Resistance, current and potential difference
Calculate resistance from two readings, compare components fairly and test when a constant-resistance model is reasonable.
Before you begin
- Current in a complete series circuit
- Potential difference across a component
- Dividing decimals and converting milliamperes
Relate resistance to current and voltage
Resistance describes how strongly a component opposes current for a given potential difference. Calculate it by dividing the potential difference across that component by the current through it. Resistance is measured in ohms (Ω): 1 Ω means 1 V per ampere. Current is charge passing per second, not a quantity consumed by resistance. A resistor transfers energy to its surroundings while charge continues around the complete circuit. Use readings for the same component and operating condition.
Compare components at the same potential difference
At the same voltage across two components, the one carrying less current has greater resistance. For example, at 3 V, a component carrying 0.10 A has 30 Ω resistance, whereas one carrying 0.30 A has 10 Ω resistance. Comparing currents measured at different voltages is not enough: calculate both ratios. Convert milliamperes before dividing: 1000 mA = 1 A, so 120 mA = 0.120 A. Ordinary conducting materials have much lower resistance than insulating materials of comparable dimensions and conditions; resistance is not simply zero for every conductor.
Measure across and through the same component
Put an ammeter in series with the resistor and a voltmeter in parallel across its ends. The ideal voltmeter draws no current, so the ammeter then measures the resistor current. Real meters can slightly affect a circuit; this introductory model neglects that effect and voltage drops in leads. Use only suitable low-voltage school apparatus under teacher supervision. Switch off before rewiring, use the correct sockets and ranges, and stay within component and supply ratings. Never place an ammeter directly across a supply or use mains electricity. A written plan does not establish practical competence.
Decide when resistance can be treated as constant
For a fixed resistor kept at constant temperature, current is approximately proportional to potential difference: doubling the voltage doubles the current, leaving V ÷ I unchanged. This is the constant-resistance model of Ohm’s law. It is not true for every component or every condition. A filament lamp heats as current increases, so its resistance changes. You can still calculate V ÷ I at a non-zero current to describe that operating point; do not assume the ratio will apply at a different voltage. At zero current, division by zero does not give a usable resistance value. Repeated readings help identify variation but do not remove systematic meter error.
Worked example
A resistor has 2.4 V across it and a current of 120 mA. Calculate its resistance. Predict its current at 3.6 V if its temperature and resistance remain unchanged.
Show the worked solution
- Convert current: 120 mA = 0.120 A.
- Divide voltage by current: R = 2.4 ÷ 0.120 = 20 Ω.
- Use the same resistance only under the stated condition: I = 3.6 ÷ 20 = 0.18 A, or 180 mA.
20 Ω; predicted current 0.18 A (180 mA).
Try it yourself
Components A and B each have 4.0 V across them. A carries 0.20 A and B carries 0.10 A. Which comparison is correct?
Common mistakes
- Convert milliamperes to amperes before using volts divided by amperes.
- Use the voltage across and current through the same component, not unrelated supply readings.
- Compare current alone only when potential difference is held fixed.
- Calculating V ÷ I at one point does not prove constant resistance at other points.
- Resistance transfers energy; it does not use up charge.
Use two measurements to test a model
Specialist review has not been completed. These original written tasks have self-check guidance, not official marks, a saved activity score or practical certification.
Choose readings and control conditions
A student puts the voltmeter into a gap in the main wire and connects the ammeter across X and Y. Explain both errors. Then plan how to compare resistance at three supply settings while limiting heating effects.
Compare my measurement plan
- The voltmeter belongs across X and Y; in series its high resistance would strongly reduce current. The ammeter belongs in the main path; across X and Y its low resistance would bypass the resistor and could cause excessive current.
- Use supervised low-voltage equipment and correct meter sockets/ranges. Switch off before changing connections and remain within component ratings.
- Keep the same resistor, leads and meters. Record voltage across R and current through R at each setting. Use low currents and brief readings, allow cooling as needed and repeat a low setting after a higher one to look for drift. These steps reduce heating but do not prove constant temperature.
Does the ratio stay the same?
These constructed values illustrate a model; they are not laboratory measurements. Calculate resistance in each row. State what the pattern supports and what it does not establish about all resistors or measurement accuracy.
| Setting | Across R (V) | Through R (A) |
|---|---|---|
| 1 | 1.5 | 0.10 |
| 2 | 3.0 | 0.20 |
| 3 | 4.5 | 0.30 |
Compare my data reasoning
The ratios are 1.5 ÷ 0.10 = 15 Ω, 3.0 ÷ 0.20 = 15 Ω and 4.5 ÷ 0.30 = 15 Ω. The values are consistent with constant resistance over this range. They do not prove every component has constant resistance, that the ratio remains fixed outside this range, or that real meters have no uncertainty. For actual measurements, check meter specifications, heating and repeatability.
Use a fresh comparison
Component P carries 80 mA with 2.0 V across it. Component Q carries 0.12 A with 6.0 V across it. Which has greater resistance at these operating points? Explain why comparing the currents alone gives the wrong conclusion. Predict P’s current at 4.0 V only if its resistance stays unchanged.
Compare my transfer reasoning
For P, 80 mA = 0.080 A, so R = 2.0 ÷ 0.080 = 25 Ω. For Q, R = 6.0 ÷ 0.12 = 50 Ω. Q has greater resistance despite its greater current because the voltages differ. If P remains at 25 Ω, its current at 4.0 V is 4.0 ÷ 25 = 0.16 A (160 mA). Without the constant-resistance condition, that prediction is not justified.
Check units, choice of readings, arithmetic and the condition behind each prediction separately. A written measurement plan does not demonstrate supervised practical competence.
Revisit current and the complete path · Revisit voltmeter connections · Later extension: combined resistances and parallel circuits
Try fresh questions
Explore the resistance mind map
Open the six-question resistance practice. Use new values and situations to check your reasoning. No practice score is saved.
Choose a question by skill
Scope and curriculum
England KS3 science includes resistance in ohms as voltage divided by current, and quantitative comparisons of conductors and insulators. Grade 8 is StudyForA’s suggested sequence, not an official grade requirement. Parallel networks and a complete electricity course remain further study.
What can you explain now?
At one operating point, a lamp has 3.0 V across it and carries 0.15 A. Find its resistance there. Can you conclude that it will carry 0.30 A at 6.0 V?
Compare with the explanation
20 Ω at that operating point. No: the 6.0 V current cannot be concluded from this reading alone.
Calculate 3.0 ÷ 0.15 = 20 Ω. Doubling current with voltage requires unchanged resistance; a filament lamp’s temperature and resistance can change. A new reading or a justified model is needed.
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