔT.
4
Add R0 to both sides: Rt = R0 + R0αΔT.
5
Taking R0 common gives Rt = R0(1 + αΔT).

5. Metals, Semiconductors and Insulators

MaterialTemperature BehaviourCoefficientConductivity
MetalsResistance increases with temperature.Positive α.Conductivity decreases.
SemiconductorsResistance decreases strongly with temperature.Negative temperature coefficient.Conductivity increases strongly.
InsulatorsVery high resistance; may conduct at high temperature.Generally negative in high-temperature conduction range.Very small but may increase.

6. High Resistivity and Low Temperature Coefficient Materials

Manganin: low α, used in resistance boxes and standard resistors.
Constantan: nearly constant resistance, useful in measuring instruments.
Nichrome: high resistivity and high melting point, used in heaters.
Tungsten: high melting point, used in filament lamps.
Copper: low resistivity, used in connecting wires.
Silver: very low resistivity but expensive, used only in special applications.

7. Resistivity and Temperature

Resistance depends on material and dimensions: R = ρL/A. If length and area remain almost constant, change in resistance mainly follows change in resistivity.

1
Rt = ρtL/A and R0 = ρ0L/A when L and A are unchanged.
2
So Rt/R0 = ρt/ρ0.
3
Therefore ΔR/R = Δρ/ρ approximately.

8. Error-Based Questions

From R = ρL/A, small relative changes give: ΔR/R = Δρ/ρ + ΔL/L - ΔA/A. If only temperature changes, ΔR/R = αΔT.

Example 1: If α = 0.004 K-1 and ΔT = 25 K, then ΔR/R = 0.1 and percentage change = 10%.
Example 2: If R0 = 20 Ω and resistance increases by 2 Ω for ΔT = 50 K, α = 2/(20 × 50) = 0.002 K-1.

9. Graphical Representation

Metal: R vs TTRpositive slope Semiconductor: R vs TTRnegative α Resistivity vs TTρ Conductivity vs TTσ Positive αmetals Negative αsemiconductors

10. Applications

Resistance thermometer: uses predictable change of resistance with temperature.
Platinum thermometer: stable, nearly linear, accurate.
Thermistors: sensitive semiconductor devices for temperature sensing.
Heating elements: nichrome is preferred because of high resistivity and heat tolerance.
Fuse wire: selected to melt safely at excessive current.
Resistance coils: alloys keep resistance nearly constant in laboratory use.

11-17. Exam Question Bank With Accordion Solutions

Click any question to open the answer and explanation.

18. Common Student Mistakes

19. Final Revision Sheet

Main formula: Rt = R0(1 + αΔT).
Coefficient: α = ΔR/(R0ΔT), unit K-1.
Metals: positive α; resistance increases with temperature.
Semiconductors: negative α; resistance decreases with temperature.
Error relation: ΔR/R = Δρ/ρ + ΔL/L - ΔA/A.
Shortcut: percentage change = αΔT × 100.

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