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IGCSE · Cambridge (CIE) · 0654

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.

Section P4.3.3 · Action and use of circuit components

Action and use of circuit components · Cambridge IGCSE 0654 · P4.3.3

Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.

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.
Original unloaded divider has upper 2 kilohms and lower 4 kilohms across 12 V. Matching resistance and voltage ratios give upper 4 V and lower-reference output 8 V. Negligible output loading is required.Match resistance and p.d. shares+−12 V2 kΩ4 kΩOutput8 VLower end is the referenceUnloaded output model4 V8 V
Original unloaded divider has upper 2 kilohms and lower 4 kilohms across 12 V. Matching resistance and voltage ratios give upper 4 V and lower-reference output 8 V. Negligible output loading is required.

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.

Sources for this lesson

Pause and explain

An unloaded 12 V divider has upper 2 kΩ and lower 4 kΩ. What is lower-reference output?

Put the idea to work

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
  1. The resistance ratio is 2:4 = 1:2.
  2. Split 12 V into three equal parts of 4 V.
  3. Upper drop is 4 V, lower drop and output 8 V.

Answer Upper 4 V; lower-reference output 8 V.

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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.

1 · Guided

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
  1. Rtotal = 6 kΩ.
  2. Vout = 12 × 4/6.
  3. Vout = 8 V.

8 V.

Look for these in your work

  • Identify the lower section.
  • Retain negligible output loading.
2 · Independent

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
  1. Upper p.d. is 10 − 4 = 6 V.
  2. Rlower/3 = 4/6.
  3. Rlower = 2 kΩ.

2 kΩ.

Look for these in your work

  • Pair resistance and voltage consistently.
  • Check shares sum to 10 V.
3 · Transfer

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
  1. Upper drop is 9 − 3 = 6 V.
  2. Rupper/2 = 6/3, so Rupper = 4 kΩ.
  3. 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.
Recall without your notes

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.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Use matched resistance and voltage ratios plus the total p.d., with output pair and loading stated.

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.

Keep learning Sciences — Co-ordinated (Double)

Course scope and assessment

This 2028–2029 pathway maps all 152 detailed subject-content sections in the published 0654 syllabus to original teaching and staged practice. It adapts teaching from the related single-science pathways and excludes clear 0610-only topics such as classification systems, human excretion, tropic responses, food supply, population dynamics, detailed pollution and biotechnology. Some explanations may include wider single-science context. A 0654-specific objective-by-objective review, Core/Supplement audit, human subject-teacher review, supervised laboratory experience and cumulative assessment remain in progress. Written practice is self-checked and does not predict an official grade. The 2025–2027 syllabus has not been reconciled here; choose the 2028–2029 syllabus only.

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