Resistance and temperature

By Elsie Porter,2014-05-09 22:08
6 views 0
Resistance and temperature

Resistance and temperature

    When a material is heated its resistance will change. This is due to the thermal motion of the atoms within the specimen

The equation for this variation is:

     2R = R[1+ αθ + βθ + . . . ] θo

     owhere R is the resistance of the specimen at some temperature θ C and R the θooresistance at 0C. In this equation β<<α and so we can express the change by the following simplified equation as long as the temperature change is not too great.

    R = R[1+ αθ] θo

Here α is called the temperature coefficient of resistance and is defined as the increase in oresistance per degree rise divided by the resistance at 0 C

    α = [R - R]Rθ θoo

Some values of the temperature coefficient of resistance (α) -4-1-4-1-4-1copper 43 x10 K tungsten 60 x10 K gold 36x10K -4-1-4-1-4-1nichrome 0.88x10 K carbon -5.1x10K steel 33x10 K

    For a metal the temperature coefficient of resistance is positive - in other words and increase in the temperature gives an increase in resistance. This can be explained by the motion of the atoms and free electrons within the solid. At low temperatures the thermal vibration is small and electrons can move easily within the lattice but at high temperatures the motion increases giving a much greater chance of collisions between the conduction electrons and the lattice and so impeding their motion. In a light bulb the filament is at oabout 2700 C when it is working and its resistance when hot is about ten times that when cold. (For a typical domestic light bulb the resistance measured at room temperature was 32 Ω and this rose to 324 Ω at its working temperature).

    Example problem ooIf the resistance of a length of copper wire is 4.5 ; at 20 C calculate its resistance at 60 C. -4Using : R = R[1+ αθ] we have 4.5 = R[1+ 43x10 x 20] therefore θ oo

    R = 4.5/1.086 = 4.14 Ω and so o

     -4R = 4.14[1+ 43x10 x 60] = 4.14x1.258 = 5.21 Ω 60

    We can also define the change in the resistivity with temperature by an equation similar to

    ρ = ρ[1 + βθ] θo

     that for resistance:


where β is the temperature coefficient of resistivity.

    We require that the variation of resistance should be small so β should be as small as

    possible for thermal stability.

    The following table gives the temperature coefficients of resistivity for a number of materials:

     -4-1-4-1Material βx10 K Material βx10 K

    Copper 43 Aluminium 38

    Lead 43 Nichrome 1.7

    Eureka 0.2 Manganin 0.2

    Iron 62 Platinum 38

    Carbon - 0.5 Tungsten 60

    However in non-metals such as semiconductors an increase in temperature leads to a drop in resistance. This can be explained by electrons gaining energy and moving into the conduction band - in fact changing from being bound to a particular atom to being able to move freely - an increase in the number of free electrons. The temperature coefficient of resistance and also that of the temperature coefficient of resistivity is therefore negative. Problems

    Where necessary use the values for the temperature coefficient of resistance quoted in the preceding


    1. Calculate the change in the resistance of a gold wire if the temperature rises from 293 K to 315 K if its resistance at 293 K was 2.5 Ω.

     oo2. A tungsten filament has a resistance of 20 Ω at 20 C. What is its resistance at 1500 C

     o3. A tungsten filament lamp has a filament whose resistance increases from 30 Ω at 20 C to 350 Ω at its

    operating temperature. Calculate the operating temperature of the lamp.

     o4. A coil of wire has a resistance of 3.5 Ω at room temperature (18 C) which rises to 6.5 W when placed

    in boiling water. Calculate the temperature coefficient of resistance of the metal of the coil.

     o5. A semiconductor used as a thermometer has a resistance of 2 kW at 15 C and this falls to 25 Ω at 100 oC. Calculate the mean temperature coefficient for the material over that range.

(See also superconductivity)


Report this document

For any questions or suggestions please email