Apply the potential-divider principle
Calculate the output p.d. from a series divider and predict how a variable resistance changes it.
Section 10 · D.C. circuits
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 10. These original circuit diagrams and calculations are study aids; full timed assessment, supervised practical competence and human subject-teacher review remain outstanding.
Before you begin
- Use current, potential difference, resistance, energy and charge with consistent circuit directions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In a two-resistor divider, what sets the output p.d.?
What you will learn
- Understand a potential divider as series components sharing supply p.d.
- Recall and use Vout = Vsupply Rout/(Rtotal)
- Predict divider output when component resistance changes
A series divider shares the supply potential difference
Two resistors R₁ and R₂ in series across a supply Vs have the same current. The p.d. across R₂ is V₂ = IR₂, and the total is Vs = I(R₁ + R₂). Dividing gives V₂ = Vs R₂/(R₁ + R₂), so when R₂ is the output section, Vout = Vs R₂/(R₁ + R₂).
The output depends on the ratio of the resistance across which it is measured to the total series resistance. Reversing which resistor is the output changes the expression; label the output terminals before substituting.
A changing resistor can control the output
If one divider resistor changes while the supply and the other resistance remain fixed, the output p.d. changes. The direction depends on which resistor is the output section.
The simple expression assumes the output is measured by a high-resistance voltmeter or otherwise unloaded. A load connected across the output can alter the effective resistance and change the divider ratio.
Pause and explain
A 10 V supply is split across equal series resistors. What is the p.d. across either one?
Worked example
A 12 V supply is across 2.0 kΩ and 4.0 kΩ in series. Find the p.d. across the 4.0 kΩ resistor.
Show the worked solution
- The output fraction is 4.0/(2.0 + 4.0).
- Vout = 12 × 4/6.
- Check the output is less than the supply.
Answer Vout = 8.0 V.
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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.
A 9.0 V supply is across 1.0 kΩ and 2.0 kΩ in series. Find output across 2.0 kΩ.
Give me a hint
Use the output resistance over total resistance fraction.
Compare my reasoning
- Vout = 9.0 × 2.0/(1.0 + 2.0).
- Vout = 6.0 V.
- Check the other resistor has the remaining 3.0 V.
6.0 V.
Look for these in your work
- Correct output resistor used
- Ratio evaluated
- Supply split checked
A 12 V divider has 3.0 kΩ above and 1.0 kΩ below the output node, with output measured across the lower resistor. Find output p.d.
Give me a hint
The lower resistor is the output section; divide it by the total.
Compare my reasoning
- Vout = 12 × 1.0/(3.0 + 1.0).
- Vout = 3.0 V.
- Keep the output reference across the lower resistor.
3.0 V.
Look for these in your work
- Output terminals identified
- Ratio correct
- Reference stated
A divider output is measured across a fixed 4.0 kΩ resistor in series with a thermistor. As temperature rises, the NTC resistance falls. Predict the output trend if the output is across the fixed resistor.
Give me a hint
Use the fixed output resistance over the sum and consider the changing denominator.
Compare my reasoning
- Vout = Vs × 4.0/(4.0 + Rthermistor).
- As temperature rises, Rthermistor decreases, so the denominator decreases.
- The fraction and the fixed-resistor output p.d. increase.
The output across the fixed resistor increases as the NTC thermistor resistance falls, for this unloaded divider arrangement.
Look for these in your work
- Output section located
- Denominator trend traced
- Output trend qualified
Common mistakes
- Using the resistor not connected across the output terminals in the numerator
- Adding parallel resistances directly in a series-divider formula
- Ignoring loading when a low-resistance load is connected to the output
What can you explain now?
In a two-resistor divider, what sets the output p.d.?
Compare with the explanation
The output resistor as a fraction of total series resistance
The same current flows through both series elements.
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Make it stick
Why does the numerator of the divider equation depend on where the output terminals are connected?
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.