Potential Dividers

A potential divider is one of the most useful and commonly tested circuits at A-Level. It allows you to obtain any output voltage between zero and the supply voltage by choosing appropriate resistor values.

How a Potential Divider Works

A potential divider consists of two (or more) resistors connected in series across a voltage supply. The output voltage is taken from the point between the resistors.

Consider two resistors, R₁ and R₂, connected in series across a supply voltage V_in. The output voltage V_out is taken across R₂.

Since the same current flows through both resistors (series circuit):

I = V_in / (R₁ + R₂)

The voltage across R₂ is:

V_out = V_in × R₂ / (R₁ + R₂)

This is the potential divider equation. Notice that the output voltage is simply the fraction R₂/(R₁ + R₂) of the input voltage.

Equivalently, the ratio of the two output voltages is:

V₁ / V₂ = R₁ / R₂

The voltage divides in the same ratio as the resistances.

Worked Example 1

A 12 V supply is connected across two resistors: R₁ = 4.0 kΩ and R₂ = 8.0 kΩ. Calculate the output voltage across R₂.

Solution: V_out = V_in × R₂/(R₁ + R₂) = 12 × 8.0/(4.0 + 8.0) = 12 × 8.0/12 = 8.0 V

Alternatively: V₁/V₂ = R₁/R₂ = 4.0/8.0 = 1/2, so V₂ = 2V₁. Since V₁ + V₂ = 12, we get V₁ = 4.0 V and V₂ = 8.0 V.

Worked Example 2: Designing a Potential Divider

A student needs an output voltage of 3.0 V from a 9.0 V supply. They have a 6.0 kΩ resistor for R₁. What value of R₂ do they need?

Solution:

V_out = V_in × R₂/(R₁ + R₂)
3.0 = 9.0 × R₂/(6.0 + R₂)
3.0(6.0 + R₂) = 9.0 × R₂
18 + 3.0R₂ = 9.0R₂
18 = 6.0R₂
R₂ = 3.0 kΩ
Check: 9.0 × 3.0/(6.0 + 3.0) = 27/9.0 = 3.0 V ✓

Potential Dividers with Sensors

The real power of potential dividers appears when one of the resistors is replaced with a sensor — either an LDR or an NTC thermistor. This creates a circuit whose output voltage varies automatically with environmental conditions.

Using a Thermistor

Replace R₂ with an NTC thermistor. As the temperature increases, the thermistor's resistance decreases.

Configuration 1: Thermistor as R₂ (bottom resistor)

V_out = V_in × R_thermistor / (R₁ + R_thermistor)

Temperature increases → R_thermistor decreases → V_out decreases
Temperature decreases → R_thermistor increases → V_out increases

Configuration 2: Thermistor as R₁ (top resistor)

V_out = V_in × R₂ / (R_thermistor + R₂)

Temperature increases → R_thermistor decreases → V_out increases
Temperature decreases → R_thermistor increases → V_out decreases

The configuration you choose depends on whether you want the output voltage to increase or decrease with temperature.

Using an LDR

Replace one resistor with an LDR. As light intensity increases, the LDR's resistance decreases.

If the LDR is R₂: more light → lower R_LDR → lower V_out

If the LDR is R₁: more light → lower R_LDR → higher V_out

Sensor Behaviour Summary Table

Sensor	Position	Condition Change	Effect on R_sensor	Effect on V_out
NTC Thermistor	R₂ (bottom)	Temperature ↑	Decreases	Decreases
NTC Thermistor	R₁ (top)	Temperature ↑	Decreases	Increases
LDR	R₂ (bottom)	Light ↑	Decreases	Decreases
LDR	R₁ (top)	Light ↑	Decreases	Increases

Exam tip: To remember which way V_out changes, think about the fraction R₂/(R₁ + R₂). If R₂ gets smaller, the fraction gets smaller, so V_out drops. If R₁ gets smaller, the denominator gets smaller while R₂ stays the same, so the fraction gets larger and V_out rises.

Worked Example 3

A potential divider uses a 10 kΩ fixed resistor (R₁) and a thermistor (R₂). The supply is 5.0 V. At 20°C, the thermistor has a resistance of 15 kΩ. At 80°C, it has a resistance of 2.0 kΩ. Calculate V_out at each temperature.

At 20°C: V_out = 5.0 × 15/(10 + 15) = 5.0 × 15/25 = 3.0 V

At 80°C: V_out = 5.0 × 2.0/(10 + 2.0) = 5.0 × 2.0/12 = 0.83 V

The output voltage has dropped from 3.0 V to 0.83 V as the temperature increased, because the thermistor's resistance has fallen dramatically.

Loading Effects

The potential divider equation V_out = V_in × R₂/(R₁ + R₂) assumes that no current is drawn from the output. In practice, if you connect a load (another component or circuit) across the output, this assumption breaks down.

When a load of resistance R_L is connected across R₂, the load and R₂ are in parallel. Their combined resistance is:

R_combined = (R₂ × R_L) / (R₂ + R_L)

This combined resistance is always less than R₂ alone, which means V_out will be lower than the unloaded value.

Worked Example 4: Loading Effect Calculation

A potential divider has R₁ = R₂ = 10 kΩ across a 10 V supply. Without load: V_out = 10 × 10/(10 + 10) = 5.0 V.

Now a 10 kΩ load is connected across R₂.

R_combined = (10 × 10)/(10 + 10) = 100/20 = 5.0 kΩ

V_out = 10 × 5.0/(10 + 5.0) = 10 × 5.0/15 = 3.3 V

The output has dropped from 5.0 V to 3.3 V. The load has significantly affected the output voltage.

Loading Effect: Quantitative Comparison

R_L relative to R₂	R_combined	V_out (if unloaded = 5.0 V, R₁ = R₂ = 10 kΩ)	% drop
100 kΩ (10× R₂)	9.09 kΩ	4.76 V	4.8%
10 kΩ (= R₂)	5.0 kΩ	3.33 V	33%
1.0 kΩ (0.1× R₂)	0.91 kΩ	0.83 V	83%

This table shows clearly why the load resistance should be at least 10 times R₂ for the potential divider equation to give a reasonable approximation.

Minimising Loading Effects

To minimise loading effects, the load resistance should be much larger than R₂ (at least 10 times greater). This ensures that the parallel combination of R₂ and R_L is approximately equal to R₂ alone.

Alternatively, using lower-value resistors in the potential divider makes it less susceptible to loading, but at the cost of greater current draw and more wasted energy.

Variable Resistors as Potential Dividers

A potentiometer (variable resistor with three terminals) acts as a continuously adjustable potential divider. The wiper contact slides along a resistive element, effectively dividing it into two resistances. By moving the wiper, you can obtain any output voltage from 0 V to V_in.

This is how volume controls in audio equipment work — the potentiometer adjusts the fraction of the signal voltage that reaches the amplifier.

Potential Dividers in Sensing Circuits

Potential dividers with sensors are widely used in:

Temperature sensing: A thermistor in a potential divider can trigger a fan or heater when the output voltage crosses a threshold.
Light sensing: An LDR in a potential divider can trigger a street lamp to switch on at dusk.
Moisture sensing: Resistance between probes changes with moisture content.

The output of the potential divider is typically connected to a comparator or transistor switch that activates when V_out exceeds (or falls below) a set threshold voltage.

Worked Example 5: Designing an Alarm Circuit

A fire alarm uses a potential divider with a fixed 5.0 kΩ resistor (R₂) and an NTC thermistor (R₁, top position). The supply is 12 V. The alarm triggers when V_out > 9.0 V. At what thermistor resistance does the alarm trigger?

Solution:

V_out = V_in × R₂/(R₁ + R₂)
9.0 = 12 × 5.0/(R₁ + 5.0)
9.0(R₁ + 5.0) = 60
9.0R₁ + 45 = 60
9.0R₁ = 15
R₁ = 1.67 kΩ
The alarm triggers when the thermistor resistance drops below 1.67 kΩ (i.e., when it gets hot enough).

Common Exam Mistakes

Mistake	Correction
Using V_out = V_in × R₁/(R₁ + R₂) when V_out is across R₂	The output is across whichever resistor you specify — check which one
Ignoring loading effects when a load is connected	Always calculate R_combined = R₂ ∥ R_L and substitute into the divider equation
Confusing which way V_out changes with a sensor	Think about the fraction: if the bottom R decreases, the fraction decreases
Applying the divider formula to a non-series circuit	The potential divider equation only works for resistors in series with no current drawn from the junction

A-Level Deep Dive: Potential Dividers

Spec mapping

Edexcel 9PH0 specification, Topic 3 — Electric circuits, sub-topic on the potential divider treats the two-resistor series network as a fixed-ratio voltage source, with output $V_\text{out} = V_\text{in} \times R_2/(R_1 + R_2)$ across the lower resistor (refer to the official Pearson Edexcel specification document for exact wording). The same sub-topic introduces sensor-driven dividers using the LDR and the negative-temperature-coefficient (NTC) thermistor as the variable resistor, and asks candidates to predict and explain how $V_\text{out}$ responds to changes in light or temperature. Although potential-divider questions appear most often on 9PH0-01 (Paper 1, Advanced Physics I), the design and interpretation work also surfaces on 9PH0-03 (Paper 3, General and Practical Principles in Physics) when sensor circuits feature in extended written-response and practical-design contexts. The formula booklet does not state the divider equation explicitly; it must be derived each time from $I = V/R$ applied to the series loop, which is itself a synoptic test of Ohm's law (3.2) and Kirchhoff's voltage law (3.6).

Worked example with full mark scheme

Question (8 marks):

A student designs a temperature-controlled switching circuit. A 12.0 V supply drives a series combination of a fixed resistor $R_1$ and an NTC thermistor whose resistance $R_T$ varies with temperature. The output voltage $V_\text{out}$ is taken across the thermistor and feeds a switching unit that energises a heater whenever $V_\text{out}$ falls below 4.0 V.

The thermistor's resistance is 18.0 kΩ at 5°C and 1.20 kΩ at 60°C. Treat the supply as ideal and assume negligible current is drawn by the switching unit.

(a) State the potential-divider equation that gives $V_\text{out}$ in terms of $V_\text{in}$ , $R_1$ and $R_T$ . (1)

(b) The student wants the heater to switch on when the temperature falls to 5°C. Show that the required value of $R_1$ is approximately 36 kΩ. (3)

(c) With this value of $R_1$ in place, calculate $V_\text{out}$ at 60°C and decide whether the heater is on or off at this temperature. (3)

(d) Explain, with reference to the thermistor's behaviour, why the chosen design correctly controls the heater. (1)

Solution with mark scheme:

(a) B1 — $V_\text{out} = V_\text{in} \times R_T/(R_1 + R_T)$ , with $V_\text{out}$ across the thermistor and $R_1$ in series with it.

(b) Step 1 — set up the equation at the switching threshold.

The heater switches on when $V_\text{out} = 4.0\ \text{V}$ , so:

$4.0 = 12.0 \times \dfrac{18\,000}{R_1 + 18\,000}$

M1 — substituting the threshold voltage and the thermistor resistance at 5°C correctly into the divider equation.

Step 2 — rearrange.

$\dfrac{4.0}{12.0} = \dfrac{18\,000}{R_1 + 18\,000} \implies \dfrac{1}{3} = \dfrac{18\,000}{R_1 + 18\,000}$

So $R_1 + 18\,000 = 54\,000$ , giving $R_1 = 36\,000\ \Omega = 36\ \text{k}\Omega$ .

M1 — algebraic rearrangement of the divider equation for $R_1$ .

A1 — $R_1 \approx 36\ \text{k}\Omega$ , matching the printed answer. Because this is a "show that", a candidate writing more decimal places ( $R_1 = 36\,000\ \Omega$ exactly) secures the A1 unambiguously; a bare "36 kΩ" with no working risks the M marks.

$V_\text{out} = 12.0 \times \dfrac{1\,200}{36\,000 + 1\,200} = 12.0 \times \dfrac{1\,200}{37\,200}$

M1 — correct substitution with the new thermistor resistance.

Step 2 — evaluate.

$1\,200/37\,200 \approx 0.0323$ , so $V_\text{out} \approx 12.0 \times 0.0323 \approx 0.39\ \text{V}$ .

A1 — $V_\text{out} \approx 0.39\ \text{V}$ to 2 s.f.

B1 — since $0.39\ \text{V} < 4.0\ \text{V}$ , the switching unit treats this as the "below threshold" condition; therefore the heater is off at 60°C only if the switching logic is "heater on when $V_\text{out}$ is below 4.0 V". A candidate who tracks the logic carefully should note that the design as written would actually leave the heater on at 60°C, which is the wrong way round — and the AO3 mark in part (d) asks them to spot this.

(d) B1 — at low temperatures the thermistor resistance is high, so $V_\text{out}$ (taken across it) is a large fraction of $V_\text{in}$ and exceeds the threshold; at high temperatures the thermistor resistance is low, so $V_\text{out}$ is small. The heater must therefore be wired to switch on when $V_\text{out}$ is above the threshold, or equivalently $V_\text{out}$ should be taken across $R_1$ rather than across the thermistor. Candidates who notice the logic mismatch and explain how to correct it score the mark; candidates who simply restate "the thermistor resistance falls" without addressing the switching logic do not.

Total: 8 marks (B1 + M1 M1 A1 + M1 A1 B1 + B1).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A potential divider consists of a 6.0 V battery (assume zero internal resistance), a fixed 4.7 kΩ resistor and a light-dependent resistor (LDR). The LDR resistance is 800 Ω in bright light and 120 kΩ in darkness. The output $V_\text{out}$ is taken across the LDR.

(a) Calculate $V_\text{out}$ in bright light. (2)

(b) Calculate $V_\text{out}$ in darkness. (2)

(c) The LDR is now connected in parallel with a load of resistance 47 kΩ. Without further calculation, explain qualitatively how the value of $V_\text{out}$ in darkness will change. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1b) — applying $V_\text{out} = 6.0 \times 800/(4\,700 + 800)$ .
A1 (AO1.1b) — $V_\text{out} \approx 0.87\ \text{V}$ (denominator $5\,500$ ).

(b)

M1 (AO1.1b) — applying $V_\text{out} = 6.0 \times 120\,000/(4\,700 + 120\,000)$ .
A1 (AO1.1b) — $V_\text{out} \approx 5.77\ \text{V}$ (or 5.8 V to 2 s.f.).

(c)

M1 (AO2.1) — recognising that the parallel combination of 120 kΩ and 47 kΩ has a resistance smaller than either ( $\approx 33.8\ \text{k}\Omega$ ), so the LDR-side resistance falls.
A1 (AO2.5) — concluding that $V_\text{out}$ in darkness will decrease compared with the unloaded case, because the divider ratio $R_2/(R_1 + R_2)$ is reduced when $R_2$ falls — this is the loading effect.

Total: 6 marks split AO1 = 4, AO2 = 2. AO2 marks reward physical reasoning about the loading effect rather than further calculation; this is a classic Edexcel pattern in which the qualitative final part is harder than the numerical opening parts and discriminates strongly at the top end.

Synoptic links

Connects to:

Topic 3 sub-topic on series and parallel circuits: the divider equation is a special case of the series-current rule $I = V/(R_1 + R_2)$ followed by Ohm's law $V_\text{out} = IR_2$ . Recognising the divider as "two-step series analysis" rather than a memorised formula keeps the physics transparent and lets candidates handle non-standard configurations (e.g. three resistors in series, or a divider with a sensor in either position).
Topic 3 sub-topic on Kirchhoff's voltage law: $V_\text{in} = V_{R_1} + V_\text{out}$ is exactly Kirchhoff's voltage law applied to the divider loop, with the supply emf equalling the sum of the two resistor p.d.s (assuming negligible internal resistance). Every divider problem is implicitly a KVL problem.