Part 1: Electromotive Force

Despite its name, emf is not a force -- it is an energy transfer per unit charge (J C$^{-1}$), which is dimensionally identical to potential difference. The key distinction is that emf refers to the total energy the source provides to each coulomb, whereas potential difference refers to the energy transferred by a component.

Here $\varepsilon$ is the emf in volts (V), $E$ is the total energy transferred by the source in joules (J), and $Q$ is the charge that flows in coulombs (C). This equation is given in the AQA data booklet.

The emf of a source can be measured by connecting a high-resistance voltmeter directly across its terminals when no current is flowing (i.e., in an open circuit). Under these conditions, no energy is lost to internal resistance, so the voltmeter reading equals the emf.

Part 2: Internal Resistance

When charge flows through a cell, electrons collide with atoms and ions inside the cell. These collisions transfer kinetic energy to the internal structure of the cell, heating it up. This means that some of the electrical energy produced by the cell is wasted before it even reaches the external circuit.

In circuit diagrams, internal resistance is represented as a small resistor of resistance $r$ drawn in series with an ideal cell of emf $\varepsilon$, as shown below. In this model, notice that the small resistor is inside the battery boundary and still in series with the external resistor, so the same current flows through both.

[DIAGRAM: asset_name: 5.1.6 - Electromotive Force and Internal Resistance - Diagram 1; asset_slug: 5.1.6 - Electromotive Force and Internal Resistance - Diagram 1; recommended_method: retained_png; description: A cell represented as an ideal emf source (labelled epsilon) in series with a small resistor (labelled r) inside a dashed box representing the real battery. The dashed box has two terminals connecting to an external circuit containing a resistor R. Current I flows clockwise around the circuit.]

Since the internal resistance $r$ is in series with the external (load) resistance $R$, the total resistance of the circuit is simply:

Applying Ohm's law to the whole circuit gives the emf equation:

This can be expanded as:

where $I$ is the current in the circuit (A), $R$ is the external resistance ($\Omega$), and $r$ is the internal resistance ($\Omega$). On the current AQA data and formulae sheet, both $\varepsilon = E/Q$ and $\varepsilon = I(R + r)$ are provided. The exam skill is recognising that $R + r$ is the total circuit resistance, then rearranging the equation correctly for the quantity you need.

This is the useful share of the emf that reaches the external circuit. The remainder is dropped inside the source itself.

From the emf equation, we can identify:

• Terminal pd: $V = IR$ (the useful pd delivered to the external circuit)
• Lost volts: $v = Ir$ (the pd wasted inside the cell)

Therefore:

This tells us that the emf is always shared between the terminal pd and the lost volts. The terminal pd is always less than the emf whenever current flows, because some voltage is always dropped across the internal resistance.

Part 3: Circuit Calculations with Internal Resistance

The emf equation $\varepsilon = I(R + r)$ can be rearranged to solve for any unknown quantity. Let us work through the key rearrangements.

Because $\varepsilon = I(R + r)$ is already supplied on the current AQA sheet, most questions are really testing whether you can decide when to treat the cell and load as one series circuit, and when to switch to the terminal-pd form.

To find the current in a circuit:

To find the internal resistance:

To find the external resistance:

Here is a worked example demonstrating the standard approach.

Worked Example 1: A battery of emf 12 V and internal resistance 1.5 $\Omega$ is connected to a 4.5 $\Omega$ resistor. Calculate (a) the current through the battery, (b) the lost pd, and (c) the terminal pd.

(a) Using $I = \dfrac{\varepsilon}{R + r}$:

(b) Lost volts: $v = Ir = 2.0 \times 1.5 = 3.0$ V

(c) Terminal pd: $V = \varepsilon - v = 12 - 3.0 = 9.0$ V

Alternatively: $V = IR = 2.0 \times 4.5 = 9.0$ V (confirming the answer).

Now consider a slightly more complex scenario involving lost volts.

Worked Example 2: A cell has an emf of 5 V and lost volts of 2 V. The external resistance is 10 $\Omega$. Find the current in the circuit.

The terminal pd is: $V = \varepsilon - v = 5 - 2 = 3$ V

Using $V = IR$: $I = \dfrac{V}{R} = \dfrac{3}{10} = 0.3$ A

Part 4: The V-I Graph Method

Rearranging $\varepsilon = V + Ir$ gives:

Compare this with the equation of a straight line $y = mx + c$:

So if you plot terminal pd ($V$) on the $y$-axis against current ($I$) on the $x$-axis, you get a straight line. The graph below is the one to picture in an experiment; notice that the $y$-intercept gives $\varepsilon$ and that the downward gradient has magnitude $r$.

• The $y$-intercept gives the emf $\varepsilon$ (because at $I = 0$, $V = \varepsilon$).
• The gradient equals $-r$, so the magnitude of the gradient gives the internal resistance.

[DIAGRAM: asset_name: 5.1.6 - Electromotive Force and Internal Resistance - Diagram 2; asset_slug: 5.1.6 - Electromotive Force and Internal Resistance - Diagram 2; recommended_method: retained_png; description: A graph with current I on the x-axis and terminal pd V on the y-axis. A straight line with negative gradient starts at the y-intercept (labelled epsilon) and slopes downward. A gradient triangle is drawn on the line, with the vertical side labelled "lost pd" and the horizontal side labelled "current". The gradient is labelled as -r.]

This graphical method is the standard experimental technique for determining both the emf and internal resistance of a cell. The experimental setup consists of a cell connected in series with a variable resistor, an ammeter, and a switch, with a voltmeter connected directly across the cell terminals. By adjusting the variable resistor, different pairs of $V$ and $I$ readings are obtained.

The internal resistance and emf can also be calculated algebraically from just two pairs of measurements. If the terminal pd is $V_1$ at current $I_1$ and $V_2$ at current $I_2$, then:

Subtracting:

The emf can then be found by substituting $r$ back into either equation.

Part 5: Using the Formula Sheet Well

On the current AQA data and formulae sheet, students are given both $\varepsilon = E/Q$ and $\varepsilon = I(R + r)$. In exam conditions, the challenge is usually not remembering the equation but choosing the right version of it for the circuit in front of you.

A reliable routine is:

1. Identify whether the question is about the whole circuit, the external resistor, or the cell terminals.
2. For the whole circuit, use $\varepsilon = I(R + r)$ or $I = \varepsilon/(R + r)$.
3. For the useful output across the external resistor, use $V = IR$ and then connect it back to the cell with $V = \varepsilon - Ir$.
4. For a graph of terminal pd against current, read $\varepsilon$ from the $y$-intercept and $r$ from the magnitude of the gradient.

A quick sense-check helps. If the current increases, the lost volts $Ir$ must increase, so the terminal pd must decrease. Any answer that makes the terminal pd bigger than the emf while current is flowing is automatically wrong.

Part 6: Bringing It All Together

The relationships between emf, internal resistance, terminal pd, and lost volts are summarised below.

Key points to remember:

• At zero current (open circuit): $V = \varepsilon$ and $v = 0$.
• As current increases: $V$ decreases and $v$ increases, but $\varepsilon$ remains constant.
• At short circuit ($R = 0$): all voltage is dropped across $r$, giving $I_{\text{max}} = \varepsilon / r$ and $V = 0$.

That single idea explains both the straight-line graph and the everyday observation that a battery voltage sags when the battery is under load.