Potential difference and energy transfer
Explain volts, connect a voltmeter across two points and check an energy balance in a series circuit.
Before you begin
- Current in a complete series circuit
- Energy measured in joules
- Dividing decimals
Explain energy transferred per charge
Potential difference, also called voltage, compares two points in a circuit. It tells us the energy transferred for each unit of charge passing between those points. Charge is measured in coulombs (C): a potential difference of 1 volt (V) means 1 joule (J) transferred per coulomb. Across a resistor, electrical energy is transferred to thermal energy. A cell transfers energy from its chemical store to the circuit. Charge is not used up in these transfers. Current measures charge passing per second; potential difference measures energy transferred per charge. They are different quantities.
Connect a voltmeter across two points
A voltmeter compares its two connections. To measure the potential difference across one resistor, connect a lead to each end of that resistor: the meter is in parallel with it. Do not replace a section of the main path with the voltmeter. An ideal voltmeter draws no current; a real high-resistance voltmeter draws a small current, so treating it as non-disruptive is an approximation. For a positive direct-current reading, connect the positive meter terminal to the higher-potential end. Use only suitable low-voltage school equipment with teacher supervision. Switch the supply off before changing connections, select the correct voltage sockets and range, and never test a mains circuit.
Account for energy around a series loop
For a steady circuit with two resistors in series and negligible voltage drops along connecting wires, the measured potential difference across the supply terminals equals the sum across the two resistors. Each coulomb receives energy from the supply and transfers it in the components. Two resistors do not necessarily share the voltage equally. Their properties determine the split. The same current flows through both series resistors in the ideal-meter model even when their potential differences differ. The supply rating alone may differ from its terminal voltage while operating; use the measured terminal voltage for a comparison.
Evaluate readings and assumptions
In a real comparison, readings may be taken at different times. Keep the circuit components and settings fixed, let readings settle and repeat rounds while checking whether the supply drifts. Small differences between a supply reading and the sum of component readings need investigation. Consider instrument accuracy, reading resolution, connections and changing conditions. A display step of 0.01 V is resolution, not proof of accuracy to 0.01 V. Repeats do not remove a systematic error. The ideal energy balance is a model with stated assumptions, not a promise that rounded measurements match exactly.
Worked example
A steady series circuit has 9.0 V across the supply terminals and 3.0 V across R1. Wires have negligible voltage drops. Find the potential difference across R2 and the energy transferred in R2 when 2.0 C passes.
Show the worked solution
- Use the series energy balance: 9.0 = 3.0 + V across R2.
- The potential difference across R2 is 6.0 V: 6.0 J transferred per coulomb.
- For 2.0 C, E = V × Q = 6.0 × 2.0 = 12 J. Charge passes through; it is not consumed.
6.0 V across R2; 12 J transferred in R2.
Try it yourself
A circuit has 6.0 V across its supply terminals and 1.5 V across R1. R1 and R2 are in series with ideal meters and negligible wire drops. Which statement is correct?
Common mistakes
- Voltage is measured across two points; current is measured through a component.
- Use coulombs for charge, joules for energy, volts for potential difference and amperes for current.
- Series resistors need not have equal potential differences; they do carry the same steady current in the ideal-meter model.
- Compare measured supply terminal voltage with component readings, not just a printed rating.
- A written measurement plan does not demonstrate supervised practical competence.
Measure across, then account for energy
Specialist review has not been completed. These are written reasoning tasks with self-check guidance, not official marks or a saved activity score.
Choose the two connections
Use the diagram to plan readings across R1, R2 and the cell terminals with one voltmeter. Explain how your connections differ from ammeter connections, and how you would keep comparisons fair.
Compare my measurement plan
- For R1, use X and Y. For R2, connect to the wire immediately before and immediately after R2. For the supply, connect to its two terminals.
- Keep the main series path intact; the voltmeter forms a parallel measuring branch. An ammeter instead goes into the main path. Never substitute an ammeter across the supply.
- Under teacher supervision, switch the supply off before moving connections, use the voltage sockets and an appropriate range, and connect the positive lead towards the higher-potential side. Only suitable low-voltage school equipment is intended.
- Keep the same supply setting, resistors and meter range. Let readings settle. Repeat complete rounds and check the supply before and after component readings to look for drift; changing the order can reveal a time effect.
Do the readings balance?
These original constructed readings are not laboratory results. One voltmeter is moved between the three pairs of points within each round; the readings are not simultaneous. Its display step is 0.01 V, but its full accuracy specification is not supplied.
| Round | Supply terminals | Across R1 | Across R2 |
|---|---|---|---|
| 1 | 5.98 | 2.40 | 3.57 |
| 2 | 6.02 | 2.42 | 3.59 |
| 3 | 6.00 | 2.39 | 3.62 |
Calculate R1 + R2 for each round, then the residual: supply − (R1 + R2). A student says any non-zero residual disproves energy conservation. Evaluate that claim and identify an improvement to the comparison.
Compare my data reasoning
- The component sums are 5.97 V, 6.01 V and 6.01 V. The residuals are +0.01 V, +0.01 V and −0.01 V respectively.
- The readings approximately follow the expected series balance, but rounded, sequential measurements do not prove exact equality. A small residual alone does not disprove energy conservation.
- Resolution is not an accuracy limit. Without an instrument specification and a full uncertainty evaluation, we cannot certify that the residuals lie within the measurement uncertainty.
- Check connections and the meter specification, monitor supply drift, keep conditions stable and repeat rounds in a different order. Consider voltage drops in connecting wires. More repeats alone cannot remove a systematic error.
Apply the model to a new circuit
Two different resistors are in a steady series circuit with 4.5 V across the supply terminals and 1.6 V across the first resistor. The current through the first resistor is 0.14 A. Assume negligible wire drops and ideal meters. Find the voltage and current for the second resistor, then the energy transferred there when 3.0 C passes. Explain why equal sharing is not justified.
Compare my transfer reasoning
The second resistor has 4.5 − 1.6 = 2.9 V across it. It carries the same 0.14 A as the first resistor. For 3.0 C, the energy transferred is 2.9 × 3.0 = 8.7 J. Two components do not imply equal voltage drops; the measured first drop already rules out a half-and-half split. Different energy transfers per coulomb are compatible with the same charge flow per second.
Self-check your connections, calculation method, units and explanation separately. Written planning does not demonstrate supervised practical competence.
Revisit current and the complete path · Connect energy transfer to power · Later extension: resistance and parallel circuits
Try fresh questions
Explore the potential-difference mind map
Open the six-question potential-difference practice. Use new situations and data to check your reasoning. No practice score is saved.
Scope and curriculum
Grade 8 is StudyForA’s suggested placement, not an official grade requirement. England KS3 science includes potential difference in volts. AQA KS3 science, section 3.2.1 includes measurement and series voltage comparisons. This is an introductory topic; resistance calculations, parallel networks and a complete electricity course remain further study.
What can you explain now?
A different component transfers 20 J when 4.0 C passes. At the same potential difference, how much energy is transferred when 2.0 C passes? Explain whether halving the charge halves the potential difference.
Compare with the explanation
The potential difference is 5.0 V, and 2.0 C transfers 10 J. The potential difference stays 5.0 V.
First divide 20 J by 4.0 C to get 5.0 J per coulomb. Then multiply 5.0 V by 2.0 C. At the stated constant voltage, the energy halves with charge, but energy per charge does not.
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