Rearrange resistance to find current or potential difference
Use I = V/R and V = IR with a supplied resistance valid at the stated operating conditions.
Section 4.2.4 · Resistance
Core and Extended; Physics 0625, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked.
Before you begin
- Use R = V/I.
- Rearrange products and use ohms.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is 2.0 kΩ in ohms?
What you will learn
- Find current or p.d. from a matching resistance.
- Retain constant or operating-point conditions rather than extrapolating a changing component.
Select the target quantity
R = V/I rearranges to I = V/R and V = IR. For a 12 Ω resistor at 6.0 V, the current is 0.50 A. For 0.30 A through a stated 20 Ω resistor, the p.d. is 6.0 V. Use the equation that isolates the unknown rather than multiplying every pair of available numbers.
Resistance in kΩ must be converted before using V and A: 2.0 kΩ = 2000 Ω. A 6.0 V p.d. then gives 0.0030 A, or 3.0 mA. Unit checks distinguish resistance, current, p.d. and the power product IV.
Use a valid resistance value
The supplied resistance must apply at the new operating conditions. A constant-resistance resistor supports changing V while retaining R. A lamp resistance measured when cool may not apply after its filament heats, so a previous quotient should not automatically be used to predict its higher-voltage current.
The resistance equation describes the stated electrical operating relation. It does not alone specify temperature, useful output efficiency or a device's safe rating. Use any supplied operating limits and do not interpret a calculated current as evidence that an experiment has been performed.
Pause and explain
6.0 V is across a constant 12 Ω resistor. What is the current?
Worked example
A constant-resistance 12 Ω resistor has p.d. 6.0 V. Find its current.
Show the worked solution
- Use I = V/R.
- 6.0/12 = 0.50.
- The current is 0.50 A.
Answer 0.50 A.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find current through a constant 12 Ω resistor at 6.0 V.
Give me a hint
Use voltage divided by the valid resistance.
Compare my reasoning
- I = V/R.
- 6.0/12 = 0.50.
- I = 0.50 A.
0.50 A.
Look for these in your work
- Use the correct quotient.
- State amperes.
Find p.d. at 0.30 A through 20 Ω, and current at 6.0 V through 2.0 kΩ.
Give me a hint
Use V = IR for the first and convert resistance for the second.
Compare my reasoning
- V = 0.30 × 20 = 6.0 V.
- 2.0 kΩ = 2000 Ω.
- I = 6.0/2000 = 0.0030 A.
6.0 V; 0.0030 A (3.0 mA).
Look for these in your work
- Choose the two rearrangements.
- Convert the resistance prefix.
A lamp has 4 Ω at a cool low-voltage state. A learner uses that value unchanged at a hotter state without new data. Assess the prediction.
Give me a hint
Check whether resistance was declared constant across the changed conditions.
Compare my reasoning
- The 4 Ω value belongs to the cool operating state.
- Filament heating can change resistance.
- The hotter-state prediction needs valid resistance or current–voltage data.
The prediction is not established; the lamp's resistance may change with heating.
Look for these in your work
- Keep the operating conditions.
- Require suitable new-state evidence.
Common mistakes
- Isolate the requested quantity.
- Convert kΩ before using V and A.
- A changing lamp may not retain a previous resistance.
What can you explain now?
What equation finds p.d. from current and resistance?
Compare with the explanation
V = IR.
Multiplying current in A by resistance in Ω gives p.d. in V.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Find current or p.d. with a resistance valid for the specified state and consistent units.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.