Derive and use combined resistance in series
Use Kirchhoff’s laws to show that series resistances add and calculate total resistance and current.
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
What is common to resistors connected in series?
What you will learn
- Derive the series-resistance equation using equal current and p.d. addition
- Use Rₛ = R₁ + R₂ + … for resistors in series
- Predict current and component p.d. in a series chain
Same current flows through each series resistor
In a series chain there is one path, so the current I is the same through every resistor. Kirchhoff's loop law says the total p.d. is V = V₁ + V₂ + … .
Using V = IR for each ohmic resistor gives IRₛ = IR₁ + IR₂ + … . Divide by the common current I to derive Rₛ = R₁ + R₂ + … .
Use the equivalent resistance for the whole chain
For two series resistors 4 Ω and 6 Ω, the equivalent is 10 Ω. The same total current would pass through this single equivalent resistor as through the original chain for the same applied p.d.
The total resistance is greater than each positive individual resistance. Once the total current is known, use V₁ = IR₁ and V₂ = IR₂ to find each p.d.; their sum should equal the source terminal p.d. in the ideal external loop.
Pause and explain
What is the equivalent resistance of 5 Ω and 7 Ω in series?
Worked example
Two resistors of 3.0 Ω and 9.0 Ω are in series across 12 V. Find equivalent resistance and current.
Show the worked solution
- Rₛ = 3.0 + 9.0 = 12 Ω.
- I = V/Rₛ = 12/12.
- Check the same current flows through both resistors.
Answer Rₛ = 12 Ω; I = 1.0 A.
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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.
Three resistors of 2.0 Ω, 4.0 Ω and 6.0 Ω are in series. Find their equivalent resistance.
Give me a hint
Add all resistances along the single current path.
Compare my reasoning
- Rₛ = 2.0 + 4.0 + 6.0.
- Rₛ = 12 Ω.
- Check the result exceeds each individual resistance.
12 Ω.
Look for these in your work
- All resistors included
- Series rule applied
- Magnitude check sensible
A 15 V source is connected across 5.0 Ω and 10 Ω resistors in series. Find the current and each resistor's p.d.
Give me a hint
Find total resistance first; the same current then flows through both.
Compare my reasoning
- Rₛ = 5.0 + 10 = 15 Ω, so I = 15/15 = 1.0 A.
- V₅ = IR = 1.0 × 5.0 = 5.0 V.
- V₁₀ = 1.0 × 10 = 10 V; the drops total 15 V.
I = 1.0 A; p.d.s are 5.0 V and 10 V.
Look for these in your work
- Equivalent found
- Shared current used
- Drops sum to source p.d.
A 4.0 Ω and an unknown resistor are in series. The total resistance is 13 Ω. Find the unknown resistance and explain the circuit check.
Give me a hint
Subtract the known series resistance from the total.
Compare my reasoning
- Runknown = 13 − 4.0 Ω.
- Runknown = 9.0 Ω.
- The total exceeds both positive series components.
9.0 Ω; 4.0 + 9.0 = 13 Ω.
Look for these in your work
- Series equation rearranged
- Unknown calculated
- Sum verified
Common mistakes
- Adding series currents as if there were a junction
- Using the parallel reciprocal formula for a series chain
- Forgetting that the source p.d. divides among series components
What can you explain now?
What is common to resistors connected in series?
Compare with the explanation
The current
There is only one path through the chain.
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Make it stick
Why do series resistances add when current is common and potential drops add around a loop?
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