Potential Dividers — AQA A-level Physics

SPEC 3.5.1.5 • ELECTRICITY

Potential Divider

Using resistor networks to supply variable and controlled voltages from a fixed power supply

What is a Potential Divider?

Definition

A potential divider is a circuit consisting of two or more resistors in series that divides the supply voltage into smaller portions.

The potential divider allows us to obtain any output voltage between zero and the supply voltage by choosing appropriate resistance values.

Key Concept

The output voltage

V_{o u t}

is taken across terminals A and B. It equals the voltage across

R_{2}

.

The Potential Divider Equation

Since the resistors are in series, the same current

I

flows through both:

Derivation

Total resistance:

R_{t o t a l} = R_{1} + R_{2}

Current:

I = \frac{V}{R _{1} + R _{2}}

Voltage across

R_{2}

:

V_{2} = I \times R_{2} = \frac{V}{R _{1} + R _{2}} \times R_{2}

Formula

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

Typical Mistake

Taking the ratio of the two resistors rather than the ratio of one resistor to the sum of both.

❌ Wrong:

V_{2} = \frac{R _{2}}{R _{1}} \times V

✓ Correct:

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

Voltage Ratio

An equivalent way to express this:

\frac{V _{1}}{V _{2}} = \frac{R _{1}}{R _{2}}

The voltages are in the same ratio as the resistances.

Worked Example01

A potential divider consists of a 3.0 kΩ resistor ( $R_{1}$ ) connected in series with a 2.0 kΩ resistor ( $R_{2}$ ). The supply voltage is 10 V. Calculate the output voltage.

1

Write down known values:

R_{1} = 3.0 kΩ

,

R_{2} = 2.0 kΩ

,

V = 10 V

2

Apply the potential divider equation:

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

3

Substitute:

V_{2} = \frac{2.0}{3.0 + 2.0} \times 10 = \frac{2.0}{5.0} \times 10

4

Calculate:

V_{2} = 0.4 \times 10 = 4.0 V

Answer: $V_{2} = 4.0 V$

Worked Example02

A 12 V supply is connected to a potential divider. Resistor $R_{1} = 4.7 kΩ$ . What value of $R_{2}$ is needed to produce an output voltage of 5.0 V?

1

Write down:

V = 12 V

,

V_{2} = 5.0 V

,

R_{1} = 4.7 kΩ

2

Rearrange for

R_{2}

:

R_{2} = \frac{V _{2} \times R _{1}}{V - V _{2}}

3

Substitute:

R_{2} = \frac{5.0 \times 4.7}{12 - 5.0} = \frac{23.5}{7.0}

4

Calculate:

R_{2} = 3.36 kΩ \approx 3.4 kΩ

Answer: $R_{2} = 3.4 kΩ$ (2 s.f.)

Worked Example03

In a potential divider circuit, the voltage across $R_{1}$ is 7.5 V and the voltage across $R_{2}$ is 2.5 V. If $R_{1} = 15 kΩ$ , calculate $R_{2}$ .

1

Write down:

V_{1} = 7.5 V

,

V_{2} = 2.5 V

,

R_{1} = 15 kΩ

2

Use the voltage ratio:

\frac{V _{1}}{V _{2}} = \frac{R _{1}}{R _{2}}

3

Rearrange:

R_{2} = \frac{V _{2} \times R _{1}}{V _{1}} = \frac{2.5 \times 15}{7.5}

4

Calculate:

R_{2} = \frac{37.5}{7.5} = 5.0 kΩ

Answer: $R_{2} = 5.0 kΩ$

The Loading Effect

When a component is connected across the output (terminals A and B), it is in parallel with

R_{2}

. This changes the effective resistance and therefore the output voltage.

Important

The load resistance in parallel with

R_{2}

gives a combined resistance less than $R_{2}$ .

This reduces the proportion of voltage dropped across the output, so

V_{2}

is less than expected.

To Calculate

1. Find combined resistance:

R_{co mbin e d} = \frac{R _{2} \times R _{l o a d}}{R _{2} + R _{l o a d}}

2. Use

R_{co mbin e d}

in place of

R_{2}

in the potential divider formula

Worked Example04

A loudspeaker (resistance 100 Ω) is connected across the output of a potential divider. If $V = 6 V$ , $R_{1} = 200 Ω$ , and $R_{2} = 500 Ω$ , calculate the p.d. across the loudspeaker.

1

The loudspeaker (100 Ω) is in parallel with

R_{2}

(500 Ω). Combined resistance:

R_{co mbin e d} = \frac{500 \times 100}{500 + 100} = \frac{50000}{600} = 83 Ω

2

Use potential divider formula with

R_{co mbin e d}

:

V_{2} = \frac{R _{co mbin e d}}{R _{1} + R _{co mbin e d}} \times V

3

Substitute:

V_{2} = \frac{83}{200 + 83} \times 6 = \frac{83}{283} \times 6

4

Calculate:

V_{2} = 0.293 \times 6 = 1.76 V \approx 1.8 V

Answer: p.d. across loudspeaker = 1.8 V

Note: Without the load, $V_{2} = \frac{500}{700} \times 6 = 4.3 V$ . Loading significantly reduces the output!

Variable Resistors (Potentiometers)

Physical Component

Circuit Diagram

How It Works

A potentiometer is a single resistor with a sliding contact (wiper) that can move along its length. The wiper divides the total resistance into two parts: R₁ (above) and R₂ (below).

Moving the wiper changes the ratio R₂/(R₁ + R₂), which changes the output voltage:

Wiper at top: R₂ = maximum, R₁ ≈ 0 → V_out ≈ V_supply

Wiper at bottom: R₂ = 0, R₁ = maximum → V_out = 0

Wiper in middle: R₁ = R₂ → V_out = V_supply/2

Worked Example05

A 10 kΩ potentiometer is connected across a 9.0 V supply. The wiper divides the potentiometer so that the resistance below the wiper is 3.5 kΩ. Calculate: (a) the resistance above the wiper, and (b) the output voltage.

1

Part (a): Total = 10 kΩ, so

R_{1} = 10 - 3.5 = 6.5 kΩ

2

Part (b): Apply potential divider:

V_{2} = \frac{3.5}{10} \times 9.0

3

Calculate:

V_{2} = 0.35 \times 9.0 = 3.15 V

Answers: (a) $R_{1} = 6.5 kΩ$ , (b) $V_{2} = 3.2 V$ (2 s.f.)

Thermistor in a Potential Divider

How It Works

An NTC (Negative Temperature Coefficient) thermistor contains a semiconductor material. As temperature increases, more charge carriers are released, so the resistance decreases.

In the potential divider: V₂ = V × R₂/(R₁ + R₂)

Temperature rises → thermistor resistance R₂ decreases → smaller fraction of V appears across R₂ → V₂ falls

Temperature falls → thermistor resistance R₂ increases → larger fraction of V appears across R₂ → V₂ rises

Application: Temperature sensors, overheat protection, fan speed controllers

Worked Example06

A thermistor is at $R_{2}$ with $R_{1} = 10 kΩ$ , $V = 6.0 V$ . At 20°C: 15 kΩ. At 50°C: 4.0 kΩ. Calculate $V_{2}$ at each temperature.

1

At 20°C (

R_{2} = 15 kΩ

):

V_{2} = \frac{15}{10 + 15} \times 6.0 = \frac{15}{25} \times 6.0 = 3.6 V

2

At 50°C (

R_{2} = 4.0 kΩ

):

V_{2} = \frac{4.0}{10 + 4.0} \times 6.0 = \frac{4.0}{14} \times 6.0 = 1.7 V

3

As temperature increased,

V_{2}

decreased from 3.6 V to 1.7 V.

Answers: 20°C: 3.6 V; 50°C: 1.7 V

LDR in a Potential Divider

How It Works

A Light Dependent Resistor (LDR) contains a semiconductor material. As light intensity increases, more electron-hole pairs are created, so the resistance decreases.

In the potential divider: V₂ = V × R₂/(R₁ + R₂)

Light intensity increases → LDR resistance R₂ decreases → smaller fraction of V appears across R₂ → V₂ falls

Light intensity decreases → LDR resistance R₂ increases → larger fraction of V appears across R₂ → V₂ rises

Application: Automatic lighting, light meters, burglar alarms

Worked Example07

An LDR at $R_{2}$ with $R_{1} = 5.0 kΩ$ , $V = 9.0 V$ . In darkness: 50 kΩ. In bright light: 500 Ω. Calculate $V_{2}$ in each condition.

1

Darkness (

R_{2} = 50 kΩ

):

V_{2} = \frac{50}{5.0 + 50} \times 9.0 = \frac{50}{55} \times 9.0 = 8.2 V

2

Bright light (

R_{2} = 0.50 kΩ

):

V_{2} = \frac{0.50}{5.0 + 0.50} \times 9.0 = \frac{0.50}{5.5} \times 9.0 = 0.82 V

3

Output changes from 8.2 V (dark) to 0.82 V (bright) — a large range for sensing.

Answers: Darkness: 8.2 V; Bright light: 0.82 V

Sensor Position: Top vs Bottom

The position of the sensor determines how

V_{2}

responds:

Sensor R ↓ → V₂ ↓

Sensor R ↓ → V₂ ↑

Important

Swapping the sensor position inverts the response! If sensor is at

R_{1}

, when sensor resistance decreases,

V_{2}

increases.

Worked Example08

A thermistor (25°C: 8.0 kΩ, 40°C: 3.0 kΩ) with 6.0 kΩ fixed resistor, 12 V supply. Compare $V_{2}$ when: (a) thermistor at $R_{2}$ , (b) thermistor at $R_{1}$ .

1

(a) Thermistor at R₂:

25°C:

V_{2} = \frac{8.0}{14} \times 12 = 6.9 V

| 40°C:

V_{2} = \frac{3.0}{9.0} \times 12 = 4.0 V

Temp ↑ → V₂ ↓

2

(b) Thermistor at R₁: (Fixed resistor = R₂ = 6.0 kΩ)

25°C:

V_{2} = \frac{6.0}{14} \times 12 = 5.1 V

| 40°C:

V_{2} = \frac{6.0}{9.0} \times 12 = 8.0 V

Temp ↑ → V₂ ↑

Conclusion: Response is inverted depending on sensor position.

Real-Life Applications

LDR at R₁: Dark → LDR R↑ → V₂↑ → Light ON

Thermistor at R₂: Hot → R₂↓ → V₂↓ → Fan ON

Design Principle

The output voltage V₂ is fed to a switching circuit (e.g., transistor, comparator, or microcontroller) that turns a device ON or OFF when V₂ crosses a threshold.

Street light: LDR at R₁ so V₂ rises in darkness → triggers light ON

Cooling fan: Thermistor at R₂ so V₂ falls when hot → triggers fan ON (or use thermistor at R₁ for V₂ to rise when hot)

Fire alarm: Thermistor at R₁ so V₂ rises rapidly when temperature increases → triggers alarm

Practice

Question 1: A potential divider has

R_{1} = 2.2 kΩ

and

R_{2} = 3.3 kΩ

connected to a 5.0 V supply. Calculate the output voltage across

R_{2}

.

1

V_{2} = \frac{R _{2}}{R _{1} + R _{2}} \times V = \frac{3.3}{2.2 + 3.3} \times 5.0

2

V_{2} = \frac{3.3}{5.5} \times 5.0 = 0.6 \times 5.0 = 3.0 V

Answer: $V_{2} = 3.0 V$

Practice

Question 2: An LDR at

R_{1}

(top) with fixed 4.7 kΩ at

R_{2}

, 6.0 V supply. LDR resistance is 20 kΩ. Calculate

V_{2}

. What happens to

V_{2}

when more light falls on the LDR?

1

V_{2} = \frac{4.7}{20 + 4.7} \times 6.0 = \frac{4.7}{24.7} \times 6.0 = 1.14 V \approx 1.1 V

2

More light → LDR resistance ↓ → Since LDR is at

R_{1}

, the fraction

R_{2} / (R_{1} + R_{2})

increases →

V_{2}

increases

Answer: 1.1 V; More light causes $V_{2}$ to increase.

Practice

Question 3: Design a potential divider that outputs 3.0 V from a 9.0 V supply using

R_{1} = 12 kΩ

. What value should

R_{2}

be?

1

Rearranged formula:

R_{2} = \frac{V _{2} \times R _{1}}{V - V _{2}}

2

R_{2} = \frac{3.0 \times 12}{9.0 - 3.0} = \frac{36}{6.0} = 6.0 kΩ

3

Check:

V_{2} = \frac{6.0}{18} \times 9.0 = 3.0 V

✓

Answer: $R_{2} = 6.0 kΩ$

Summary

Key Formulas

FormulaDescription

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

Potential divider equation

\frac{V _{1}}{V _{2}} = \frac{R _{1}}{R _{2}}

Voltage ratio = resistance ratio

R_{co mb} = \frac{R _{2} \times R _{l o a d}}{R _{2} + R _{l o a d}}

Loading effect (parallel)

Key Exam Skills

Calculate output voltages using the potential divider equation

Design circuits to achieve specific output voltages

Analyse loading effect when components connect across output

Explain how sensor position affects circuit response

Remember

Output is across

R_{2}

• Loading reduces

V_{2}

• Swap sensor position to invert response