# Potential difference and e.m.f.

> Physics · CIE A-Level
> Source: https://www.owlsprep.com/study/cie-9702-u9-potential-difference-and-e-m/

This sub-topic clarifies the core distinction between potential difference (p.d.) and electromotive force (e.m.f.), two foundational concepts for circuit analysis. You will learn how both relate to energy transfer in electrical circuits.

**Prerequisites:** [Electric current and charge](https://www.owlsprep.com/study/cie-9702-u9-electric-current-charge/); [Work and energy transfer](https://www.owlsprep.com/study/cie-9702-u2-work-energy-power/)

## Learning objectives

- Distinguish between potential difference (p.d.) and e.m.f. in terms of energy transfer in circuits
- Calculate p.d. and e.m.f. using the energy-charge relationship
- Apply the relationship between e.m.f. and terminal p.d. for sources with internal resistance

## Potential Difference: Definition and Calculation

Potential difference (p.d.) is measured between two points in a circuit, and describes the amount of electrical energy converted to other usable or wasted forms of energy (like heat, light, or kinetic energy) when charge passes through a component.

**Potential Difference** — Potential difference is defined as: $V = \frac{W}{Q}$, where $W$ is work done (energy converted) by charge $Q$ moving through the component.

*Notation:* V

*Example:* A 1.5 V torch bulb converts 1.5 J of electrical energy to heat and light per coulomb of charge passing through it.

**Worked example:** A 500 mA current flows through a lamp for 2 minutes, and 360 J of energy is dissipated as light and heat. Calculate the p.d. across the lamp.

1. First calculate total charge passing through the lamp using $Q = It$:
2. $$Q = (500 \times 10^{-3} \text{ A}) \times (2 \times 60 \text{ s}) = 60 \text{ C}$$
3. Substitute into the definition of p.d.:
4. $$V = \frac{W}{Q} = \frac{360 \text{ J}}{60 \text{ C}} = 6 \text{ V}$$
5. Final answer: p.d. across the lamp = 6 V

> **tip**
>
> Remember that 1 volt = 1 joule per coulomb ($1 \text{ V} = 1 \text{ J C}^{-1}$). This definition is commonly tested in short structured exam questions.

*Calculator:* allowed

## Electromotive Force: What It Actually Is

Unlike p.d. (which describes energy leaving electrical form), e.m.f. describes energy being converted into electrical energy from other sources. Examples include chemical energy converted in a battery, or mechanical energy converted in a generator.

**Electromotive Force (e.m.f.)** — E.m.f. is the total energy transferred per unit charge by a source, given by: $\mathcal{E} = \frac{W_{\text{total}}}{Q}$, where $W_{\text{total}}$ is the total energy converted from non-electrical to electrical form.

*Notation:* \mathcal{E}

*Example:* A 12 V car battery converts 12 J of chemical energy to electrical energy per coulomb of charge that passes through it.

**Worked example:** A battery converts 1800 J of chemical energy to electrical energy when 150 C of charge flows through it. Calculate the e.m.f. of the battery.

1. Substitute directly into the definition of e.m.f.:
2. $$\mathcal{E} = \frac{1800 \text{ J}}{150 \text{ C}} = 12 \text{ V}$$
3. Final answer: e.m.f. of the battery = 12 V

> **warning**
>
> E.m.f. is *not* a force! The name is a historical misnomer. It is always measured in volts, not newtons. This is a common multiple choice trick question.

*Calculator:* allowed

## E.m.f. and Terminal P.d. in Real Circuits

All real sources have internal resistance, meaning some energy is wasted as heat inside the source itself. By conservation of energy, the total e.m.f. of the source equals the sum of the useful terminal p.d. (across the external circuit) and the p.d. lost across the internal resistance: $\mathcal{E} = V_{\text{terminal}} + v_{\text{lost}}$.

**Worked example:** A cell of e.m.f. 1.5 V dissipates 0.2 J of energy as heat inside the cell when 1 C of charge flows through it. Calculate the terminal p.d. available to the external circuit.

1. First calculate the p.d. lost across the internal resistance, using $v_{\text{lost}} = \frac{W_{\text{lost}}}{Q}$:
2. $$v_{\text{lost}} = \frac{0.2 \text{ J}}{1 \text{ C}} = 0.2 \text{ V}$$
3. Rearrange the energy conservation relationship to solve for terminal p.d.:
4. $$V_{\text{terminal}} = \mathcal{E} - v_{\text{lost}} = 1.5 \text{ V} - 0.2 \text{ V} = 1.3 \text{ V}$$
5. Final answer: terminal p.d. = 1.3 V

**Check your understanding**

Test your understanding of the core difference between p.d. and e.m.f.:

1. Which statement correctly describes e.m.f.?

   - Electrical energy is converted to heat per unit charge
   - Chemical energy is converted to electrical energy per unit charge
   - Kinetic energy is converted to heat per unit charge
   - Electrical energy is converted to light per unit charge

   *Answer:* Chemical energy is converted to electrical energy per unit charge

   *Why:* E.m.f. describes conversion of non-electrical energy to electrical energy, while p.d. describes the opposite conversion of electrical energy to other forms.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Calling e.m.f. a force and stating it has units of newtons.
  - Why it fails: The name 'electromotive force' is misleading, e.m.f. is not a force.
  - Correct: Always define e.m.f. as energy per unit charge with units of volts (J C⁻¹).
- **Wrong:** Treating p.d. and e.m.f. as interchangeable terms because both are measured in volts.
  - Why it fails: They describe opposite directions of energy transfer in a circuit, so they are distinct concepts.
  - Correct: Always distinguish them: p.d. = energy out of electrical form, e.m.f. = energy into electrical form.
- **Wrong:** Assuming terminal p.d. of a cell is always equal to its e.m.f.
  - Why it fails: All real cells have internal resistance, so energy is lost when current flows.
  - Correct: Only state e.m.f. equals terminal p.d. when the circuit is open (zero current) or internal resistance is explicitly neglected.
- **Wrong:** Writing the definition of p.d. as $V = Q/W$ instead of $V = W/Q$.
  - Why it fails: Mixing up the order, p.d. is work *per unit charge*, not charge per work.
  - Correct: Remember the phrase: 'work per unit charge' → work divided by charge.

## Cheatsheet

| Quantity | Symbol | Definition | Energy Transfer |
| --- | --- | --- | --- |
| Potential difference | $V$ | $V = \frac{W}{Q}$ | Electrical → other forms (heat/light) |
| E.m.f. | $\mathcal{E}$ | $\mathcal{E} = \frac{W_{\text{total}}}{Q}$ | Other forms → electrical (chemical/mechanical) |
| Circuit relationship | - | $\mathcal{E} = V_{\text{terminal}} + v_{\text{lost}}$ | For sources with internal resistance |
| Unit | V | $1 \text{ V} = 1 \text{ J C}^{-1}$ | Same for p.d. and e.m.f. |

## What's next

The distinction between potential difference and e.m.f. is the foundation for all circuit analysis in CIE A-Level Physics. This concept underpins calculations of resistance, power, internal resistance, and complex circuit problems using Kirchhoff's laws. You will apply these definitions to solve real-world circuit problems, from simple series-parallel combinations to potential divider circuits used in sensors and measurement systems.

- [Resistance and resistivity](https://www.owlsprep.com/study/cie-9702-u9-resistance-and-resistivity/)
- [I-V Characteristics](https://www.owlsprep.com/study/cie-9702-u9-i-v-characteristics/)
- [Temperature dependence of resistance](https://www.owlsprep.com/study/cie-9702-u9-temperature-dependence-of-resistance/)

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