3.8 Temperature Dependence of Resistivity

The resistivity of a material is found to be dependent on the temperature. Different materials do not exhibit the same dependence on temperatures. Over a limited range of temperatures, that is not too large, the resistivity of a metallic conductor is approximately given by,
\(
\rho_{ T }=\rho_0\left[1+\alpha\left(T-T_0\right)\right] \dots(3.26)
\)
where \(\rho_{ T }\) is the resistivity at a temperature \(T\) and \(\rho_0\) is the same at a reference temperature \(T_0 . \alpha\) is called the temperature coefficient of resistivity, and from Eq. (3.26), the dimension of \(\alpha\) is (Temperature) \({ }^{-1}\). For metals, \(\alpha\) is positive.

The relation of Eq. (3.26) implies that a graph of \(\rho_{ T }\) plotted against \(T\) would be a straight line. At temperatures much lower than \(0^{\circ} C\), the graph, however, deviates considerably from a straight line (Fig. 3.8).
Equation (3.26) thus, can be used approximately over a limited range of \(T\) around any reference temperature \(T_0\), where the graph can be approximated as a straight line.

Some materials like Nichrome (which is an alloy of nickel, iron, and chromium) exhibit a very weak dependence of resistivity with temperature (Fig. 3.9). Manganin and constantan have similar properties. These materials are thus widely used in wire-bound standard resistors since their resistance values would change very little with temperatures.

Unlike metals, the resistivities of semiconductors decrease with increasing temperatures. A typical dependence is shown in Fig. 3.10.
We can qualitatively understand the temperature dependence of resistivity, in the light of our derivation of Eq. \(\sigma=\frac{n e^2}{m} \tau\). From this equation, the resistivity of a material is given by
\(
\rho=\frac{1}{\sigma}=\frac{m}{n e^2 \tau} \dots(3.27)
\)
\(\rho\) thus depends inversely both on the number \(n\) of free electrons per unit volume and on the average time \(\tau\) between collisions. As we increase temperature, the average speed of the electrons, which act as the carriers of current, increases resulting in more frequent collisions. The average time of collisions \(\tau\), thus decreases with temperature.

In a metal, \(n\) is not dependent on temperature to any appreciable extent, and thus the decrease in the value of \(\tau\) with rise in temperature causes \(\rho\) to increase as we have observed. For insulators and semiconductors, however, \(n\) increases with temperature. This increase more than compensates for any decrease in \(\tau\) in Eq.\(\sigma=\frac{n e^2}{m} \tau\) so that for such materials, \(\rho\) decreases with temperature.

Temperature For a conductor

Resistance \(\propto\) Temperature.
\(
\begin{aligned}
& \text { If } \quad R_0=\text { resistance of conductor at } 0^{\circ} C \\
& R _{ t }=\text { resistance of conductor at } t ^{\circ} C \\
&
\end{aligned}
\)
\(
\text { and } \alpha=\text { temperature co-efficient of resistance then } R_t=R_0(1+\alpha t)
\)
If \(R_1\) and \(R_2\) are the resistance at \(t_1{ }^{\circ} C\) and \(t_2{ }^{\circ} C\) respectively then \(\frac{R_1}{R_2}=\frac{1+\alpha t_1}{1+\alpha t_2}\) The value of \(\alpha\) is different at different temperature rage \(t_1{ }^{\circ} C\) to \(t_2{ }^{\circ} C\) is given by \(\alpha=\frac{R_2-R_1}{R_1\left(t_2-t_1\right)}\) which given \(R_2=R_1\left[1+\alpha\left(t_2-t_1\right)\right]\). This formula gives an approximate value.
In general, we can write, Resistance corresponding to temperature difference \((\Delta T )\) is given as \(R _t=R_0(1+\alpha \Delta T )\)

Resistance of the conductor decreases linearly with decrease in temperature and becomes zero at a specific temperature. This temperature is called critical temperature. At this temperature conductor becomes a superconductor.

Example 21: An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature \(\left(27.0^{\circ} C \right)\) is found to be \(75.3 \Omega\). When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element? The temperature coefficient of resistance of nichrome averaged over the temperature range involved, is \(1.70 \times 10^{-4}{ }^{\circ} C ^{-1}\).

Solution: When the current through the element is very small, heating effects can be ignored and the temperature \(T_1\) of the element is the same as room temperature. When the toaster is connected to the supply, its initial current will be slightly higher than its steady value of 2.68 A. But due to the heating effect of the current, the temperature will rise. This will cause an increase in resistance and a slight decrease in current. In a few seconds, a steady state will be reached when the temperature will rise no further, and both the resistance of the element and the current drawn will achieve steady values. The resistance \(R_2\) at the steady temperature \(T_2\) is
\(
R_2=\frac{230 V}{2.68 A}=85.8 \Omega
\)
Using the relation
\(
R_2=R_1\left[1+\alpha\left(T_2-T_1\right)\right]
\)
with \(\alpha=1.70 \times 10^{-4}{ }^{\circ} C ^{-1}\), we get
\(
T_2-T_1=\frac{(85.8-75.3)}{(75.3) \times 1.70 \times 10^{-4}}=820^{\circ} C
\)
that is, \(T_2=(820+27.0)^{\circ} C =847^{\circ} C\)
Thus, the steady temperature of the heating element (when the heating effect due to the current equals heat loss to the surroundings) is \(847^{\circ} C\).

Example 22: The resistance of the platinum wire of a platinum resistance thermometer at the ice point is \(5 \Omega\) and at steam, point is \(5.23 \Omega\). When the thermometer is inserted in a hot bath, the resistance of the platinum wire is \(5.795 \Omega\). Calculate the temperature of the bath.

Solution: \(R_0=5 \Omega, R_{100}=5.23 \Omega\) and \(R_t=5.795 \Omega\)
Now, \(\quad t=\frac{R_t-R_0}{R_{100}-R_0} \times 100, \quad R_t=R_0(1+\alpha t)\)
\(
\begin{aligned}
& =\frac{5.795-5}{5.23-5} \times 100 \\
& =\frac{0.795}{0.23} \times 100=345.65^{\circ} C
\end{aligned}
\)

Example 23: (a) At room temperature \(\left(0^{\circ} C \right)\) the resistance of a heating element is \(100 \Omega\). Calculate the temperature of the element if the resistance is found to be \(117 \Omega\) (the temperature coefficient of resistance of the material is \(1.7 \times 10^{-4}{ }^{\circ} C ^{-1}\) )

(b) The power of a heater is 500 W at \(800^{\circ} C\). What will be its power at \(200^{\circ} C\). If \(\alpha=4 \times 10^{-4}\) per \({ }^{\circ} C\) ?

Solution:(a)

\(
\begin{array}{ll}
\because & R _\theta= R _0(1+\alpha \theta) \therefore 117=100(1+\alpha \theta) \\
\Rightarrow & \theta=\frac{117-100}{100 \alpha}=\frac{17}{100 \times 1.7 \times 10^{-4}}=1000^{\circ} C
\end{array}
\)

(b)

\(
P =\frac{ V ^2}{ R } \quad \therefore \quad \frac{ P _{200}}{ P _{800}}=\frac{ R _{800}}{ R _{200}}=\frac{ R _0\left(1+4 \times 10^{-4} \times 800\right)}{ R _0\left(1+4 \times 10^{-4} \times 200\right)} \quad \Rightarrow \quad P _{200}=\frac{500 \times 1.32}{1.08}=611 W
\)