3.5.1.4ElectricityCore

Potential Dividers

Most sensors — thermistors, LDRs, microphones — don’t output a voltage directly; they only change their resistance. A $potential divider$ is the simple trick that turns that changing resistance into a changing $voltage$ your circuit can actually use, and it’s built from nothing more than two resistors in series.

35 min Video by Science Shorts 3.5.2 DC circuits

Series & Parallel - Potential Divider Circuits — GCSE & A-level PhysicsWatch the full walkthrough before the notes below.

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What you'll be able to do

Use the potential divider equation to find the output voltage of two series resistors
Explain how a variable resistor (potentiometer) gives a continuously adjustable output
Design a sensor circuit using a thermistor or LDR to give an increasing or decreasing output
Explain the effect of connecting a load across the output of a potential divider

The potential divider equation

Two resistors, $R_{1}$ and $R_{2}$ , connected in series across a supply of voltage $V_{in}$ , "divide" that voltage between them in proportion to their resistance (since the same current flows through both, and $V = I R$ ). Taking the output across $R_{2}$ gives a fraction of the input voltage set entirely by the ratio of the two resistances.

V_{out} = V_{in} \times \frac{R _{2}}{R _{1} + R _{2}}

Potential divider equation (output taken across R₂).

1Use

V_{out} = V_{in} \times \frac{R _{2}}{R _{1} + R _{2}}

V_{out} = 9.0 \times \frac{4.0}{2.0 + 4.0} = 9.0 \times \frac{4.0}{6.0}

Answer

V_{out} = 6.0

Tip — A single variable resistor (a potentiometer) can act as both R₁ and R₂ at once — sliding its wiper continuously changes the ratio, giving a smoothly adjustable output voltage from a fixed supply.

Sensor circuits with a thermistor or LDR

Replacing one of the fixed resistors with a $thermistor$ or an $LDR$ makes the output voltage respond automatically to temperature or light. Whether the output voltage $increases$ or $decreases$ as the sensor’s resistance falls depends entirely on which position — $R_{1}$ or $R_{2}$ — the sensor occupies in the divider.

If the sensor is $R_{2}$ (the resistor you’re taking the output across) and its resistance falls (say, an NTC thermistor warming up), $V_{out}$ falls too, since a smaller fraction of the total resistance is on the output side. If the sensor is instead $R_{1}$ , a fall in its resistance makes $V_{out}$ $rise$ , since relatively more of the fixed resistance is now on the output side.

1In darkness:

V_{out} = 6.0 \times \frac{20}{5.0 + 20} = 6.0 \times \frac{20}{25} = 4.8

2In bright light:

V_{out} = 6.0 \times \frac{1.0}{5.0 + 1.0} = 6.0 \times \frac{1.0}{6.0} = 1.0

3Since

V_{out}

is high (4.8 V) in darkness and low (1.0 V) in light, this arrangement is suitable for a circuit that switches something on when it gets dark.

AnswerV_out = 4.8 V (dark), 1.0 V (light) — suitable for a "switch on at night" sensor.

Tip — To design a sensor circuit, first decide whether you want the output to rise or fall as the sensed quantity increases — then place the sensor as R₁ or R₂ accordingly, since swapping its position reverses the response.

The effect of a load

The potential divider equation above assumes nothing else draws current from the output. If a load (such as a voltmeter with finite resistance, or another circuit) is connected across $R_{2}$ , it effectively places a resistor in parallel with $R_{2}$ , lowering the resistance on that side of the divider — and so lowering $V_{out}$ below the value the simple equation predicts.

This effect is smaller the larger the load’s resistance is compared with $R_{2}$ — in practice, potential dividers are designed so any load connected has a resistance much greater than the divider’s own resistors, to minimise this "loading effect".

Tip — An ideal voltmeter has infinite resistance precisely so that connecting it to measure V_out doesn’t itself change the reading it’s trying to measure.

Equation recap

V_{out} = V_{in} \times \frac{R _{2}}{R _{1} + R _{2}}

Potential divider equation (output across R₂).

Common mistakes to avoid

Assuming the sensor should always be placed as R₁, regardless of the desired output behaviour.

Whether output voltage rises or falls with the sensed quantity depends on whether the sensor is R₁ or R₂ — always check which arrangement gives the behaviour you actually want.

Forgetting that connecting a low-resistance load across the output changes the potential divider’s behaviour.

A load in parallel with R₂ lowers the effective resistance there, pulling V_out below the simple no-load prediction — significant loading only becomes negligible when the load’s resistance is much larger than R₂.

Using the input voltage V_in in place of the supply’s terminal p.d. when the source has significant internal resistance.

If the supply itself has internal resistance, some voltage is already "lost" before reaching the divider — use the actual terminal p.d. delivered to the divider, not the labelled e.m.f.

Key takeaways

A potential divider splits a supply voltage between two series resistors: Vout = Vin × R2/(R1+R2).
A potentiometer (single variable resistor) gives a continuously adjustable output from a fixed supply.
Replacing R1 or R2 with a thermistor or LDR creates a sensor circuit; which position gives a rising or falling output depends on which resistor the sensor replaces.
A load connected across the output draws current and lowers Vout below the simple no-load prediction — minimised by using a load with much greater resistance than R2.

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Potential Dividers

Most sensors — thermistors, LDRs, microphones — don’t output a voltage directly; they only change their resistance. A

potential divider

is the simple trick that turns that changing resistance into a changing

voltage

your circuit can actually use, and it’s built from nothing more than two resistors in series.

35 min Video by Science Shorts 3.5.2 DC circuits