# Practical potential divider circuits

> Physics · CIE A-Level AS
> Source: https://www.owlsprep.com/study/cie-9702-u12-practical-potential-divider-circuits/

This module covers practical applications of potential divider circuits, including variable voltage output with potentiometers and sensing physical quantities using resistive sensors. You will learn to calculate output voltages and identify common practical errors.

**Prerequisites:** [Basic potential divider theory](https://www.owlsprep.com/study/cie-9702-u12-basic-potential-divider-theory/); [Resistance variation in thermistors and LDRs](https://www.owlsprep.com/study/cie-9702-u11-resistive-sensors/)

## Learning objectives

- Recall the structure and operating principle of practical potential dividers
- Use potential dividers for voltage adjustment and sensor-based measurement
- Solve problems involving potentiometer and sensor divider circuits
- Identify common sources of error in practical divider setups

## 1. Operating Principle of Practical Potential Dividers

**Practical potential divider** — A series circuit connected across a fixed input voltage, with output taken between one terminal and a tap between the two series resistors. Output voltage depends on the ratio of resistances.

*Notation:* V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}

*Example:* For 12V input, R1=1kΩ, R2=2kΩ, Vout = 8V

Unlike ideal theoretical circuits, practical dividers must account for the effect of any load connected across the output. When an external load (such as a voltmeter or bulb) is connected in parallel with the lower resistor, the combined resistance of the parallel pair drops, which changes the output voltage from the ideal no-load value.

**Worked example:** A 10V input is connected across a 10kΩ fixed upper resistor and 10kΩ fixed lower resistor. A 10kΩ voltmeter is connected across the lower resistor to measure Vout. What voltage does the voltmeter read? Calculate the error compared to the no-load output.

1. Step 1: Calculate the ideal no-load output voltage

   $$V_{out(ideal)} = 10 \times \frac{10}{10 + 10} = 5 \text{ V}$$
2. Step 2: Calculate the combined parallel resistance of the lower 10kΩ resistor and 10kΩ voltmeter load

   $$R_{parallel} = \frac{10 \times 10}{10 + 10} = 5 \text{ kΩ}$$
3. Step 3: Calculate the actual output voltage with load connected

   $$V_{out(actual)} = 10 \times \frac{5}{10 + 5} = 3.33 \text{ V}$$
4. Step 4: Calculate the percentage measurement error

   $$\text{Percentage error} = \frac{|5 - 3.33|}{5} \times 100 = 33.4\%$$

> **Exam tip:** Always check if a load is connected across the output; CIE examiners regularly test for the effect of load resistance.

## 2. Potentiometers as Variable Voltage Dividers

**Potentiometer** — A three-terminal variable resistor with a sliding tap that moves along a uniform resistive track, producing a continuously adjustable output voltage from 0V to the full input voltage.

Potentiometers are used in practical circuits wherever an adjustable voltage is needed, from volume controls in audio equipment to adjustable reference voltages for physics experiments. For a uniform resistive track, resistance is proportional to length, so output voltage is directly proportional to the position of the tap along the track.

**Worked example:** A 5.0 cm long uniform potentiometer is connected across a 6.0 V DC supply. What is the output voltage when the tap is 1.5 cm from the negative end of the track?

1. For a uniform potentiometer, resistance is proportional to track length, so the divider rule simplifies to:
2. $$V_{out} = V_{in} \times \frac{\text{length of lower section}}{\text{total track length}}$$
3. Substitute the values given:

   $$V_{out} = 6.0 \times \frac{1.5}{5.0} = 1.8 \text{ V}$$
4. Note: If the tap was 1.5 cm from the positive end, output would be 4.2 V, so always confirm which end you are measuring from.

> **tip**
>
> A 3-terminal potentiometer used as a divider gives output from 0V to full $V_{in}$, unlike a 2-terminal variable resistor which can only reduce voltage to a non-zero minimum.

## 3. Potential Dividers for Resistive Sensing

Potential dividers are the most common practical arrangement for converting resistance changes of sensors into proportional voltage changes that can be measured by a voltmeter or read by a microcontroller. Common sensors used this way include LDRs (light-dependent resistors) for light intensity, and NTC thermistors for temperature.

**Sensor Divider** — A potential divider with one fixed resistor and one variable resistive sensor. Output voltage changes as the physical quantity being measured changes the sensor's resistance.

**Worked example:** An LDR with resistance 100Ω in bright light and 10kΩ in the dark is connected in a divider with a 1kΩ fixed resistor, across a 9V supply. Output is taken across the fixed resistor. Calculate output voltage in bright light and dark, and comment on the result.

1. Step 1: Bright light, $R_{LDR} = 100 \Omega$. Use the divider rule for output across the fixed 1kΩ resistor:

   $$V_{out} = 9 \times \frac{1000}{1000 + 100} = 8.18 \text{ V}$$
2. Step 2: Dark, $R_{LDR} = 10 k\Omega$. Recalculate output voltage:

   $$V_{out} = 9 \times \frac{1000}{1000 + 10000} = 0.82 \text{ V}$$
3. Step 3: Comment: Output voltage decreases as light intensity decreases, so higher voltage corresponds to brighter light. This arrangement can be used to trigger an automatic light when it gets dark.

**Check your understanding**

Check your understanding of this set-up:

1. An NTC thermistor (resistance decreases with increasing temperature) is in the upper position of a divider, with output taken across the lower fixed resistor. What happens to $V_{out}$ when temperature increases?

   - Vout increases
   - Vout decreases
   - Vout stays the same

   *Why:* Correct: NTC resistance drops, so total circuit resistance drops, current increases, so $V_{out} = I \times R_{fixed}$ increases.

## Common pitfalls

- **Wrong:** Ignoring the effect of load resistance when calculating output voltage
  - Why it fails: Ideal no-load voltage is only correct when load resistance is much larger than divider resistance. If load is comparable, output drops significantly.
  - Correct: Always calculate the combined parallel resistance of the lower divider resistor and load before applying the divider rule.
- **Wrong:** Mixing up which resistor output is taken across when applying the divider formula
  - Why it fails: Putting the wrong resistance in the numerator gives an output larger than input or the opposite ratio to the correct value.
  - Correct: Always label your circuit: the resistance that output is measured across goes in the numerator.
- **Wrong:** Assuming NTC thermistor resistance increases with temperature
  - Why it fails: NTC stands for Negative Temperature Coefficient, so resistance decreases when temperature increases, reversing all predictions of output voltage change.
  - Correct: Remember: NTC = Negative, resistance goes Down when Temperature goes Up.
- **Wrong:** Choosing a fixed resistor far from the midpoint of the sensor's resistance range
  - Why it fails: Output voltage changes will be concentrated at only one extreme of the input range, giving poor resolution for most measurements.
  - Correct: Choose a fixed resistor with resistance close to the midpoint of the sensor's working resistance range for your measurement.

## Cheatsheet

| Circuit Type | Formula | Key Property |
| --- | --- | --- |
| Ideal no-load divider | $V_{out} = V_{in} \frac{R_2}{R_1 + R_2}$ | Output depends only on resistance ratio |
| Uniform potentiometer | $V_{out} = V_{in} \frac{l}{L}$ | Output proportional to tap position |
| Sensor divider (output across fixed R) | $V_{out} = V_{in} \frac{R_{fixed}}{R_{sensor} + R_{fixed}}$ | Output changes with sensed physical quantity |
| With load $R_L$ | $V_{out} = V_{in} \frac{R_{parallel}}{R_1 + R_{parallel}}$, $R_{parallel} = \frac{R_2 R_L}{R_2 + R_L}$ | Output lower than ideal no-load value |

## What's next

Practical potential dividers are a core building block for all measurement circuits in CIE AS and A-level Physics, for both written exam questions and practical assessment. Mastering the effect of load resistance and sensor output calculation will help you analyse almost any resistive sensing circuit you encounter in exams and your own practical work. The concepts you learned here also underpin the use of potentiometers for comparing electromotive forces of cells, which is the next key topic in this unit on practical circuits. You will also apply these principles when designing sensor circuits for your practical endorsement assessment.

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