Calculate the resistance and voltage shares of a divider
Use R1/R2 = V1/V2 and the total p.d. without applying an unloaded divider rule to an arbitrary loaded circuit.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Use common current in two series resistors.
- Add their p.d.s.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What p.d. sum holds for the divider sections?
What you will learn
- Recall and use the resistance–voltage ratio.
- Calculate output and infer a resistance share.
Use matching pairs
For two series resistors used as an unloaded potential divider, their common current gives V1 = IR1 and V2 = IR2. Dividing gives R1/R2 = V1/V2. Pair each resistor with the p.d. across that same resistor; reversing only one ratio gives an incorrect result.
With 2 kΩ above and 4 kΩ below across 12 V, the p.d. ratio is 1:2. The shares are 4 V above and 8 V below. Output measured from the junction to the lower end is the lower share. Both resistors use the same units, so kΩ cancels consistently in their ratio.
Find shares and missing data
For the lower output, Vout = Vsupply Rlower/(Rupper + Rlower). This follows from the ratio and the sum, not a different rule. Across 10 V, an output of 4 V with Rupper = 3 kΩ leaves 6 V above, so Rlower/3 = 4/6 and Rlower = 2 kΩ.
The two original resistors must share the same current for their original resistance ratio to give their voltage ratio. An appreciable load across the lower resistor makes current split there, so first replace that parallel pair by its effective resistance. Endpoint zero-resistance limits can be interpreted directly; avoid dividing by zero in a ratio.
Pause and explain
An unloaded 12 V divider has upper 2 kΩ and lower 4 kΩ. What is lower-reference output?
Worked example
A 12 V unloaded divider has 2 kΩ above and 4 kΩ below. Find both drops and the lower-reference output.
Show the worked solution
- The resistance ratio is 2:4 = 1:2.
- Split 12 V into three equal parts of 4 V.
- Upper drop is 4 V, lower drop and output 8 V.
Answer Upper 4 V; lower-reference output 8 V.
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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 the lower-reference output for 12 V with 2 kΩ upper and 4 kΩ lower, unloaded.
Give me a hint
Use the lower resistance fraction of the total.
Compare my reasoning
- Rtotal = 6 kΩ.
- Vout = 12 × 4/6.
- Vout = 8 V.
8 V.
Look for these in your work
- Identify the lower section.
- Retain negligible output loading.
An unloaded 10 V divider has upper 3 kΩ and output 4 V across the lower resistor. Find the lower resistance.
Give me a hint
Find the upper p.d. first, then use the matched ratio.
Compare my reasoning
- Upper p.d. is 10 − 4 = 6 V.
- Rlower/3 = 4/6.
- Rlower = 2 kΩ.
2 kΩ.
Look for these in your work
- Pair resistance and voltage consistently.
- Check shares sum to 10 V.
A 9 V unloaded divider should give 3 V across its lower 2 kΩ resistor. Find the upper resistance and explain whether the same calculation holds with a low-resistance load attached.
Give me a hint
Use the upper drop and then inspect the equal-current assumption.
Compare my reasoning
- Upper drop is 9 − 3 = 6 V.
- Rupper/2 = 6/3, so Rupper = 4 kΩ.
- An appreciable load changes the lower effective resistance and requires recalculation.
Upper 4 kΩ for the unloaded model; an appreciable load requires a new equivalent.
Look for these in your work
- State the original assumption.
- Do not guarantee an unchanged loaded output.
Common mistakes
- Pair each voltage with its own resistance.
- Use the defined output section.
- Loading can destroy equal current in the original pair.
What can you explain now?
What ratio applies to two unloaded series divider resistors?
Compare with the explanation
R1/R2 = V1/V2.
The same current cancels when the matching voltage and resistance equations are divided.
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Make it stick
Use matched resistance and voltage ratios plus the total p.d., with output pair and loading stated.
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