Potential Dividers

A potential divider is a simple but powerful circuit that uses two or more resistors in series to produce an output voltage that is a fraction of the input voltage. Potential dividers are used extensively in sensor circuits and electronic control systems.

Spec mapping: This lesson maps to AQA 7408 specification section 3.5.1.5 — Potential divider, covering the basic two-resistor divider equation V_out = V_in × R₂/(R₁+R₂), the use of variable resistors (potentiometers) to produce continuously variable outputs, and the use of LDRs and thermistors in divider circuits as light and temperature sensors. The qualitative behaviour of loaded versus unloaded dividers is also implicit in the spec coverage. (Refer to the official AQA specification document for exact wording.)

Synoptic links:

• 3.5.1.5 — Series circuits: the potential-divider equation is a direct application of the voltage-sharing rule in series circuits, V_i/V_total = R_i/R_total. Lesson 6 builds the rule; this lesson applies it.
• 3.5.1.3 — Current/voltage characteristics: the non-linear behaviour of LDRs and thermistors investigated in Lesson 4 is the mechanism that makes them useful in sensor dividers. The monotonic R(T) or R(intensity) relation is what turns environmental change into an output voltage change.
• 3.10 — Engineering applications (synoptic): potential dividers underpin most sensor instrumentation in real engineering — automotive temperature sensors, smart-home light sensors, audio volume controls (the potentiometer in a domestic amplifier), and the input stages of analog-to-digital converters in microcontrollers.

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The Potential Divider Principle

When two resistors R₁ and R₂ are connected in series across a supply voltage V_in, the voltage is divided between them in proportion to their resistances.

Circuit description:

• V_in is applied across R₁ and R₂ in series
• V_out is measured across R₂ (the output voltage)

The Potential Divider Equation

$V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$Vout​=Vin​×R1​+R2​R2​​

Derivation

The current through both resistors (they are in series) is:

I = V_in / (R₁ + R₂)

The voltage across R₂ is:

V_out = IR₂ = V_in × R₂ / (R₁ + R₂)

Similarly, the voltage across R₁ is:

V₁ = IR₁ = V_in × R₁ / (R₁ + R₂)

Note: V₁ + V_out = V_in (conservation of energy) ✓

Key Point: The potential divider equation only works when no current is drawn from the output (i.e., the output is connected to a device with very high resistance, or is unloaded). If a load is connected in parallel with R₂, the effective resistance of the bottom section changes and the simple formula no longer applies.

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Worked Examples

Worked Example 1 — Basic Potential Divider

A potential divider consists of a 4.0 kΩ resistor (R₁) and a 6.0 kΩ resistor (R₂) connected in series across a 10 V supply. Calculate V_out across R₂.

Solution:

V_out = V_in × R₂ / (R₁ + R₂) = 10 × 6000 / (4000 + 6000) = 10 × 6000 / 10000 = 10 × 0.60 = 6.0 V

Worked Example 2 — Designing a Potential Divider

You need to produce a 3.0 V output from a 9.0 V supply. If R₂ = 10 kΩ, calculate the required value of R₁.

Solution:

V_out = V_in × R₂ / (R₁ + R₂)

3.0 = 9.0 × 10000 / (R₁ + 10000)

3.0(R₁ + 10000) = 90000

3.0R₁ + 30000 = 90000

3.0R₁ = 60000

R₁ = 20 kΩ

Check: V_out = 9.0 × 10000 / (20000 + 10000) = 9.0 × 10000/30000 = 9.0 × 1/3 = 3.0 V ✓

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Sensor Circuits Using Potential Dividers

By replacing one of the fixed resistors with a sensor (LDR or thermistor), the output voltage varies automatically in response to environmental changes.

Thermistor in a Potential Divider

Configuration 1: Thermistor as R₂ (bottom resistor)

V_out = V_in × R_thermistor / (R_fixed + R_thermistor)

• Temperature increases → R_thermistor decreases (NTC) → V_out decreases

Configuration 2: Thermistor as R₁ (top resistor)

V_out = V_in × R_fixed / (R_thermistor + R_fixed)

• Temperature increases → R_thermistor decreases (NTC) → denominator decreases → V_out increases

Worked Example 3 — Thermistor Sensor Circuit

A thermistor (NTC) is connected as R₁ in a potential divider with a fixed 10 kΩ resistor as 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.

Solution:

At 20°C: R₁ = 15 kΩ, R₂ = 10 kΩ

V_out = 5.0 × 10000 / (15000 + 10000) = 5.0 × 10000 / 25000 = 2.0 V

At 80°C: R₁ = 2.0 kΩ, R₂ = 10 kΩ

V_out = 5.0 × 10000 / (2000 + 10000) = 5.0 × 10000 / 12000 = 4.17 V

The output voltage increases from 2.0 V to 4.17 V as the temperature rises. This voltage change could be used to trigger a warning system or control a cooling fan.

LDR in a Potential Divider

Configuration 1: LDR as R₂ (bottom resistor)

• Light intensity increases → R_LDR decreases → V_out decreases

Configuration 2: LDR as R₁ (top resistor)

• Light intensity increases → R_LDR decreases → denominator decreases → V_out increases

Worked Example 4 — LDR Sensor Circuit

An LDR is used as R₁ with a fixed 4.7 kΩ resistor as R₂, powered by 6.0 V. In daylight, R_LDR = 500 Ω. In darkness, R_LDR = 200 kΩ. Calculate V_out in each condition.

Solution:

Daylight: R₁ = 500 Ω, R₂ = 4700 Ω

V_out = 6.0 × 4700 / (500 + 4700) = 6.0 × 4700 / 5200 = 5.42 V

Darkness: R₁ = 200 000 Ω, R₂ = 4700 Ω

V_out = 6.0 × 4700 / (200000 + 4700) = 6.0 × 4700 / 204700 = 0.138 V

The output drops dramatically from 5.42 V to 0.14 V when it goes dark. This could be used to switch on automatic lighting.

Exam Tip: Always think about what happens at the extremes. If R₂ → 0, then V_out → 0. If R₂ → ∞, then V_out → V_in. If R₁ → 0, then V_out → V_in. If R₁ → ∞, then V_out → 0. This helps you predict how the output changes.

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Using a Variable Resistor (Potentiometer)

A potentiometer is a three-terminal variable resistor that acts as a continuously adjustable potential divider.

How it works:

• The total resistance of the track is R_total
• A sliding contact (wiper) divides the track into two parts
• Moving the wiper changes the ratio R₂/R_total
• V_out varies continuously from 0 to V_in

Advantage over a simple variable resistor: A potentiometer can give any output voltage from zero up to V_in, whereas a variable resistor in series can only reduce the current (and cannot give 0 V across a component in series with it).

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Loaded Potential Dividers

When a load resistor R_L is connected in parallel with R₂, the effective resistance of the lower section changes.

The effective lower resistance is the parallel combination of R₂ and R_L:

$R_{eff} = \frac{R_2 R_L}{R_2 + R_L}$Reff​=R2​+RL​R2​RL​​

The new output voltage is:

$V_{out} = V_{in} \times \frac{R_{eff}}{R_1 + R_{eff}}$Vout​=Vin​×R1​+Reff​Reff​​

Since R_eff < R₂ (parallel combination is always less than either individual resistance), the loaded output voltage is always less than the unloaded output.

Worked Example 5 — Loaded Potential Divider

A potential divider has R₁ = 10 kΩ and R₂ = 10 kΩ connected to a 12 V supply. Calculate the output voltage (a) unloaded and (b) when a 10 kΩ load is connected across R₂.

Solution:

(a) Unloaded:

V_out = 12 × 10000 / (10000 + 10000) = 12 × 0.5 = 6.0 V

(b) Loaded with R_L = 10 kΩ:

R_eff = (10000 × 10000) / (10000 + 10000) = 100 × 10⁶ / 20000 = 5000 Ω = 5.0 kΩ

V_out = 12 × 5000 / (10000 + 5000) = 12 × 5000 / 15000 = 12 × 1/3 = 4.0 V

The output has dropped from 6.0 V to 4.0 V — a significant reduction. This shows that loading a potential divider can substantially alter the output voltage.

Common Misconception: Many students assume the potential divider equation always gives the correct output. It only works for an unloaded divider. If a load is connected, you must account for the parallel combination. In practice, for the load to have negligible effect, R_L should be much greater than R₂ (at least 10 times greater).

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Worked Example: Light-Sensitive Switch Using an LDR Potential Divider

An automatic outdoor light is built around a potential-divider sensor. The supply is V_in = 9.0 V. A fixed resistor R₁ = 1.0 kΩ sits in the upper position; the lower component is a light-dependent resistor (LDR) whose resistance is approximately 1.0 MΩ in complete darkness and 1.0 kΩ in bright daylight. The output voltage V_out is measured across the LDR and fed to the base of a transistor switch that activates a lamp when V_out exceeds about 5 V. (a) Calculate V_out in complete darkness. (b) Calculate V_out in bright daylight. (c) Determine whether the lamp switches ON or OFF in each case, and explain how the design achieves "lights-on at dusk, lights-off at dawn". (d) Suggest one limitation of this simple design and one improvement.

Solution — Part (a): Output in darkness.

In darkness the LDR resistance is R_LDR = 1.0 MΩ = 1.0 × 10⁶ Ω. The potential-divider equation gives the voltage across the lower component:

V_out = V_in × R_LDR / (R₁ + R_LDR)

V_out = 9.0 × (1.0 × 10⁶) / (1.0 × 10³ + 1.0 × 10⁶) = 9.0 × (1.0 × 10⁶) / (1.001 × 10⁶) ≈ 9.0 × 0.999 ≈ 8.99 V

So in darkness, V_out ≈ 9.0 V — essentially the full supply voltage. This is because the LDR is now a thousand times larger than R₁, so almost all the voltage drops across it.

Part (b): Output in daylight.

In bright daylight the LDR resistance falls to R_LDR = 1.0 kΩ = 1.0 × 10³ Ω. The two resistances are now equal, so the divider gives:

V_out = 9.0 × (1.0 × 10³) / (1.0 × 10³ + 1.0 × 10³) = 9.0 × 0.5 = 4.5 V

Part (c): Lamp switching behaviour.

The switching threshold is 5.0 V — the lamp turns ON when V_out exceeds 5 V.

• Darkness: V_out ≈ 8.99 V > 5.0 V → lamp ON ✓
• Daylight: V_out = 4.5 V < 5.0 V → lamp OFF ✓

The circuit therefore behaves correctly as a dusk-to-dawn switch. The physical mechanism is straightforward: as light intensity falls, the photoconductive material in the LDR loses charge carriers, its resistance rises, and a larger fraction of the supply voltage appears across it. The transistor base voltage crosses the switching threshold at some intermediate light level (corresponding to twilight), turning the lamp on.

Part (d): Limitation and improvement.

Limitation: the transition between "lamp on" and "lamp off" is gradual — there is a range of twilight intensities at which V_out hovers near 5.0 V, causing the transistor to operate in its linear region (partially on, dissipating heat) rather than as a clean switch. This can cause flickering or burn out the transistor over time.

Improvement: add positive feedback via a Schmitt-trigger configuration (or use a comparator IC with hysteresis), so that the switching thresholds for turning on and turning off differ. Typically the lamp would turn on at one light level (e.g. V_out > 5.5 V) and turn off only at a noticeably brighter level (e.g. V_out < 4.5 V). This eliminates flickering during twilight and gives the circuit clear noise immunity.

Note: This worked example illustrates how a single potential divider plus a non-linear sensor (LDR or NTC thermistor) forms the front end of almost every consumer light-sensitive or temperature-sensitive switch — from streetlights to fridge thermostats to garden solar lamps. Examiners frequently set this exact "configure-the-divider" problem and expect candidates to (i) compute V_out at the two extreme sensor states, (ii) identify which state triggers the switch, and (iii) comment intelligently on a limitation. Skipping step (iii) typically costs the AO3 mark.