Electromotive force and potential difference

1. Overview

This topic explores how energy is transferred into and out of an electrical circuit. Understanding the difference between electromotive force (e.m.f.) and potential difference (p.d.) is essential for calculating how much energy a power source provides and how much energy components like lamps or motors consume.

Key Definitions

• Electromotive Force (e.m.f.): The electrical work done by a source (such as a cell or generator) in moving a unit charge around a complete circuit.
• Potential Difference (p.d.): The work done by a unit charge passing through a specific component.
• Volt (V): The unit of measurement for both e.m.f. and p.d. One volt is equivalent to one joule of work done per coulomb of charge ($1 V = 1 J/C$).
• Voltmeter: A device used to measure the p.d. between two points or the e.m.f. of a source.

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Core Content

Understanding e.m.f. vs. p.d.

While both are measured in Volts, they represent different sides of energy transfer:

• e.m.f. represents the energy supplied to the charges by the battery/cell.
• p.d. represents the energy used (dissipated) by the charges as they pass through components like resistors or bulbs.

Measuring Voltage

To measure the p.d. across a component, a voltmeter must be connected in parallel with that component.

• Analogue Voltmeters: Use a needle and scale. They require the user to look directly at the scale to avoid parallax errors and to choose the correct range (e.g., 0-5V or 0-15V) to get an accurate reading.
• Digital Voltmeters: Provide a direct numerical readout. They are generally more accurate and easier to read, as they automatically handle range adjustments in many cases (auto-ranging).

Voltmeters with Different Ranges

• When using a voltmeter, always start with the highest range if the voltage is unknown to prevent damaging the meter.
• If the reading is small, switch to a lower range to increase the precision of the measurement.

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Extended Content (Extended Curriculum Only)

Calculating e.m.f. and p.d.

The relationship between work done ($W$), charge ($Q$), and voltage ($V$ or $\varepsilon$) is defined by the amount of energy transferred per unit of charge.

The Equation for e.m.f.: $$\varepsilon = \frac{W}{Q}$$ Where:

• $\varepsilon$ = electromotive force (V)
• $W$ = Work done/Energy transferred to the charge by the source (J)
• $Q$ = Charge (C)

The Equation for p.d.: $$V = \frac{W}{Q}$$ Where:

• $V$ = potential difference (V)
• $W$ = Work done/Energy transferred by the charge to the component (J)
• $Q$ = Charge (C)

Worked Example

Question: A battery does 12 J of work to move 2 C of charge around a circuit. Calculate the e.m.f. of the battery. Solution:

1. Identify the variables: $W = 12\text{ J}$, $Q = 2\text{ C}$
2. Use the formula: $\varepsilon = W / Q$
3. Calculate: $\varepsilon = 12 / 2 = 6$
4. Answer: $6\text{ V}$

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Key Equations

Equation	Symbols	Units
$\varepsilon = \frac{W}{Q}$	$\varepsilon$ = e.m.f., $W$ = Work, $Q$ = Charge	$V$ (Volts), $J$ (Joules), $C$ (Coulombs)
$V = \frac{W}{Q}$	$V$ = p.d., $W$ = Work, $Q$ = Charge	$V$ (Volts), $J$ (Joules), $C$ (Coulombs)
$V = I \times R$	$V$ = p.d., $I$ = Current, $R$ = Resistance	$V$ (Volts), $A$ (Amperes), $\Omega$ (Ohms)

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Common Mistakes to Avoid

• ❌ Wrong: Using Joules ($J$) as the unit for e.m.f. or p.d.
• ✅ Right: Always use Volts ($V$). Joules is the unit for Work Done/Energy, not the potential difference itself.
• ❌ Wrong: Thinking that if you decrease the resistance of one component in a loop, the total current stays the same.
• ✅ Right: Decreasing the resistance of any component in a series circuit increases the total current in the entire loop.
• ❌ Wrong: Dividing resistance by voltage ($R/V$) to find current.
• ✅ Right: Use the formula $I = V / R$. Always check your algebraic rearrangement.
• ❌ Wrong: Forgetting to convert milli-units (like $mV$) into standard units ($V$) before calculating.
• ✅ Right: Always convert to standard SI units (e.g., $100\text{ mV} = 0.1\text{ V}$) before plugging numbers into equations.

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Exam Tips

1. Work vs. Voltage: If a question asks for the definition of e.m.f. or p.d., ensure you include the phrase "per unit charge." Simply saying "work done" will not get the mark.
2. Voltmeter Placement: In circuit diagrams, always draw the voltmeter in a separate "loop" around the component you are measuring (parallel). If you draw it in the main line (series), the circuit will not work because voltmeters have very high resistance.
3. Energy Transfer Direction: Remember that e.m.f. is about energy entering the circuit (chemical $\rightarrow$ electrical) and p.d. is about energy leaving the circuit (electrical $\rightarrow$ heat/light).

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A student connects a battery to a resistor using connecting wires.

(a) Define electromotive force (e.m.f.) in terms of energy and charge. [2]

(b) State the name of the instrument used to measure potential difference. [1]

(c) The potential difference across the resistor is 6.0 V, and a charge of 3.0 C passes through it. Calculate the electrical energy transferred to the resistor. [2]

Worked Solution:

(a)

1. Electromotive force is defined as the work done per unit charge. $e.m.f. = \frac{W}{Q}$ This shows the relationship between work, charge, and emf.

2. Therefore, e.m.f. is the electrical work done by a source in moving a unit charge around a complete circuit.

How to earn full marks:

• State that e.m.f. is work done per unit charge.
• State that the charge moves around a complete circuit.

(b)

1. Voltmeter

How to earn full marks:

• Correct spelling is required.

(c)

1. Recall the formula relating energy, charge, and potential difference. $E = VQ$ This is the equation to use.

2. Substitute the values into the equation. $E = 6.0 \text{ V} \times 3.0 \text{ C} = 18 \text{ J}$

How to earn full marks:

• Correctly recall the formula $E = VQ$.
• Correctly substitute the values into the equation.
• State the final answer as $\boxed{18 \text{ J}}$ with the correct unit.

Common Pitfall: Remember that electromotive force (e.m.f.) refers to the energy supplied by a source, while potential difference (p.d.) refers to the energy used by a component. Don't confuse the two definitions, and always remember to include the correct units in your calculations.

Exam-Style Question 2 — Extended Response [8 marks]

Question:

A mobile phone contains a rechargeable battery. During charging, the phone is connected to a charger that provides a potential difference.

(a) Define potential difference in terms of work done and charge. [2]

(b) The battery has an e.m.f. of 3.7 V. Explain what is meant by this statement. [2]

(c) When the phone is in use, a current of 0.20 A flows through the phone's circuitry for 15 minutes. Calculate:

(i) the charge that flows through the circuitry. [2] (ii) the electrical energy supplied by the battery during this time. [2]

Worked Solution:

(a)

1. Potential difference is defined as the work done per unit charge. $V = \frac{W}{Q}$ This shows the relationship between work, charge, and potential difference.

2. Therefore, potential difference is the work done by a unit charge passing through a component.

How to earn full marks:

• State that potential difference is work done per unit charge.
• State that the charge moves through a component.

(b)

1. The e.m.f. is the energy supplied by the battery to move 1 Coulomb of charge around the entire circuit.

2. This means the battery does 3.7 J of work to move 1 C of charge around the complete circuit.

How to earn full marks:

• State that the e.m.f. represents energy supplied per unit charge.
• Correctly use the value 3.7 V in your explanation.

(c)(i)

1. Recall the formula relating charge, current, and time. $Q = It$ This is the equation to use.

2. Convert time to seconds. $t = 15 \text{ min} \times 60 \text{ s/min} = 900 \text{ s}$

3. Substitute the values into the equation. $Q = 0.20 \text{ A} \times 900 \text{ s} = 180 \text{ C}$

How to earn full marks:

• Correctly recall the formula $Q = It$.
• Correctly convert minutes to seconds.
• State the final answer as $\boxed{180 \text{ C}}$ with the correct unit.

(c)(ii)

1. Recall the formula relating energy, charge, and potential difference. $E = VQ$ This is the equation to use.

2. Substitute the values into the equation. $E = 3.7 \text{ V} \times 180 \text{ C} = 666 \text{ J}$

How to earn full marks:

• Correctly recall the formula $E = VQ$.
• Use your answer from part (c)(i) (ECF allowed).
• State the final answer as $\boxed{666 \text{ J}}$ with the correct unit.

Common Pitfall: Always convert time to seconds before using it in calculations. Also, remember to use the correct formula for calculating energy, charge, and potential difference. A common mistake is to mix up the formulas or forget to include the correct units in your answers.

Exam-Style Question 3 — Short Answer [6 marks]

Question:

A student sets up a circuit with a battery, a resistor, and an ammeter.

(a) State what a voltmeter measures in a circuit. [1]

(b) Draw a circuit diagram showing a battery, a resistor, an ammeter to measure the current through the resistor, and a voltmeter to measure the potential difference across the resistor. [3]

(c) The voltmeter reads 4.5 V and the ammeter reads 0.50 A. Calculate the resistance of the resistor. [2]

Worked Solution:

(a)

1. Potential difference

How to earn full marks:

• Correct spelling is required.

(b)

1. 📊A series circuit consisting of a battery (long and short parallel lines), a resistor (rectangle), and an ammeter (circle with an A inside). A voltmeter (circle with a V inside) is connected in parallel across the resistor. Connecting wires are straight lines. All components are clearly labelled.

How to earn full marks:

• Battery symbol correct.
• Resistor symbol correct and in series with the battery and ammeter.
• Ammeter in series with the resistor.
• Voltmeter in parallel with the resistor.

(c)

1. Recall Ohm's Law. $V = IR$ This relates voltage, current, and resistance.

2. Rearrange the formula to solve for resistance. $R = \frac{V}{I}$

3. Substitute values and calculate. $R = \frac{4.5 \text{ V}}{0.50 \text{ A}} = 9.0 \Omega$

How to earn full marks:

• Recall Ohm's Law or its rearrangement.
• Correctly substitute the values into the equation.
• State the final answer as $\boxed{9.0 \Omega}$ with the correct unit.

Common Pitfall: When drawing circuit diagrams, make sure to use the correct symbols for each component. Also, remember that a voltmeter is always connected in parallel, while an ammeter is always connected in series. Don't forget to include the correct units in your final answer.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A car windscreen contains a heating element to demist the screen. The heating element consists of several parallel wires connected to the car's 12 V battery.

(a) State the relationship between electromotive force (e.m.f.) and potential difference (p.d.) when there is no current drawn from the battery. [1]

(b) When the heating element is switched on, a total current of 5.0 A flows from the car's 12 V battery. Calculate:

(i) the power supplied to the heating element. [2] (ii) the energy transferred to the heating element in 2.0 minutes. [2]

(c) The total resistance of the parallel wires in the heating element is 2.4 $\Omega$. Explain why the actual potential difference across the battery terminals will be slightly less than 12 V when the heating element is switched on. Include the concept of internal resistance in your explanation. [4]

Worked Solution:

(a)

1. The electromotive force (e.m.f.) is equal to the potential difference (p.d.).

How to earn full marks:

• State that the e.m.f. equals the p.d.

(b)(i)

1. Recall the formula relating power, potential difference, and current. $P = VI$ This is the equation to use.

2. Substitute values and calculate. $P = 12 \text{ V} \times 5.0 \text{ A} = 60 \text{ W}$

How to earn full marks:

• Correctly recall the formula $P = VI$.
• Correctly substitute the values into the equation.
• State the final answer as $\boxed{60 \text{ W}}$ with the correct unit.

(b)(ii)

1. Recall the formula relating energy, power, and time. $E = Pt$ This is the equation to use.

2. Convert time to seconds. $t = 2.0 \text{ min} \times 60 \text{ s/min} = 120 \text{ s}$

3. Substitute values and calculate. $E = 60 \text{ W} \times 120 \text{ s} = 7200 \text{ J}$

How to earn full marks:

• Correctly recall the formula $E = Pt$.
• Correctly convert minutes to seconds.
• Use your answer from part (b)(i) (ECF allowed).
• State the final answer as $\boxed{7200 \text{ J}}$ with the correct unit.

(c)

1. The battery has internal resistance.

2. When a current flows, there is a potential drop across the internal resistance.

3. This is because some of the battery's e.m.f. is used to drive the current through the internal resistance ($V = Ir$).

4. Therefore, the potential difference across the terminals is the e.m.f. minus the potential drop across the internal resistance, making it less than 12 V.

How to earn full marks:

• State that the battery has internal resistance.
• State that there is a potential drop across the internal resistance when a current flows.
• Explain that the potential drop is due to the current flowing through the internal resistance.
• State that the terminal potential difference is less than the e.m.f. because of the internal potential drop.

Common Pitfall: Remember that real batteries have internal resistance, which causes a voltage drop when current flows. This means the potential difference across the terminals is less than the e.m.f. when the circuit is complete. Don't forget to convert minutes to seconds when calculating energy.