Advanced Physics Topics like quantum mechanics and relativity have revolutionized our understanding of the universe.

## Factors Determining Resistivity of Materials

It has been found by experiments that :

(i) The resistance of a given conductor is directly proportional to its length. That is :

R ∝ l …………… (1)

(ii) The resistance of a given conductor is inversely proportional to its area of cross-section. That is :

R ∝ \(\frac{1}{A}\) ……………….. (2)

By combining the relations (1) and (2), we get :

R ∝ \(\frac{l}{A}\)

or R = \(\frac{\rho \times l}{A}\) …………………… (3)

where ρ (rho) is a constant known as resistivity of the material of the conductor. Resistivity is also known as specific resistance.

From equation (3), it is clear that for a given conductor having a specified length l and area of cross-section A, the resistance R is directly proportional to its resistivity ρ. So, if we change the material of a conductor to one whose resistivity is two times, then the resistance will also become two times. And if we change the material of a conductor to one whose resistivity is three times, then the resistance will also become three times.

If we rearrange equation (3), we can write it as :

Resistivity, ρ = \(\frac{R \times A}{l}\) ……………….. (4)

where R = resistance of the conductor

A = area of cross-section of the conductor

and l = length of the conductor

This formula for calculating the resistivity of the material of a conductor should be memorised because it will be used to solve numerical problems. By using this formula, we will now obtain the definition of resistivity. Let us take a conductor having a unit area of cross-section of 1 m^{2} and a unit length of 1 m. So, putting A = 1 and l = 1 in equation (4), we get :

Resistivity, ρ = R

Thus, the resistivity of a substance is numerically equal to the resistance of a rod of that substance which is 1 metre long and 1 square metre in cross-section. Since the length is 1 metre and the area of cross-section is 1 square metre, so it becomes a 1 metre cube. So, we can also say that the resistivity of a substance is equal to the resistance between the opposite faces of a 1 metre cube of the substance. We will now find out the unit of resistivity.

We have just seen that:

Resistivity, ρ = \(\frac{R \times A}{l}\)

Now, to get the unit of resistivity p we should put the units of resistance R, area of cross-section A and length l in the above equation. We know that :

The unit of resistance R is ohm The unit of area of cross-section A is (metre)^{2}

And, The unit of length l is metre

So, putting these units in the above equation, we get :

Unit of resistivity, ρ = \(\frac{\text { ohm } \times(\text { metre })^2}{\text { metre }}\)

= ohm-metre (or Ω m)

Thus, the SI unit of resistivity is ohm-metre which is written in symbols as Ω m.

Please note that the resistivity of a substance does not depend on its length or thickness. It depends on the nature of the substance and temperature. The resistivity of a substance is its characteristic property. So, we can use the resistivity values to compare the resistances of two or more substances.

Another point to be noted is that just as when we talk of resistance in the context of electricity, it actually means electrical resistance, in the same way, when we talk of resistivity, it actually means electrical resistivity. The resistivities of some of the common substances (or materials) are given.

Resistivities of Some Common Substances (at 20°C)

From the above table we find that the resistivity of copper is 1.69 × 10^{-8} ohm-metre. Now, by saying that the resistivity of copper is 1.69 × 10^{-8} ohm-metre, we mean that if we take a rod of copper metal 1 metre long and 1 square metre in area of cross-section, then its resistance will be 1.69 × 10^{-8} ohms. Please note that a good conductor of electricity should have a low resistivity and a poor conductor of electricity will have a high resistivity.

From the above table we find that of all the metals, silver has the lowest resistivity (of 1.60 × 10^{-8} Ω m), which means that silver offers the least resistance to the flow of current through it. Thus, silver metal is the best conductor of electricity. It is obvious that we should make electric wires of silver metal.

But silver is a very costly metal. We use copper and aluminium wires for the transmission of electricity because copper and aluminium have very low resistivities (due to which they are very good conductors of electricity). From this discussion we conclude that silver, copper and aluminium are very good conductors of electricity.

The resistivities of alloys are much more higher than those of the pure metals (from which they are made). For example, the resistivity of manganin (which is an alloy of copper, manganese and nickel) is about 25 times more than that of copper; and the resistivity of constantan (which is an alloy of copper and nickel) is about 30 times more than that of copper metal.

It is due to their high resistivities that manganin and constantan alloys are used to make resistance wires (or resistors) used in electronic appliances to reduce the current in an electrical circuit. Another alloy having a high resistivity is nichrome. This is an alloy of nickel, chromium, manganese and iron having a resistivity of about 60 times more than that of copper.

The heating elements (or heating coils) of electrical heating appliances such as electric iron and toaster, etc., are made of an alloy rather than a pure metal because (i) the resistivity of an alloy is much higher than that of pure metal, and (ii) an alloy does not undergo oxidation (or burn) easily even at high temperature, when it is red hot. For example, nichrome alloy is used for making the heating elements of electrical appliances such as electric iron, toaster, electric kettle, room heaters, water heaters (geysers), and hair dryers, etc., because :

- nichrome has very high resistivity (due to which the heating element made of nichrome has a high resistance and produces a lot of heat on passing current).
- nichrome does not undergo oxidation (or bum) easily even at high temperature. Due to this nichrome wire can be kept red-hot without burning or breaking in air.

The resistivity of conductors (like metals) is very low. The resistivity of most of the metals increases with temperature. On the other hand, the resistivity of insulators like ebonite, glass and diamond is very high and does not change with temperature.

The resistivity of semi-conductors like silicon and germanium is in-between those of conductors and insulators, and decreases on increasing the temperature. Semi-conductors are proving to be of great practical importance because of their marked change in conducting properties with temperature, impurity, concentration, etc. Semi-conductors are used for making solar cells and transistors. We will now solve some problems based on resistivity.

**Example Problem 1.**

A copper wire of length 2 m and area of cross-section 1.7 × 10^{-6} m^{2} has a resistance of 2 × 10^{-2} ohms. Calculate the resistivity of copper.

**Solution:**

The formula for resistivity is :

Resistivity, ρ = \(\frac{R \times A}{l}\)

Here, Resistance, R = 2 × 10^{-2} Ω

Area of cross-section, A = 1.7 × 10^{-6} m^{2}

And, Length, l = 2 m

So, putting these values in the above formula, we get :

ρ = \(\frac{2 \times 10^{-2} \times 1.7 \times 10^{-6}}{2}\)

= 1.7 × 10^{-8} Ω m

Thus, the resistivity of copper is 1.7 × 10^{-8} ohm-metre.

**Example Problem 2.**

A copper wire has a diameter of 0.5 mm and resistivity of 1.6 × 10^{-8} Ω m.

(a) What will be the length of this wire to make its resistance 10 Ω ?

(b) How much does the resistance change if the diameter is doubled ?

**Solution:**

(a) First of all we will calculate the area of cross-section of the copper wire. Here the diameter of copper wire is 0.5 mm, so its radius (r) will be \(\frac{0.5}{2}\) mm or 0.25 mm. This radius of 0.25 mm will be equal to \(\frac{0.25}{2}\) m or 0.25 × 10^{-3} m. Thus, the radius r of this copper wire is 0.25 × 10^{-3} m. We will now find out the area of cross-section of the copper wire by using this value of the radius. So,

Area of cross-section of wire, A = πr^{2}

= \(\frac{22}{7}\) × (0.25 × 10^{-3})^{2}

= 0.1964 × 10^{-6} m^{2}

Resistivity, ρ = 1.6 × 10^{-8} Ω m

Resistance, R = 10 Ω

And, Length, l = ? (To be calculated)

Now, putting these values in the formula :

ρ = \(\frac{R \times A}{l}\)

We get : 1.6 × 10^{-8} = \(\frac{10 \times 0.1964 \times 10^{-6}}{l}\)

So, l = \(\frac{10 \times 0.1964 \times 10^{-6}}{1.6 \times 10^{-8}}\)

l = \(\frac{1964}{16}\)

l = 122.7 m

Thus, the length of copper wire required to make 10 Ω resistance will be 122.7 metres.

(b) The resistance of a wire is inversely proportional to the square of its diameter. So, when the diameter of the wire is doubled (that is, made 2 times), then its resistance will become \(\left(\frac{1}{2}\right)^2\) or \(\frac{1}{4}\) (one-fourth).

**Example Problem 3.**

A 6 Ω resistance wire is doubled up by folding. Calculate the new resistance of the wire.

**Solution:**

Suppose the length of 6 Ω resistance wire is l, its area of cross-section is A and its resistivity is ρ. Then :

R = \(\frac{\rho \times l}{A}\)

or 6 = \(\frac{\rho \times l}{A}\) ……………. (1)

Now, when this wire is doubled up by folding, then its length will become half, that is, the length will become \(\frac{l}{2}\). But on doubling the wire by folding, its area of cross-section will become double, that is, the area of cross-section will become 2A. Suppose the new resistance of the doubled up wire (or folded wire) is R. So,

R = \(\frac{\rho \times l}{2 \times 2 A}\)

or R = \(\frac{\rho \times l}{4 A}\) …………….. (2)

Now, dividing equation (2) by equation (1), we get:

\(\frac{R}{6}\) = \(\frac{\rho \times l \times A}{4 A \times \rho \times l}\)

or \(\frac{R}{6}\) = \(\frac{1}{4}\)

4R = 6

R = \(\frac{6}{4}\)

R = 1.5 Ω

Thus, the new resistance of the doubled up wire is 1.5 Ω.