4.3ElectricityCore

Potential Dividers

A thermistor or LDR only ever changes its resistance — never directly producing a usable voltage signal on its own. A potential divider is the simple two-resistor trick that converts that changing resistance into a changing voltage, turning a passive sensor into an active part of a circuit.

35 min Video by Physics Online Module 4.3 Electrical circuits

Potential Divider Circuits — 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
Explain how a potentiometer gives a continuously variable output
Design a sensor circuit using a thermistor or LDR to give a rising or falling output
Explain the effect of a load connected across the output

The potential divider equation

Two resistors in series across a supply divide the supply voltage in proportion to their resistance, since the same current flows through both.

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} = 12 \times \frac{9.0}{3.0 + 9.0} = 12 \times \frac{9.0}{12.0}

Answer

V_{out} = 9.0

Tip — A single variable resistor (a potentiometer) acts as both R₁ and R₂ at once — sliding its wiper continuously adjusts the output voltage from a fixed supply.

Sensor circuits

Replacing $R_{1}$ or $R_{2}$ with a thermistor or LDR makes the output voltage respond to temperature or light. Whether the output rises or falls as the sensor’s resistance falls depends entirely on whether the sensor occupies the $R_{1}$ or $R_{2}$ position.

1Low temperature:

V_{out} = 8.0 \times \frac{16}{4.0 + 16} = 8.0 \times \frac{16}{20} = 6.4

2High temperature:

V_{out} = 8.0 \times \frac{2.0}{4.0 + 2.0} = 8.0 \times \frac{2.0}{6.0} \approx 2.67

AnswerV_out = 6.4 V (cold), 2.67 V (hot) — output falls as temperature rises.

Tip — To design a sensor circuit for a specific behaviour, first decide whether you want a rising or falling output, then place the sensor as R₁ or R₂ accordingly — swapping its position reverses the response.

The effect of a load

The simple potential divider equation assumes no current is drawn from the output. A load connected across $R_{2}$ acts in parallel with it, lowering the effective resistance there and pulling $V_{out}$ below the value predicted — an effect minimised by using a load with a much higher resistance than $R_{2}$ .

Equation recap

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

Potential divider equation.

Common mistakes to avoid

Placing a sensor in the wrong position for the intended output behaviour.

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

Ignoring the effect of a low-resistance load connected to the output.

A load in parallel with R2 pulls Vout below the simple no-load prediction — significant unless the load resistance is much larger than R2.

Key takeaways

Potential divider: Vout = Vin × R2/(R1+R2).
A potentiometer (single variable resistor) gives a continuously adjustable output.
Replacing R1 or R2 with a thermistor or LDR creates a sensor circuit; sensor position determines whether output rises or falls with the sensed quantity.
A load connected across the output lowers Vout below the ideal no-load prediction.

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Potential Dividers

35 min Video by Physics Online Module 4.3 Electrical circuits