### Resistance

• Define resistance (R) and know how it can be altered.

• Use Ohm’s law to design circuits.

• Calculate resistance from resistivity (conductivity)

## Electrical Resistance:

Electrical resistance is a measure of the difficulty to pass an electric current through that conductor. $R = \dfrac{V}{I}$ or more commonly written as: $V = IR$(known as Ohm’s Law), where R = electric resistance (this is the resistance to flow of charge through a device).
An object or device with greater resistance will require a greater voltage to produce a certain amount of current.
SI unit for resistance is Ohm:
1 Ohm = 1 Volt/1 Ampere or W = $\dfrac{V}{A}$
(The number of ohms indicates how many volts are required to produce 1 Ampere of current.)

## Resistivity:

Resistivity is a measure of the resisting power of a specified material to the flow of an electric current.
For a conductor of length L, electrical resistivity $\rho$, and cross-sectional area A, the resistance is: $R = \dfrac{\rho L}{A}$

## Internal Resistance:

Internal resistance is the resistance within a battery, or other voltage source, that causes a drop in the source voltage when there is a current. Thus, the voltage $V$ of the battery is related to its emf $E$ and internal resistance $r$ via $V = {E} - I\,r.$

### Solved Example:

#### 18-4-01

Internal resistance of cell when there is current of 0.40 A when a battery of 6.0 V is connected to a resistor of 13.5 $\Omega$ is:

Solution:
Total resistance of the circuit, including internal resistance, can be calculated by: $R = \dfrac{V}{I} = \dfrac{6}{0.4} = 15\ \Omega$ This resistance is including internal resistance. Out of this 15 $\Omega$ , 13.5 $\Omega$ is external resistance.
Internal resistance, $r = R_{Total} - R_{External} = 15 - 13.5 = 1.5\ \Omega$

### Solved Example:

#### 18-4-02

Maximum current a battery of e.m.f. 3.0 V and internal resistance 1.0 $\Omega$ is:

Solution:
Maximum current will flow, when the ends of the battery are short-circuited, i.e. the external resistance will be zero.
In such case, the electrons will encounter only internal resistance of the battery. $i = \dfrac{V}{R} = \dfrac{3}{1} = 3\ A$

### Solved Example:

#### 18-4-03

The resistance of an electric bulb drawing 1.2 A current at 6.0 V is ______________.

Solution:
$R = \dfrac{V}{I} = \dfrac{6}{1.2} = 5\ \Omega$

### Solved Example:

#### 18-4-04

The unit of resistivity is _________________.

### Solved Example:

#### 18-4-05

Specific resistance of a wire can be measured by formula:

Solution:

Resistance of a wire depends as follows:

1. Directly proportional to the length of the wire,

2. Inversely proportional to the cross-sectional area of the wire

$R \propto \dfrac{L}{A} = \rho \cdot \dfrac{L}{A}$ Hence, $\rho = \dfrac {RA}{L}$

### Solved Example:

#### 18-4-06

Current of 0.75 A, when a battery of 1.5 V is connected to wire of 5 m having cross sectional area 2.5 $\times$ 10$^{-7}$ m$^2$, will have resistivity, in $\Omega$.m:

Solution:
$R = \dfrac{V}{I} = \dfrac{1.5}{0.75} = 2\ \Omega$ Also, \begin{align*} R &= \dfrac{\rho \cdot L }{ A}\\ 2 &= \dfrac{\rho \cdot 5}{2.5 \times 10^{-7}}\\ \rho &= \dfrac{2 \times 2.5 \times 10^{-7}}{5} = 1 \times 10^{-7}\ \Omega.m \end{align*}

### Solved Example:

#### 18-4-07

Resistivity of a wire depends on:

Solution:

textbfResistance of a wire depends upon three factors.

1. The length of the wire,

2. The cross-sectional area of the wire, and

3. The resistivity of the material composing the wire.

Resistivity should not be confused with resistance of a wire.
Resistivity is a specific property of a material. It depends only upon the material, and NOT on other geometrical factors such as length or cross-sectional area.