Resistance

Resistance is a property of conductor due to which it resists or opposes the flow of electric current through it.

The way a surface offers resistance to movement by offering friction, conductors oppose the flow of current. Component that is used to resist the flow of electric current in a circuit is called resistor.

In practical applications, resistors are used to increase or decrease the electric current.

Variable Resistance: The component of an electric circuit which is used to regulate the current, without changing the voltage from the source; is called variable resistance.

Rheostat: This is a device which is used in a circuit to provide variable resistance.

Cause of Resistance in a Conductor:

You have read that the flow of electrons in a conductor is electric current. The particles of conductor stick together due to force of attraction between them. Due to this, the particles create hindrance to flow of electrons. This hindrance is the cause of resistance to the flow of electricity.

Resistance in a conductor depends on nature, length and area of cross section of the conductor.

Nature of material

Some materials create least hindrance and hence are called good conductors. Silver is the best conductor of electricity. While some other materials create more hindrance to the flow of electric current, i.e. flow of electrons through them. Such materials are called bad conductors. Bad conductors are also known as insulators. Hard plastic is one of the best insulators of electricity.

Length of conductor

Resistance R is directly proportional to the length of the conductor. This means, Resistance increases with increase in length of the conductor. This is the cause that long electric wires create more resistance to electric current.

Thus, Resistance (R) ∝ length of conductor (l)

Or `R ∝ l` --------(i)

Area of cross section

Resistance R is inversely proportional to the area of cross section ( A) of the conductor. This means R will decrease with increase in the area of conductor and vice versa. More area of conductor facilitates the flow of electric current through bigger area and thus decreases the resistance. This is the cause that thick copper wire creates less resistance to the electric current.

Thus, `text(resistance) ∝ (1)/(text(Area))` of cross section of conductor (A)

Or, `R∝1/A` ---------(ii)

From equation (i) and (ii)

`R∝l/A`

Or, `R=ρ\l/A` -----------(iii)

Where ρ (rho) is the proportionality constant. It is called the electrical resistivity of the material of conductors.

From equation (iii)

`RA=ρl`

Or, `ρ=(RA)/(l)` ----------(iv)

The SI unit of resistivity: Since, the SI unit of R is Ω, SI unit of Area is m² and SI unit of length is m. Hence

Unit of resistivity (ρ) `=(Ωxxm^2)/(m)=Ωm`

Thus, SI unit of resistivity (ρ) is Ω m

Materials having resistivity in the range of 10⁻⁸ Ω m to 10⁻⁶ Ω m are considered as very good conductors. Silver has resistivity equal to 1.60 X 10⁻⁸ Ω m and copper has resistivity equal to 1.62 X 10⁻⁸ Ω m.

Rubber and glass are very good insulators. They have resistivity in the order of 10¹² Ω m to 10¹⁷ Ω m.

Resistivity of materials varies with temperature.

Example 1: What will be the resistivity of a metal wire of 2 m length and 0.6 mm in diameter, if the resistance of the wire is 50 Ω.

Solution: Given, Resistance ( R ) = 50 Ω, Length ( l ) = 2 m

Diameter = 0.6 mm

Hence, radius `= 0.3  mm = 3 xx 10^-4  m`

Resistivity (ρ) = ?

Now, area of cross section of wire `= π r^2`

Or, `A = 3.14 xx (3 xx 10^-4)^2`

Or, `A = 28.26 xx 10^-8  m^2`

`= 2.826 xx 10^-7  m^2`

We know that

`ρ=(RA)/(l)`

Or, `ρ=(50Ωxx2.826xx10^-7m^2)/(2m)`

Or, `ρ=25xx2.826xx10^-7Ωm`

`=70.65xx10^-7Ωm`

Or, `ρ=7.065xx10^-6Ωm`

Example 2: The resistance of an electric wire of an alloy is 10 Ω. If the thickness of wire is 0.001 meter, and length is 1 m, find its resistivity.

Solution: Given, Resistance ( R ) = 10 Ω, Length ( l ) = 1 m

Diameter = 0.001 m

Therefore, radius = 0.0005 m

Resistivity (ρ) =?

Now, area of cross section of wire `= π r^2`

Or, `A = 3.14 xx (0.005)^2 m^2`

Or, A = 0.00007850 m²

We know that

`ρ=(RA)/(l)`

Or, `ρ=(10Ωxx0.0000785m^2)/(1m)`

`=10Ωxx0.0000785 m`

`=0.000785 Ωm`

`=7.85xx10^-4Ωm`

Example 3: The resistivity of a metal wire is 10 x 10⁻⁸ Ω m at 20°C. Find the resistance of the same wire of 2 meter length and 0.3 mm thickness.

Solution: Given, Resistivity (ρ) = 10 x 10⁻⁸ Ω m, Length ( l ) = 2 m, Diameter = 0.3 mm

Resistance (R) =?

Now, Radius of wire `= text(Diameter)/(2) = (0.3 mm) / (2) = 0.15 mm = 1.5 xx 10^-5 m`

Now, area of cross section of wire `= π r^2`

Or, `A = 3.14 xx (1.5 xx 10^-5)^2`

Or, `A = 70.65 xx 10^-10m^2`

We know that

`R=ρ(l)/(A)`

Or, `R=(10xx10^-8Ω\xx2m)/(7.65xx10^-9m^2)`

`=(10xx2Ω)/(7.65xx10)`

`=(2Ω)/(7.65)=0.26Ω`

Example 4: The area of cross section of wire becomes half when its length is stretched to double. How the resistance of wire is affected in new condition?

Solution: Let the area of cross section of wire = A

Let length of wire before stretching = L

Let Resistance of wire = R

After stretching of wire, let

Area of cross section = A / 2

Length = 2L

Resistance = R1

Thus, ratio of resistance before stretching to resistance after stretching can be given as follows:

Or, `R:R_1=(ρL)/(A): (ρ2L)/(A/2)`

Or, `R:R_1=(ρL)/(A): (4ρ L)/(A)`

Or, `R:R_1=1:4`

This means R = 1 and R1 = 4

Thus, resistance increases four times after stretching of wire.