Potential dividersAQA A-Level Physics: Revision notes

Section 1

The potential divider principle

A potential divider is two (or more) resistors in series across a supply. The same current flows through both, so the supply pd is shared in proportion to the resistances:

$V_{out} = \frac{R_2}{R_1 + R_2} \times V_{in}$

where $R_2$ is the resistor across which the output is taken. The larger a resistor, the larger its share of the pd. A potential divider supplies a constant pd when the resistors are fixed and a variable pd when one of them can change.

To derive it: $I = V_{in}/(R_1 + R_2)$ and $V_{out} = IR_2$ .

Key termspotential divider

Exam tip

Always identify which resistor the output is taken across, then put that resistance on the top line.

Section 2

Variable resistors as dividers

A variable resistor (rheostat or potentiometer track) can be used in two ways. With one end of the track and the slider in the circuit it just varies resistance. With both ends of the track across the supply and the output taken between the slider and one end, it acts as a potential divider that supplies any pd from 0 V up to the supply pd.

A variable divider is used for dimmers, volume controls and to set test voltages.

Key termssliderrheostat

Common mistake

Connecting only one end of the track to the supply makes a variable resistor, not a divider, and the pd cannot reach 0 V.

Section 3

Thermistors and LDRs in dividers

An NTC thermistor has a resistance that decreases as temperature rises. An LDR has a resistance that decreases as light intensity rises. Either can replace one resistor in a divider so the output depends on temperature or light.

Example: an LDR on top with a fixed resistor below. In the dark the LDR's resistance is large, so it takes most of the pd and the output across the fixed resistor is small. In bright light the LDR's resistance is small and the output is large. Swapping the positions of the sensor and the fixed resistor reverses the behaviour.

The output is often fed to a switching circuit, which turns a device on when the output crosses a set value.

Key termsthermistorLDR

Common mistake

Saying the sensor 'lets more current through' without saying how this changes the pd share. Always argue in terms of shares of the supply pd.

Section 4

Loading the output

If a load (a lamp, motor or meter) is connected across the output, it is in parallel with the output resistor. The combined resistance is lower, so the output pd falls below its unloaded value. The effect is small when the load resistance is much larger than the output resistor. Switching circuits and voltmeters have very high resistance, so their effect is usually negligible.

To work out a loaded output, combine the load with the output resistor in parallel first, then use the divider formula.

Key termsload

Section 5

Designing a divider

To design a divider choose the resistances so the output has the required value.

Worked example: from 6.0 V, an output of 2.0 V is needed across a resistor $R$ in series with a 4.0 kΩ thermistor. $\frac{R}{R + 4.0} = \frac{2.0}{6.0}$ , so $3R = R + 4.0$ and $R = 2.0$ kΩ.

Check by recalculating: $6.0 \times 2.0/6.0 = 2.0$ V.

Must Know

$V_{out} = \frac{R_2}{R_1 + R_2} V_{in}$
Both ends of a variable track across the supply gives 0 V to $V_{in}$
Thermistor: R falls as temperature rises; LDR: R falls as light rises
Larger resistance in series takes the larger share of pd
A load in parallel with the output lowers the output pd

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Potential dividers

A potential divider consists of a 6.0 kΩ resistor in series with a 3.0 kΩ resistor connected across a 9.0 V supply of negligible internal resistance. The output is taken across the 3.0 kΩ resistor.
Explain why connecting the load across the output reduces the output pd.2 marks
A light-dependent resistor (LDR) is connected in series with a fixed 2.2 kΩ resistor across a 5.0 V supply of negligible internal resistance, forming a potential divider that controls a street lamp. The output is taken across the fixed resistor. The resistance of the LDR is 500 Ω in bright light and 20 kΩ in darkness.
Calculate the output pd in darkness.2 marks
A designer needs to supply a motor with a variable pd from 0 V to 12 V from a 12 V supply of negligible internal resistance. A variable resistor with a track of total resistance 100 Ω and a sliding contact is available.
Describe how the variable resistor should be connected so that it acts as a potential divider supplying a pd that can be varied from 0 V to 12 V.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).