Electromotive force and internal resistance

What you'll learn:

The true definition of electromotive force (emf) and how it differs from potential difference (pd).
Why batteries get warm and seem to "lose" voltage when connected to a circuit.
How to calculate internal resistance and terminal pd using the equation $ε=I(R+r)\varepsilon = I(R + r)$ .
The theory behind Required Practical 6: determining emf and internal resistance from a graph.

What is Electromotive Force?

If you pick up a standard AA battery, the label will tell you it provides 1.5 volts. You might think of this as the "push" the battery gives to the electrons. In Physics, we call this theoretical maximum voltage the electromotive force, or emf for short.

However, the name is slightly misleading! Emf is not actually a force measured in newtons. It is a measure of energy transferred per unit of charge.

Definition

Electromotive force (emf, $ε$ )

Electromotive force is the electrical energy transferred by a power supply to each coulomb of charge passing through it.

\varepsilon = \frac{E}{Q}

Where:

$ε\varepsilon$ is the electromotive force in volts ( $V\text{V}$ )
$E$ is the energy transferred to the charge in joules ( $J\text{J}$ )
$Q$ is the charge in coulombs ( $C\text{C}$ )

You already know the formula for potential difference ( $\frac{W}{Q}$ ). Notice how similar the equation for emf is? The difference lies in energy transfer:

Emf ( $ε\varepsilon$ ) is the energy gained by charges as they pass through the power supply (chemical energy converting to electrical energy).
Potential difference ( $V$ ) is the energy lost or transferred by charges as they pass through components like resistors or bulbs (electrical energy converting to heat or light).

Example

Calculating emf

A battery transfers 45 joules of chemical energy to electrical energy when a charge of 3.0 coulombs passes through it. Calculate the electromotive force of the battery.

State the known values: $45\text{ J}$ and $3.0\text{ C}$ .
State the formula for electromotive force:

\varepsilon = \frac{E}{Q}

Substitute the values and calculate the result:

\varepsilon = \frac{45}{3.0} = 15\text{ V}

The electromotive force is 15 volts.

Internal Resistance and "Lost Volts"

Here is a strange but true fact: if you connect a voltmeter across an unconnected 9 volt battery, it reads 9 volts. But the moment you connect that same battery to a circuit and allow current to flow, the voltage across the battery drops. It might read 8.5 volts instead.

Where did the missing 0.5 volts go? The answer is internal resistance.

Real batteries are not perfect. They are filled with chemicals, pastes, and metal electrodes. When charge flows through the battery itself, these materials resist the flow.

Definition

Internal resistance ( $r$ )

The opposition to the flow of electrical current inside a power source itself, causing a loss of energy (as heat) before the current even leaves the source. It is measured in ohms ( $Ω\Omega$ ).

Because the battery has its own resistance, some of the electrical energy it generates is used up just pushing the charge through the battery itself. This is why batteries get warm when you use them!

We can picture a real battery as a perfect, ideal power source (providing the emf, $ε\varepsilon$ ) in series with a tiny internal resistor ( $r$ ).

Circuit Diagram showing Internal Resistance

The dashed box represents the physical battery. $R$ is the external load resistance. Notice how the internal resistor $r$ is inescapably in series with the ideal emf source $ε\varepsilon$ .

Terminal potential difference

Because some energy is wasted inside the battery, the voltage available to the rest of the circuit is always less than the emf. We call this available voltage the terminal potential difference (or terminal pd, $V$ ).

The "missing" voltage is called the lost volts. By Ohm's law, the voltage lost across the internal resistance $r$ when a current $I$ flows is $I r$ .

This gives us a crucial relationship:

\text{Terminal pd} = \text{Emf} - \text{Lost volts}

\varepsilon - Ir

Common Mistake

When does terminal pd equal emf?

The terminal pd $V$ is only exactly equal to the emf $ε\varepsilon$ when there is absolutely no current flowing ( $I = 0$ ), because then the lost volts ( $I r$ ) will be zero. This is called an "open circuit".

The Main Circuit Equation

If we look at the whole circuit, the terminal pd ( $V$ ) is simply the voltage across the external load resistor ( $R$ ). So, we can replace $V$ with $I R$ .

If we substitute $I R$ into our equation:

\varepsilon - Ir

Rearranging this to make $ε\varepsilon$ the subject gives us the main equation you will use in the AQA exam:

Key Idea

The Emf Circuit Equation

\varepsilon = I(R + r)

Where:

$ε\varepsilon$ = electromotive force in volts ( $V\text{V}$ )
$I$ = current in the circuit in amps ( $A\text{A}$ )
$R$ = total external resistance in ohms ( $Ω\Omega$ )
$r$ = internal resistance of the supply in ohms ( $Ω\Omega$ )

This is fundamentally a statement of the Conservation of Energy. The total energy supplied per coulomb ( $ε\varepsilon$ ) equals the energy used per coulomb in the external circuit ( $I R$ ) plus the energy wasted per coulomb inside the battery ( $I r$ ).

Example

Finding the internal resistance

A cell of electromotive force 1.5 volts is connected to an external resistor of 4.5 ohms. A current of 0.30 amps flows through the circuit. Calculate the internal resistance of the cell.

State the known variables: $V\varepsilon = 1.5\text{ V}$ , $4.5\text{ }\Omega$ , $0.30\text{ A}$ .
Write down the equation linking these variables:

\varepsilon = I(R + r)

Substitute the known values into the equation:

1.5 = 0.30 (4.5 + r)

Expand the bracket (or divide both sides by the current):

1.5 = 1.35 + 0.30 r

Rearrange to find the lost volts:

1.5 - 1.35 = 0.30 r

0.15 = 0.30 r

Solve for the internal resistance, $r$ :

\frac{0.15}{0.30} = 0.50\text{ }\Omega

Required Practical 6: The V-I Graph

You are required to know how to investigate the emf and internal resistance of a cell practically. The standard method involves connecting a cell to a variable resistor, measuring the terminal pd ( $V$ ) and the current ( $I$ ) for various resistance settings.

When you plot the terminal pd ( $V$ ) on the y-axis against the current ( $I$ ) on the x-axis, you get a straight line with a negative gradient. Let's see why mathematically.

Take our terminal pd equation:

\varepsilon - Ir

Rearrange it slightly to match the equation of a straight line, $y = m x + c$ :

\varepsilon

Looking at this matching pattern:

$y$ is the terminal pd ( $V$ ).
$x$ is the current ( $I$ ).
$m$ (the gradient) is $- r$ .
$c$ (the y-intercept) is $ε\varepsilon$ .

V-I Graph for a Cell

The y-intercept gives the electromotive force, and the negative of the gradient gives the internal resistance.

Tip

Analysing the graph quickly

If an exam question gives you a $V$ - $I$ graph:

To find the emf, just read the value where the line crosses the vertical $V$ axis.
To find the internal resistance, calculate the gradient ( $I\frac{\text{change in } V}{\text{change in } I}$ ) and ignore the negative sign.

Example

Calculating constants from a V-I graph

A student plots a graph of terminal pd against current for a particular cell. The line crosses the y-axis at 6.0 volts. At a current of 2.0 amps, the terminal pd is 4.5 volts. Calculate the internal resistance of the cell.

Identify the emf from the y-intercept:

\varepsilon = 6.0\text{ V}

Identify two points on the line to find the gradient. We have $(0, 6.0)$ and $(2.0, 4.5)$ .
Calculate the gradient ( $\frac{\Delta y}{\Delta x}$ ):

\frac{4.5 - 6.0}{2.0 - 0} = \frac{-1.5}{2.0} = -0.75

Equate the gradient to $- r$ :

- r = - 0.75

State the final internal resistance:

0.75\text{ }\Omega

Common Mistake

Forgetting the variable resistor

When drawing the circuit diagram for this required practical from memory, students often forget how to vary the current. You must include a variable resistor in series with the cell and ammeter, otherwise you will only get one reading for $V$ and $I$ and won't be able to plot a line!

Exam technique

In the exam

Be careful with definitions: If asked to define emf, explicitly state "energy transferred per unit charge". Saying "voltage of the battery" will score zero marks.
Watch out for power: The exam often ties this topic to electrical power. If you multiply the entire equation $\varepsilon - Ir$ by current $I$ , you get $\varepsilon I - I^2 r$ . This translates to: Power output to circuit = Total power generated - Power wasted as heat inside the battery.
Multiple cells: If a question puts two identical cells in series, their emfs add up ( $2ε2\varepsilon$ ) and their internal resistances add up ( $2 r$ ). If they are in parallel, the total emf is just $ε\varepsilon$ , but the total internal resistance halves ( $r2\frac{r}{2}$ ).

Self review

Check yourself

Can you explain why the terminal pd of a battery is usually lower than its printed emf?
What physical property of the battery does the gradient represent on a $V$ - $I$ graph?
If a battery is short-circuited (the external resistance $R$ is zero), what happens to the terminal pd?
Can you write down the equation linking emf, terminal pd, current, and internal resistance without looking at the notes above?

Electromotive force and internal resistance

What you'll learn:

The true definition of electromotive force (emf) and how it differs from potential difference (pd).
Why batteries get warm and seem to "lose" voltage when connected to a circuit.
How to calculate internal resistance and terminal pd using the equation $ε=I(R+r)\varepsilon = I(R + r)$ .
The theory behind Required Practical 6: determining emf and internal resistance from a graph.

What is Electromotive Force?

However, the name is slightly misleading! Emf is not actually a force measured in newtons. It is a measure of energy transferred per unit of charge.

Definition

Electromotive force (emf, $ε$ )

Electromotive force is the electrical energy transferred by a power supply to each coulomb of charge passing through it.

\varepsilon = \frac{E}{Q}

Where:

$ε\varepsilon$ is the electromotive force in volts ( $V\text{V}$ )
$E$ is the energy transferred to the charge in joules ( $J\text{J}$ )
$Q$ is the charge in coulombs ( $C\text{C}$ )

You already know the formula for potential difference ( $\frac{W}{Q}$ ). Notice how similar the equation for emf is? The difference lies in energy transfer:

Emf ( $ε\varepsilon$ ) is the energy gained by charges as they pass through the power supply (chemical energy converting to electrical energy).
Potential difference ( $V$ ) is the energy lost or transferred by charges as they pass through components like resistors or bulbs (electrical energy converting to heat or light).

Example

Calculating emf

A battery transfers 45 joules of chemical energy to electrical energy when a charge of 3.0 coulombs passes through it. Calculate the electromotive force of the battery.

State the known values: $45\text{ J}$ and $3.0\text{ C}$ .
State the formula for electromotive force:

\varepsilon = \frac{E}{Q}

Substitute the values and calculate the result:

\varepsilon = \frac{45}{3.0} = 15\text{ V}

The electromotive force is 15 volts.

Internal Resistance and "Lost Volts"

Where did the missing 0.5 volts go? The answer is internal resistance.

Real batteries are not perfect. They are filled with chemicals, pastes, and metal electrodes. When charge flows through the battery itself, these materials resist the flow.

Definition

Internal resistance ( $r$ )

The opposition to the flow of electrical current inside a power source itself, causing a loss of energy (as heat) before the current even leaves the source. It is measured in ohms ( $Ω\Omega$ ).

We can picture a real battery as a perfect, ideal power source (providing the emf, $ε\varepsilon$ ) in series with a tiny internal resistor ( $r$ ).

Circuit Diagram showing Internal Resistance

The dashed box represents the physical battery. $R$ is the external load resistance. Notice how the internal resistor $r$ is inescapably in series with the ideal emf source $ε\varepsilon$ .

Terminal potential difference

The "missing" voltage is called the lost volts. By Ohm's law, the voltage lost across the internal resistance $r$ when a current $I$ flows is $I r$ .

This gives us a crucial relationship:

\text{Terminal pd} = \text{Emf} - \text{Lost volts}

\varepsilon - Ir

Common Mistake

When does terminal pd equal emf?

The Main Circuit Equation

If we look at the whole circuit, the terminal pd ( $V$ ) is simply the voltage across the external load resistor ( $R$ ). So, we can replace $V$ with $I R$ .

If we substitute $I R$ into our equation:

\varepsilon - Ir

Rearranging this to make $ε\varepsilon$ the subject gives us the main equation you will use in the AQA exam:

Key Idea

The Emf Circuit Equation

\varepsilon = I(R + r)

Where:

$ε\varepsilon$ = electromotive force in volts ( $V\text{V}$ )
$I$ = current in the circuit in amps ( $A\text{A}$ )
$R$ = total external resistance in ohms ( $Ω\Omega$ )
$r$ = internal resistance of the supply in ohms ( $Ω\Omega$ )

Example

Finding the internal resistance

A cell of electromotive force 1.5 volts is connected to an external resistor of 4.5 ohms. A current of 0.30 amps flows through the circuit. Calculate the internal resistance of the cell.

State the known variables: $V\varepsilon = 1.5\text{ V}$ , $4.5\text{ }\Omega$ , $0.30\text{ A}$ .
Write down the equation linking these variables:

\varepsilon = I(R + r)

Substitute the known values into the equation:

1.5 = 0.30 (4.5 + r)

Expand the bracket (or divide both sides by the current):

1.5 = 1.35 + 0.30 r

Rearrange to find the lost volts:

1.5 - 1.35 = 0.30 r

0.15 = 0.30 r

Solve for the internal resistance, $r$ :

\frac{0.15}{0.30} = 0.50\text{ }\Omega

Required Practical 6: The V-I Graph

When you plot the terminal pd ( $V$ ) on the y-axis against the current ( $I$ ) on the x-axis, you get a straight line with a negative gradient. Let's see why mathematically.

Take our terminal pd equation:

\varepsilon - Ir

Rearrange it slightly to match the equation of a straight line, $y = m x + c$ :

\varepsilon

Looking at this matching pattern:

$y$ is the terminal pd ( $V$ ).
$x$ is the current ( $I$ ).
$m$ (the gradient) is $- r$ .
$c$ (the y-intercept) is $ε\varepsilon$ .

V-I Graph for a Cell

The y-intercept gives the electromotive force, and the negative of the gradient gives the internal resistance.

Tip

Analysing the graph quickly

If an exam question gives you a $V$ - $I$ graph:

To find the emf, just read the value where the line crosses the vertical $V$ axis.
To find the internal resistance, calculate the gradient ( $I\frac{\text{change in } V}{\text{change in } I}$ ) and ignore the negative sign.

Example

Calculating constants from a V-I graph

Identify the emf from the y-intercept:

\varepsilon = 6.0\text{ V}

Identify two points on the line to find the gradient. We have $(0, 6.0)$ and $(2.0, 4.5)$ .
Calculate the gradient ( $\frac{\Delta y}{\Delta x}$ ):

\frac{4.5 - 6.0}{2.0 - 0} = \frac{-1.5}{2.0} = -0.75

Equate the gradient to $- r$ :

- r = - 0.75

State the final internal resistance:

0.75\text{ }\Omega

Common Mistake

Forgetting the variable resistor

Exam technique

In the exam

Be careful with definitions: If asked to define emf, explicitly state "energy transferred per unit charge". Saying "voltage of the battery" will score zero marks.
Watch out for power: The exam often ties this topic to electrical power. If you multiply the entire equation $\varepsilon - Ir$ by current $I$ , you get $\varepsilon I - I^2 r$ . This translates to: Power output to circuit = Total power generated - Power wasted as heat inside the battery.
Multiple cells: If a question puts two identical cells in series, their emfs add up ( $2ε2\varepsilon$ ) and their internal resistances add up ( $2 r$ ). If they are in parallel, the total emf is just $ε\varepsilon$ , but the total internal resistance halves ( $r2\frac{r}{2}$ ).

Self review

Check yourself

Can you explain why the terminal pd of a battery is usually lower than its printed emf?
What physical property of the battery does the gradient represent on a $V$ - $I$ graph?
If a battery is short-circuited (the external resistance $R$ is zero), what happens to the terminal pd?
Can you write down the equation linking emf, terminal pd, current, and internal resistance without looking at the notes above?

Electromotive force and internal resistance

What is Electromotive Force?

Internal Resistance and "Lost Volts"

Terminal potential difference

The Main Circuit Equation

Required Practical 6: The V-I Graph

Electromotive force and internal resistance Revision Guide

Electromotive force and internal resistance

What is Electromotive Force?

Internal Resistance and "Lost Volts"

Terminal potential difference

The Main Circuit Equation

Required Practical 6: The V-I Graph

Electromotive force and internal resistance Revision Guide