Physics LIMITATIONS OF OHM’S LAW ,RESISTIVITY OF VARIOUS MATERIALS ,TEMPERATURE DEPENDENCE OF RESISTIVITY FOR CBSE-NCERT-2

### Topic covered

color{blue}{star} LIMITATIONS OF OHM’S LAW
color{blue}{star} RESISTIVITY OF VARIOUS MATERIALS
color{blue}{star} TEMPERATURE DEPENDENCE OF RESISTIVITY

### LIMITATIONS OF OHM’S LAW

color{blue} ✍️ Although Ohm’s law has been found valid over a large class of materials, there do exist materials and devices used in electric circuits where the proportionality of V and I does not hold.

color{blue} ✍️ The deviations broadly are one or more of the following types:

color{blue} {(a)} V ceases to be proportional to I (Fig. 3.5).

color{blue}{(b)} The relation between V and I depends on the sign of V.

color{blue} ✍️ In other words, if I is the current for a certain V, then reversing the direction of V keeping its magnitude fixed, does not produce a current of the same magnitude as I in the opposite direction (Fig. 3.6). This happens, for example, in a diode which we will study in Chapter 14.

color{blue} { (c)} The relation between V and I is not unique, i.e., there is more than one value of V for the same current I (Fig. 3.7).

color{blue} ✍️ A material exhibiting such behaviour is GaAs. Materials and devices not obeying Ohm’s law in the form of Eq. (3.3) are actually widely used in electronic circuits. In this and a few subsequent chapters, however, we will study the electrical currents in materials that obey Ohm’s law.

### RESISTIVITY OF VARIOUS MATERIALS

color{blue} ✍️ The resistivities of various common materials are listed in Table 3.1.

color{blue} ✍️ The materials are classified as conductors, semiconductors and insulators depending on their resistivities, in an increasing order of their values. Metals have low resistivities in the range of 10^(–8) Ωm to 10^(–6) Ωm.

color{blue} ✍️ At the other end are insulators like ceramic, rubber and plastics having resistivities 10^(18) times greater than metals or more. In between the two are the semiconductors.

color{blue} ✍️ These, however, have resistivities characteristically decreasing with a rise in temperature. The resistivities of semiconductors are also affected by presence of small amount of impurities.

color{blue} ✍️ This last feature is exploited in use of semiconductors for electronic devices. Commercially produced resistors for domestic use or in laboratories are of two major types: wire bound resistors and carbon resistors. Wire bound resistors are made by winding the wires of an alloy, viz., manganin, constantan, nichrome or similar ones.

color{blue} ✍️ The choice of these materials is dictated mostly by the fact that their resistivities are relatively insensitive to temperature. These resistances are typically in the range of a fraction of an ohm to a few hundred ohms. Resistors in the higher range are made mostly from carbon.

color{blue} ✍️ Carbon resistors are compact, inexpensive and thus find extensive use in electronic circuits. Carbon resistors are small in size and hence their values are given using a colour code.

color{blue} ✍️ The resistors have a set of co-axial coloured rings on them whose significance are listed in Table 3.2. The first two bands from the end indicate the first two significant figures of the resistance in ohms.

color{blue} ✍️ The third band indicates the decimal multiplier (as listed in Table 3.2). The last band stands for tolerance or possible variation in percentage about the indicated values. Sometimes, this last band is absent and that indicates a tolerance of 20% (Fig. 3.8). For example, if the four colours are orange, blue, yellow and gold, the resistance value is 36 × 10^4 Ω, with a tolerence value of 5%.

### TEMPERATURE DEPENDENCE OF RESISTIVITY

color{blue} ✍️ The resistivity of a material is found to be dependent on the temperature.

color{blue} ✍️ 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,

color{purple}{p_T =P_O [1+alpha(T-T_0)]}........(3.26)

color{blue} ✍️ where ρ_T is the resistivity at a temperature T and ρ_0 is the same at a reference temperature T_0. α is called the temperature co-efficient of resistivity, and from Eq. (3.26), the dimension of α is ("Temperature")^(–1.)

color{blue} ✍️ For metals, α is positive and values of α for some metals at T_0 = 0°C are listed in Table 3.1. The relation of Eq. (3.26) implies that a graph of ρT plotted against T would be a straight line. At temperatures much lower than 0°C, the graph,

color{blue} ✍️ However, deviates considerably from a straight line (Fig. 3.9). 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.

color{blue} ✍️ Some materials like Nichrome (which is an alloy of nickel, iron and chromium) exhibit a very weak dependence of resistivity with temperature (Fig. 3.10). 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.

color{blue} ✍️ Unlike metals, the resistivities of semiconductors decrease with increasing temperatures. A typical dependence is shown in Fig. 3.11.

color{blue} ✍️ We can qualitatively understand the temperature dependence of resistivity, in the light of our derivation of Eq. (3.23). From this equation, resistivity of a material is given by

rho = 1/sigma = m/(n e^2 tau)........(3.27)

color{blue} ✍️ ρ thus depends inversely both on the number n of free electrons per unit volume and on the average time τ between collisions. As we increase temperature, average speed of the electrons, which act as the carriers of current, increases resulting in more frequent collisions. The average time of collisions τ, thus decreases with temperature.

color{blue} ✍️ In a metal, n is not dependent on temperature to any appreciable extent and thus the decrease in the value of τ with rise in temperature causes ρ to increase as we have observed.

color{blue} ✍️ For insulators and semiconductors, however, n increases with temperature. This increase more than compensates any decrease in τ in Eq.(3.23) so that for such materials,ρ decreases with temperature.
Q 3144091853

An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature (27.0 °C) is found to be 75.3 Ω. 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 × 10^(–4) °C^(–1.)
Class 12 Chapter example 3
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 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 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 = (230V)/(2.68A) = 85.8Ω
Using the relation
R_22 = R_1 [1 + α (T_2 – T_1)]
with α = 1.70 × 10^(–4) °C^(–1), we get

T_2 - T_1 = (85.8-75.3)/(75.3)xx1.70xx10^(-4) = 820°C

that is, T_2 = (820 + 27.0) °C = 847 °C
Thus, the steady temperature of the heating element (when heating
effect due to the current equals heat loss to the surroundings) is
847 °C.
Q 3164091855

The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 Ω and at steam point is 5.23 Ω. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is 5.795 Ω. Calculate the temperature of the bath.
Class 12 Chapter example 4
Solution:

R_0 = 5 Ω, R_(100) = 5.23 Ω and R_t = 5.795 Ω
Now t = (R_i-R_0)/(R_(100) -R_0 ) xx100
R_t= R_0 (1+alphat)

= (5.795-5)/(5.23-5) xx100

= (0.795)/(0.23)xx100= 345.65°C