Series, parallel and potential-divider networks
Key idea: H2 Physics lessons on circuit representation, material behaviour, networks and RC transients.
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The core idea
Build the idea
Learn the idea
Big question: How do conservation laws determine currents and voltages in networks?
At a junction, current entering equals current leaving. Around a complete loop, potential rises equal potential drops. Series resistors carry one current and add; parallel resistors share a p.d. and add by reciprocals. A potential divider uses series resistance to provide a chosen fraction of the supply p.d.
Reduce networks without losing the physics
Series components carry the same current and their p.d.s add, so their resistances add. Parallel branches share p.d. and branch currents add, giving the reciprocal rule.
Simplify one recognisable group at a time, redraw if necessary, then work back. A parallel equivalent must be smaller than its smallest branch resistance; this is a useful error check.
Check your understanding: Why is current not necessarily shared equally at a junction?
Parallel branches have the same p.d., but different resistances produce different currents.
Make a sensor divider respond in the intended direction
For an unloaded divider, Vout = VinRout/(Rtop + Rout), where Rout is the resistance directly across which output is measured. Sensor position determines whether output rises or falls.
A load across Rout is in parallel with it. This lowers the effective lower resistance and usually changes the output, so the unloaded formula cannot be used unchanged.
Check your understanding: An LDR is the lower resistor and Vout is across it. What happens to Vout as light increases?
LDR resistance falls, so it takes a smaller fraction of the input and Vout falls.
Fixed and variable potential-divider circuits
The upper circuit has resistors R1 and R2 in series across a supply, with output voltage measured across the lower resistor R2. The lower circuit has a three-terminal potentiometer across a supply, with output measured from its slider to the zero-volt end.
View figure data
| Circuit | Connection |
|---|---|
| Fixed divider | R₁ and R₂ are in series; output is across R₂ |
| Variable divider | The potentiometer track is across the supply; output is taken from its wiper |
Key ideas to keep
- Current divides at junctions; it is not used up by components.
- Only components connected across the same two nodes share p.d.
- An unloaded divider relation changes when a load draws current.
See the reasoning
Worked example
Analyse a loaded potential divider
Question: A 12 V divider has 3.0 kΩ on top and 6.0 kΩ below. A 6.0 kΩ load is connected across the lower resistor. Find unloaded and loaded output p.d.s.
Step 1: Find unloaded output
Why: Before loading, the lower resistor alone determines the output fraction.
Working: Vout = 12[6.0/(3.0 + 6.0)] = 8.0 V.
Step 2: Replace the loaded section
Why: The load and lower resistor share both nodes and are parallel.
Working: Rlower,eff = 6.0 || 6.0 = 3.0 kΩ.
Step 3: Apply the divider again
Why: The source now sees equal upper and effective lower resistances.
Working: Vout,loaded = 12[3.0/(3.0 + 3.0)] = 6.0 V.
Answer: Unloaded output is 8.0 V; the load reduces it to 6.0 V.
Check: Adding a parallel load lowers the lower resistance, so a smaller output fraction is expected.
Another worked model
Question
A 10 V divider uses 8.0 kΩ above 4.0 kΩ; a 6.0 kΩ load is placed across the lower resistor. Calculate the loaded output.
Check the worked solution
The lower branch becomes 4.0 || 6.0 = 2.4 kΩ. It is in series with 8.0 kΩ, so Vout = 10[2.4/(8.0 + 2.4)] = 2.31 V, below the unloaded 3.33 V.
Use a hint if needed
Practise with support
Try this
Find the equivalent resistance of 4.0 Ω and 12 Ω in parallel, then in series with 2.0 Ω.
Hint: Reduce one network block at a time.
Check your answer
Parallel resistance is (4 × 12)/(4 + 12) = 3.0 Ω; total is 5.0 Ω.
Now work without the hint
Practise independently
Your turn
Derive the series and parallel resistor rules, then explain how an NTC thermistor changes divider output when heated.
Check your answer
Series components share current and their p.d.s add, giving Rtotal = ΣR. Parallel branches share p.d. and their currents add, giving 1/Rtotal = Σ(1/R). Heating lowers an NTC resistance; whether Vout rises or falls depends on whether output is across the NTC or the other arm.
Avoid these traps
Common mistakes
Common mistake
A divider ratio is unchanged when a load is connected.
What is wrong with this reasoning?
Show better thinking
A finite load changes the effective resistance of the arm it parallels and therefore changes Vout.
Write for the examiner
Exam guidance
Mark nodes and current directions first, then write junction and loop equations before combining resistors.
Exam-style practice [8 marks]
A 9.0 V temperature alarm uses a 4.0 kΩ fixed resistor above an NTC thermistor, with Vout measured across the thermistor. Its resistance falls from 8.0 kΩ to 2.0 kΩ when hot. Find both outputs. The alarm needs a rising voltage when hot: redesign the divider and explain your choice. Then find the hot output if a 4.0 kΩ alarm input loads the output.
Plan before you answer
- Calculate the stated divider first.
- Swap sensor position for the opposite response.
- For loading, replace the output section by a parallel equivalent.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Initially Vcold = 9[8/(4+8)] = 6.0 V and Vhot = 9[2/(4+2)] = 3.0 V, so this arrangement falls with temperature. Put the NTC on top and measure across the 4.0 kΩ fixed resistor below; then warming reduces the top resistance and the output rises to 9[4/(2+4)] = 6.0 V. With a 4.0 kΩ input across the lower 4.0 kΩ resistor, their equivalent is 2.0 kΩ, so the hot loaded output is 9[2/(2+2)] = 4.5 V.
Come back in three days
Check what stayed with you
Recall question 1
What is shared by parallel branches?
Check the answer
Potential difference.
Recall question 2
What is shared by series components?
Check the answer
Current.
Recall question 3
How is a load across a divider output combined?
Check the answer
In parallel with the output resistor.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 16 states no explicit exclusions. I–V graph gradient language always names the plotted axes; resistance is V/I at an operating point rather than automatically a graph gradient. Metal temperature dependence is explained through drift velocity and NTC behaviour through carrier number density. Terminal p.d. is E − Ir while a source supplies current. Capacitor rules are not copied from resistor rules. The exponential forms assume constant R and C, an ideal source or isolated discharge loop, and τ = RC.
- GCE A-Level H2 PhysicsTopic 16(h) / Topic 16(i) / Topic 16(j) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027