Series, parallel and potential-divider networks

Key idea: Circuit networks are solved by combining two conservation ideas: charge flow at junctions and energy transfer around loops.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 3 of 3

Check your understanding

By the end of this lesson, you should be able to

Find combined resistance for series and parallel groups.
Solve mixed networks from one source.
Analyse unloaded and loaded potential dividers, including NTC thermistors and LDRs.

Learn the idea

Big question: How do charge conservation and energy conservation guide every series, parallel and divider calculation?

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.

A fixed two-resistor potential divider beside a three-terminal variable potential divider — Output voltage must be defined between two points. Here it is measured from the junction or slider to the 0 V end of the divider.

Potential-divider topology
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

Potential-divider topology
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

Same current in series; same p.d. in parallel.
Equivalent resistance is not found by averaging branch values.
Define exactly which resistor the output is measured across.

Relationships to know

Rseries = R₁ + R₂ + …
1/Rparallel = 1/R₁ + 1/R₂ + …
Vout = Vin Rout/(Rtop + Rout) for an unloaded divider

Follow 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.

Now try it with support

Practise with support

A 4.0 Ω resistor is in series with a parallel pair of 6.0 Ω and 3.0 Ω. Find the total resistance.

Hints

Combine the parallel pair first.
Then add the series resistance.

View the guided answer

Rparallel = (6.0 × 3.0)/(6.0 + 3.0) = 2.0 Ω. Rtotal = 4.0 + 2.0 = 6.0 Ω.

Your turn

Practise independently

A 4.0 kΩ–6.0 kΩ potential divider is connected to 10 V, then a 6.0 kΩ load is placed across the lower resistor. Calculate the loaded output voltage.

Check your answer

The 6.0 kΩ lower resistor in parallel with the 6.0 kΩ load has equivalent resistance 3.0 kΩ. The loaded divider is therefore 4.0 kΩ above 3.0 kΩ, so Vout = 10[3.0/(4.0 + 3.0)] = 4.29 V.

Common mistakes and exam guidance

Watch out for

Dividing p.d. equally between unequal series resistors.
Using the unloaded divider formula after attaching a finite load.

In an exam

Redraw a loaded divider to show the parallel pair before calculating.
Check limiting behaviour: as an NTC thermistor warms its resistance falls; decide how that changes the measured fraction.

Put the ideas together

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.

View the marking points and model answer

Marking points

Cold output = 6.0 V.
Hot output = 3.0 V.
Places NTC on top and fixed resistor below for rising hot output.
Explains warming lowers top resistance/increases lower voltage fraction.
Finds hot unloaded redesigned output = 6.0 V.
Finds lower loaded resistance 4.0 || 4.0 = 2.0 kΩ.
Finds loaded hot output = 4.5 V.
Recognises loading reduces the expected output.

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.

Finish from memory

Three-question recap

What is shared by parallel branches?
Check
Potential difference.
What is shared by series components?
Check
Current.
How is a load across a divider output combined?
Check
In parallel with the output resistor.

Try this next

Try one divider with the sensor on top and one with it below so you can predict opposite output trends.

Check the whole topic Return to Circuits Review Resistance, resistivity and component behaviour

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027