# Electromotive force and internal resistance

> IB Physics SL · IB Physics SL
> Source: https://www.owlsprep.com/study/ib-physics-sl-u2-electromotive-force-and-internal-resistance/

This guide covers emf definitions, internal resistance of real cells, the core circuit formula, V-I graph interpretation, and common exam calculation problems.

**Prerequisites:** [Ohm's Law and series circuit rules](https://www.owlsprep.com/study/ib-physics-sl-u2-ohms-law-series-circuits/); [Potential difference and electrical power](https://www.owlsprep.com/study/ib-physics-sl-u2-potential-difference-power/)

## Learning objectives

- Distinguish between electromotive force (emf) and terminal potential difference
- Derive the relationship between emf, internal resistance, load resistance and terminal voltage
- Solve quantitative problems involving internal resistance for real batteries
- Interpret V-I graphs for a discharging cell to extract emf and internal resistance

## Core Definitions of emf and Internal Resistance

Ideal voltage sources have zero internal resistance, so their terminal voltage never drops no matter how much current is drawn. All real cells, batteries, and generators have a small but non-zero internal resistance that causes a voltage drop inside the source itself when current flows.

**Electromotive Force (emf)** — The work done per unit charge by non-electrical forces inside the source to separate positive and negative charges, creating a potential difference across the open terminals of the cell.

*Notation:* $\varepsilon$

*Example:* A standard 1.5 V AA alkaline cell has an emf of 1.5 V when no current is drawn.

> **warning**
>
> Emf is not a force: it is a potential difference, measured in volts, not newtons. This is a very common exam trick question.

**Check your understanding**

Test your understanding of the definitions below:

1. When a cell is on open circuit (no current flowing), what is the relationship between its terminal voltage and its emf?

   - Terminal voltage is zero
   - Terminal voltage equals emf
   - Terminal voltage is less than emf
   - Terminal voltage is greater than emf

   *Why:* No current flows so there is no voltage drop across the internal resistance, so the full emf appears across the terminals.

## Derivation of the emf Circuit Relationship

**Derivation:** Derive the relationship between emf, terminal voltage, current and internal resistance

*Starting from:* Conservation of energy for a complete circuit with a cell of emf $\varepsilon$, internal resistance $r$, connected to an external load resistor $R$

1. Total energy supplied per unit charge by the cell is equal to the sum of energy dissipated per unit charge across all resistances in the circuit
2. $$\varepsilon = V_{\text{load}} + V_{\text{internal}}$$
3. Apply Ohm's Law to the internal resistance: the voltage drop across internal resistance is $V_{\text{internal}} = Ir$
4. Apply Ohm's Law to the external load: $V_{\text{load}} = IR = V$, the terminal voltage across the cell
5. $$\varepsilon = V + Ir$$

*Conclusion:* This rearranges to the standard formula for terminal voltage: $V = \varepsilon - Ir$. As current $I$ drawn from the cell increases, the terminal voltage $V$ decreases linearly.

**Worked example:** A cell of emf 12 V has internal resistance 0.5 Ω. It is connected to a 5.5 Ω external resistor. Calculate the terminal voltage across the cell.

1. First calculate total circuit resistance: $R_{\text{total}} = R + r = 5.5 + 0.5 = 6.0\ \Omega$
2. $$Calculate total current in the circuit using Ohm's Law: $I = \frac{\varepsilon}{R_{\text{total}}} = \frac{12}{6} = 2.0\ \text{A}$$$
3. Use the emf formula to find terminal voltage: $V = \varepsilon - Ir$
4. $$V = 12 - (2.0 \times 0.5) = 12 - 1 = 11\ \text{V}$$
5. Verify by calculating voltage across external resistor: $V = IR = 2.0 \times 5.5 = 11\ \text{V}$, which matches.

## V-I Graph for a Discharging Cell

When you measure terminal voltage $V$ across a cell for different values of drawn current $I$, you get a straight line graph with negative gradient, following the equation $V = -rI + \varepsilon$, which matches the standard straight line form $y = mx + c$.

| Graph Feature | Physical Quantity |
| --- | --- |
| Y-intercept (at I = 0) | Emf $\varepsilon$ of the cell |
| Gradient (slope magnitude) | Internal resistance $r$ of the cell |
| X-intercept (at V = 0) | Short-circuit current of the cell |

**Exam command terms**

IB exam questions often ask you to describe what you can obtain from a V-I graph for a cell, watch for these command cues:

- **Determine from the graph** — You must read values directly off the plotted line, no calculation from given numbers

- **Show that the internal resistance is equal to the gradient** — You must rearrange $V = \varepsilon - Ir$ to match $y = mx + c$ and state that the gradient is negative r

## Common pitfalls

- **Wrong:** Confusing emf with terminal voltage for a cell that is supplying current
  - Why it fails: Students often use the emf value as the terminal voltage when current is flowing, ignoring the internal voltage drop
  - Correct: Always use $V = \varepsilon - Ir$ to calculate terminal voltage when the cell is not on open circuit
- **Wrong:** Forgetting that the gradient of the V-I graph is negative and taking the sign into account for internal resistance
  - Why it fails: Students sometimes calculate a negative value for internal resistance, which is physically impossible
  - Correct: Take the magnitude of the negative gradient to get the positive internal resistance value
- **Wrong:** Treating internal resistance as an external resistor in series when calculating the terminal voltage
  - Why it fails: This leads to double-counting the voltage drop across internal resistance
  - Correct: The internal resistance is inside the cell, so the terminal voltage is the voltage across the external load only
- **Wrong:** Stating that emf is the force that pushes charges around the circuit
  - Why it fails: Emf is a potential difference, measured in volts, not a force measured in newtons, this is a mark-losing definition error
  - Correct: Define emf as work done per unit charge by non-electrical forces inside the source
- **Wrong:** Calculating short circuit current as $\varepsilon R$ instead of $\varepsilon / r$
  - Why it fails: Short circuit means external load resistance R = 0, so total resistance is only r
  - Correct: Short circuit current $I_{\text{sc}} = \varepsilon / r$, no external resistance is present

## Cheatsheet

| Quantity | Formula | Units |
| --- | --- | --- |
| Electromotive force | $\varepsilon = V + Ir$ | Volts (V) |
| Terminal voltage | $V = \varepsilon - Ir$ | Volts (V) |
| Internal resistance | $r = \frac{\varepsilon - V}{I}$ | Ohms ($\Omega$) |
| Short circuit current | $I_{\text{sc}} = \frac{\varepsilon}{r}$ | Amperes (A) |
| Maximum load power | $P_{\text{max}} = \frac{\varepsilon^2}{4r}$ | Watts (W) |

## What's next

You will frequently combine these concepts to solve multi-loop circuit problems that appear on Paper 2 section A and extended response questions. Understanding how real batteries behave under load also helps you connect theory to practical lab work, where you will perform a investigation you might design for your IA to measure internal resistance by varying external resistance and plotting a V-I graph. This practical is commonly assessed in your IA, so make sure you can explain sources of uncertainty and error for this experiment clearly.

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ib-physics-sl-u2-electromotive-force-and-internal-resistance/