# Electric Current, e.m.f. and Potential Difference

> Physics · CIE IGCSE 0625
> Source: https://www.owlsprep.com/study/cie-0625-u4-electric-current-e-m-f/

This guide covers Core and Extended content for electric current, e.m.f. and potential difference aligned to CIE IGCSE Physics 0625 syllabus points 4.2.2 and 4.2.3, including definitions, formula applications and exam-style worked examples.

**Prerequisites:** [Basic circuit components and circuit diagrams](https://www.owlsprep.com/study/cie-0625-u4-introduction-to-circuits/)

## Learning objectives

- Define electric current as rate of charge flow, recall and apply $I = Q/t$
- Distinguish between e.m.f. and potential difference using energy transfer principles
- Solve structured exam questions for both Core and Extended tiers
- Identify and avoid common mark-loss pitfalls for this topic

## 1. Electric Current (Core)

**Electric Current** — Electric current is related to the flow of electric charge around a circuit. In a metal, the current is due to the movement of free (delocalised) electrons. Current is measured in amperes (A).

*Example:* When a lamp lights up, charge is flowing through it — the larger the current, the greater the rate of flow of charge.

Current is measured using an **ammeter** connected in **series** with the component being tested, so the same current passes through both the ammeter and the component. In a metal conductor the charge is carried by free electrons moving through the fixed lattice of positive ions.

> **note**
>
> **Core vs Extended:** At Core you only need to know that current is related to the flow of charge, describe conduction in metals as the movement of free electrons, and use an ammeter. The equation $I = Q/t$ and the convention for current direction are **Extended only** — see Section 4.

> **Exam tip:** An ammeter must be connected in series with the component. Connecting it in parallel effectively short-circuits the component and can damage the meter.

## 2. Potential Difference (Core)

**Potential Difference (p.d.)** — The energy transferred (or work done) per unit of charge passing between two points in a circuit. 1 V = 1 J/C.

*Example:* A p.d. of 5 V across a bulb means 5 J of energy is transferred to light and heat for every 1 C of charge passing through the bulb.

Potential difference is measured using a voltmeter connected in parallel across the component you are testing. Voltmeters have very high resistance so they do not draw significant current from the circuit.

> **tip**
>
> Remember: voltmeters go in parallel, ammeters go in series. This is one of the most frequent mark loss points on both multiple choice and structured papers. Using the equation $V = W/Q$ to calculate a p.d. is **Extended only** — see Section 4.

## 3. Electromotive Force (e.m.f.) (Core)

**Electromotive Force (e.m.f.)** — The total energy supplied per unit of charge by a power source (e.g. cell, battery) to move charge around a complete circuit. It is measured in volts (V).

*Example:* A cell with e.m.f. 1.5 V supplies 1.5 J of energy to every coulomb of charge that passes through it.

A common Core exam question asks you to distinguish between e.m.f. and potential difference. The key difference is the direction of energy transfer: e.m.f. adds energy to the circuit (energy supplied per unit charge by the source), while p.d. is energy transferred from the charge to a component. Both are defined as work done per unit charge and are measured in volts (V).

> **note**
>
> **Core vs Extended:** At Core you define e.m.f. and p.d. in words and know both are measured in volts. Using the equations $E = W/Q$ (e.m.f.) and $V = W/Q$ (p.d.) to calculate values is **Extended only** — see Section 4.

> **Exam tip:** You do not need to explain internal resistance for Core tier questions, only state the difference between e.m.f. and p.d. in terms of energy transfer.

## 4. Extended Only: Equations for Current, Charge, e.m.f. and p.d.

> **warning**
>
> **Extended only.** Everything in this section is Supplement content. Core (Paper 1/3) learners are not required to define current as charge per unit time, use these equations, or state the direction of conventional current, and may skip this section.

**Electric Current (Extended definition)** — Electric current is the charge passing a point per unit time. Rearranging gives the charge transferred as $Q = I \times t$, where $I$ is current in amperes (A), $Q$ is charge in coulombs (C) and $t$ is time in seconds (s).

*Notation:* $I = \dfrac{Q}{t}$

*Example:* A current of 2 A flowing for 3 s transfers $Q = 2 \times 3 = 6$ C of charge.

**Worked example:** Calculate the current flowing in a circuit if 18 C of charge passes a point in 1 minute.

1. Step 1: Convert time to SI units (seconds): 1 minute = 60 s
2. $$I = \frac{Q}{t}$$
3. Step 2: Substitute values into the formula
4. $$I = \frac{18}{60} = 0.3 \text{ A}$$

**Conventional current direction (Extended):** Conventional current is taken to flow from the **positive** terminal of the source, around the external circuit, to the **negative** terminal. In a metal the free electrons actually flow the opposite way — from **negative** to **positive**.

The Supplement also requires you to recall and use the defining equations for e.m.f. and potential difference. Both are 'work (energy) done per unit charge':

- **e.m.f.:** $E = \dfrac{W}{Q}$, so the energy supplied by a source is $W = E \times Q$ (often written $E_{supplied} = V \times Q$).
- **potential difference:** $V = \dfrac{W}{Q}$, so the energy transferred in a component is $W = V \times Q$.

**Worked example:** A 4 V bulb transfers 12 J of energy to light and heat. Calculate the total charge that passed through the bulb.

1. Step 1: Rearrange $V = \frac{W}{Q}$ to solve for charge: $Q = \frac{W}{V}$
2. Step 2: Substitute given values
3. $$Q = \frac{12}{4} = 3 \text{ C}$$

**Worked example:** A 9 V battery powers a torch circuit. Calculate the total energy supplied to 2 C of charge passing through the battery.

1. Step 1: Use the e.m.f. equation rearranged for energy: $W = E \times Q$
2. $$W = 9 \times 2 = 18 \text{ J}$$

When a component carries a current $I$ for a time $t$, combining $Q = It$ with $W = VQ$ gives the energy transferred as $E = VIt$, where $V$ is the p.d. or e.m.f. in volts, $I$ is current in amperes and $t$ is time in seconds.

**Worked example:** A heater connected to a 230 V mains supply draws a current of 5 A for 2 minutes. Calculate the total energy transferred by the heater.

1. Step 1: Convert time to SI units: 2 minutes = 120 s
2. Step 2: Select the appropriate formula for the given values: $E = VIt$
3. $$E = 230 \times 5 \times 120 = 138000 \text{ J} = 138 \text{ kJ}$$

> **Exam tip:** Always convert time to seconds before substituting into $I=Q/t$, $Q=It$ or $E=VIt$, and check that $V$ is in volts and $I$ in amperes.

## Common pitfalls

- **Wrong:** Mixing up ammeter and voltmeter connections (series vs parallel)
  - Why it fails: Ammeters have low resistance so they must be in series to measure current; parallel connection will cause a short circuit and give invalid readings.
  - Correct: Connect ammeters in series with components, voltmeters in parallel across components.
- **Wrong:** Stating that e.m.f. and p.d. are the same quantity
  - Why it fails: e.m.f. measures energy supplied to charge by a source, while p.d. measures energy used by charge passing through a component.
  - Correct: Explicitly reference energy direction (supplied vs transferred to components) when distinguishing the two terms.
- **Wrong:** Using time in minutes instead of seconds in $I=Q/t$ or $E=VIt$ calculations
  - Why it fails: SI units are required for all physics calculations; using non-SI units will give an incorrect final answer.
  - Correct: Convert all time values to seconds before substituting into formulas.
- **Wrong:** Confusing conventional current direction with electron flow
  - Why it fails: Conventional current is defined as flow of positive charge from positive to negative terminal, while electrons (negative) flow the opposite direction.
  - Correct: Use conventional current direction for all circuit diagrams and calculations unless explicitly asked for electron flow.
- **Wrong:** Core students including internal resistance explanations when distinguishing e.m.f. and p.d.
  - Why it fails: Internal resistance is Extended-only content, and Core questions only require an energy-focused distinction.
  - Correct: Core tier answers should only reference energy supplied vs energy transferred, not internal resistance.

## Cheatsheet

| Quantity | Symbol | Unit | Definition | Formula |
| --- | --- | --- | --- | --- |
| Electric Current | I | Ampere (A) | Related to the rate of flow of charge | $I = Q/t$ (Extended) |
| Charge | Q | Coulomb (C) | Total charge flowing past a point | $Q = I \times t$ (Extended) |
| Potential Difference | V | Volt (V) | Energy transferred per unit charge across a component | $V = E/Q$ (Extended) |
| e.m.f. | E/V | Volt (V) | Energy supplied per unit charge by a power source | $E_{supplied} = V \times Q$ (Extended) |
| Energy (Extended) | E | Joule (J) | Total energy transferred in circuit | $E = VIt$ |

## What's next

Now that you have mastered electric current, e.m.f. and potential difference, you are ready to move on to resistance and Ohm's Law, the next key topic in the CIE IGCSE Physics 0625 Electricity unit. You will apply the definitions you learned here to calculate resistance from current and potential difference, and analyze series and parallel circuit behavior. Practicing structured questions on this topic will also prepare you for upcoming topics on electrical power and mains electricity, which build directly on the energy transfer relationships covered in this guide. Make sure to review the common pitfalls before attempting past paper questions to avoid easily avoidable mark losses.

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