This is because; the temperature increases

the number of collisions between the

charge carriers and the atoms in a

conductor as a result the resistance

increases.

2.4 FACTORS AFFECTING RESISTANCE

OF A CONDUYCTOR

1. Cross Sectional Area:

푅 ∝ 푡 ………………… (3)

The wire with large cross – section area

has low resistance than of small cross –

section area

4. Material of the conductor:

Resistance also depends on the material

used to make the conductor

This is because; a conductor with large

cross section area has more charge

carriers to carry the electrical current than

of small cross section area.

Example a steel wire has high resistance

than a copper wire of identical dimension

at the same temperature.

1

푅 ∝ ……………………… (1)

On combining expression (1) and (2)

Assumption made, temperature is kept

constant

퐴

Points to note

The greater the diameter, the lower the

resistance i.e. the thicker it is, the less is the

resistance.

ꢀ

푅 = 휌 _퐴…………………… (4)

Whereby

휌

is

the

constant

of

Doubling the diameter or thickness of the

wire reduces its resistance to quarter.

proportionality called resistivity of the

material

2. Length of the Conductor:

RESISTIVITY

Resistivity is the measure of the ability of a

material to oppose the flow of an electric

current.

When the length of a conductor is

increased, while other factors remain

constant, the resistance of the conductor

also increases

The SI – unit of resistivity is the ohm –

metre (Ωm)

This is because charges collide with more

atoms in a long conductor than a short

conductor

RESISTIVITY

MATERIALS

OF

SOME

푅 ∝ 푙 ……………..……….. (2)

Material

Resistivity in (Ω푚) at

20^oC

3. Temperature:

Aluminum

Chromium

Copper

Iron

2.7 x 10^-8

When the temperature of a conductor is

increased, the resistance of a conductor

also increases.

1.3 x 10^-7

1.68x 10^-8

9.71 x 10^-8

2.1 x 10^-8

Lead

Given that

Silver

1.6 x 10^-8

Constantan

Manganin

Nichrome

Glass

4.9 x 10^-7

푙푒푛푔푡ℎ, 푙 = 2.3푚,

푑 = 2 × 10⁻⁵푚, 푟 = 1 × 10⁻⁵

푚

4.8 x 10^-7

1.0 x 10^-6

1 x 10⁹- 1 x 10¹³

1 x 10¹³- 1 x 10¹⁵

7.5 x 10¹⁷

Step 1: To find the area of the wire

ꢁ푟푒푎 = 휋푟²

ꢁ푟푒푎 = 3.14 (1 × 10⁻⁵푚)²

ꢁ푟푒푎 = 3.14 × 10⁻¹⁰푚²

Rubber

(

)

Quarts

푟푒푠푖푠푡푖푣푖푡푦, 휌 = 10.5 × 10⁻⁸Ω푚

푎푟푒푎 × 푟푒푠푖푠푡푎푛푐푒

푙푒푛푔푡ℎ

푟푒푠푖푠푡푖푣푖푡푦, 휌 =

휌 =

Step 2: To find the resistance of the

wire

ꢁ × 푅

퐿

휌 × 퐿

푅 =

ꢁ

(10.5 × 10⁻⁸) × 2.3

푅 =

Example 01

3.14 × 10⁻¹⁰

What is the resistance of a copper wire of

length 20m and diameter of 0. 080cm? The

resistivity of copper is 1.68 × 10⁻⁸Ω푚

푅 = 768.72Ω

Example 03

Solution

What is the resistance of the wire if its

length is doubled?

Given that

Solution

푙 = 20푚, 푑 = 0.080푐푚, 푟 = 0.004푚

ꢁ푟푒푎 = 휋푟²

Let

퐿₂= 2퐿₁

2

(

)

ꢁ푟푒푎 = 3.14 (0.004)

ꢁ푟푒푎 = 5.024 × 10⁻⁷푚²

푅 ∝ 퐿

푅 = 퐾퐿

푟푒푠푖푠푡푖푣푖푡푦, 휌 = 1.68 × 10⁻⁸Ω푚

푅₁

퐾퐿₁

From

=

푅₂

퐾퐿₂

휌 × 퐿

ꢁ

푅 =

퐿₂

푅₂= ( )푅₁

퐿₁

2퐿₁

(1.68 × 10⁻⁸) × 20

푅 =

푅₂= (

)푅₁

5.024 × 10⁻⁷

퐿₁

푅 = 0.67Ω

푅₂= 2푅₁

The resistance is doubled and the ratio

is 푅₂: 푅₁= 2: 1

Example 02

A steel bar has a length of 2.3m and a

diameter of 2 × 10⁻⁵m. What is the

resistance of the bar? The resistivity of steel

is 10.5 × 10⁻⁸Ω푚.

Example 04

How will the resistance of the wire be

affected if its diameter is doubled?

Solution

푅_푋(푑_푋)²

푅_푌(푑_푌)²

퐿_푋

퐿_푌

Let 푑₂= 2푑₁

=

1

푑²

푅 ∝

푑²푅 = 퐾

2

푑_푌

⁄

푅_푋(

)

2퐿_푌

퐿_푌

2

=

푅_푌(푑_푌)²

2

푅₁푑₁= 푅₂푑₂

Since the wire of the same material,휌 =

푐표푛푠푡푎푛푡

2

푑₁

푅₂=

× 푅₁

2

푑₂

푅_푋

푅_푌

1

×

= 2

푑²

(2푑)²

4

× 푅₁

푅_푋

푅_푌

8

=

1

푅₂=

× 푅₁

4

푅_푋: 푅_푌= 8: 1

1

The resistance will be reduced to

of

4

its original value and the ratio 푅₂: 푅₁=

1: 4

Example 06

Calculate the length of wire of 1.0mm in

diameter and 5.0 × 10⁻⁶Ω푚 resistivity that

would have a resistance of 5. 0Ω.

Solution

Example 05

Calculate the ratio of the resistance of the

wire X which has twice the length and half

thickness of Y.

휌 × 퐿

푅 =

ꢁ

휋푑²푅

퐿 =

Solution

Let 푑_푥= ¹푑_푌, 퐿_푋= 2퐿_푌

4휌

2

(3.14) 1 × 10^{−3 2}× 5

(

)

From

퐿 =

4 × 5.0 × 10⁻⁶

1

푑²

( )

… … … … 푖

푅 ∝

퐿 = 0.79푚 ≈ 79푐푚

푅 ∝ 퐿 … … … . . (푖푖)

The length of the wire is 79cm.

on combining the two equations

퐿

푅 ∝

푑²

Example 07

Two wires A and B are such that the radius

of A is twice that of B and the length of B is

twice that of A. If the two are of material,

푅 푑²= 퐾퐿

푅_푋(푑_푋)²= 퐿_푋… … … . . . (푖푖푖)

푅_푌(푑_푌)²= 퐿_푌… … … . . . (푖푣)

determine the ratio ^{퐫퐞퐬퐢퐬퐭퐚퐧퐜퐞 퐨퐟 퐀}

퐫퐞퐬퐢퐬퐭퐚퐧퐜퐞 퐨퐟 퐁

current is related to the resistance as

follows:

Solution

p. d

current =

ꢁ푅

휌 =

resistance

퐿

1

current ∝

… … … (ii)

ꢁ_퐴푅_퐴

=

ꢁ_퐵푅_퐵

퐿_퐵

resistance

퐿_퐴

Increase in resistance, lowers the current

passing through the given material.

퐿_퐵= 2퐿_퐴,

ꢁ = 휋푟²

푟 = 2푟_퐵

퐴

Therefore, more current flows in wire A

than wire B, because A has low resistance

compared to B.

L_A× (¹₂r_A)²

R_A

R_B

=

2

2L_A× r_A

R_A

R_B

1

8

Example 08

Two wires A and B of the same material and

length have cross sectional area in the ratio

2:1. If the same potential difference is

applied across each wire, comment on the

amount of current flowing between the two

wires A and B.

2.5 HEATING EFFECTS OFELECTRIC

CURRENT

Current flow is accompanied by conversion

of electric energy into other forms of energy

like light, sound and heat. In all these cases,

parts of electric energy are converted to heat

due to the resistance of the circuit

components.

Solution

Given that: 휌_퐴= 휌_퐵, 퐿_퐴= 퐿_퐵,

Area ratios,ꢁ_퐴: ꢁ_퐵= 2: 1

Since are made of the same material,

the two wires have the same resistivity:

퐴 ꢂ

The heating effect of electric current is

observed in electric kettles, electric irons

and electric fire.This explains why the steel

wool strand getshot as the current flows

through it, this is because; electrical energy

is converted to heat energy in the steel

wool strand.

From the equation;휌 =

ꢃ

( ꢁ푅)_퐴= (ꢁ푅)_퐵

2 × 푅_퐴= 1 × 푅_퐵

푅_퐵= 2푅_퐴… … … (푖)

This equation shows that, wire B has

higher resistance than wire A. Since the

potential difference is the same, the

Note