Potential Dividers: Principles and Applications Revision Notes | A Level AQA Physics

1. Why Potential Dividers Matter

Here is the problem: your power supply gives you 12 V, but your sensor circuit needs 4 V. You cannot change the supply — so how do you get the voltage you need? That is exactly the problem a potential divider solves.

Potential dividers appear throughout real electronics — from temperature-sensing circuits in thermostats to automatic street lights that switch on at dusk. Mastering this topic means you can analyse any sensor circuit the examiner puts in front of you.

2. The Core Principle: Sharing Voltage in a Series Circuit

Think back to what you know about series circuits: the total voltage across the supply is shared between all the components. The share each component receives depends on its resistance relative to the total. A resistor with a larger resistance claims a larger share of the voltage.

A potential divider uses this sharing principle deliberately. Two resistors, $R_1$ and $R_2$, are connected in series across a supply voltage $V_{in}$. The output voltage $V_{out}$ is measured across $R_2$. By choosing the ratio of $R_1$ to $R_2$, you control exactly what fraction of $V_{in}$ appears at the output.

Picture the circuit: $V_{in}$ at the top, $R_1$ connected first, then $R_2$ to the bottom (0 V). The output terminals sit either side of $R_2$. Current flows through both resistors in series, so the same current passes through $R_1$ and $R_2$. That shared current is the key to deriving the equation.

The derivation is worth seeing once. The current through the series chain is $I = \frac{V_{in}}{R_1 + R_2}$. The voltage across $R_2$ is $V_{out} = I \times R_2 = \frac{V_{in}}{R_1 + R_2} \times R_2$, which rearranges to the equation above. Every step follows directly from Ohm's law and the series circuit rule — no new physics required.

3. Calculating Output Voltage: Worked Examples

The equation is straightforward, but exam questions often disguise the numbers or ask you to find a resistance rather than a voltage. Work through the following example carefully, noting where each mark would be awarded.

Worked example: Calculating the Output Voltage of a Potential Divider

A potential divider consists of $R_1 = 6000\ \Omega$ and $R_2 = 3000\ \Omega$ connected in series across a 9 V supply. Calculate the output voltage $V_{out}$ taken across $R_2$. [3 marks]

1. $V_{out} = \frac{R_2}{R_1 + R_2} \times V_{in}$ — [M1] Correct formula stated. Always write the equation first — this secures the method mark even if arithmetic goes wrong later.
2. $V_{out} = \frac{3000}{6000 + 3000} \times 9$ — [M1] Values substituted with units consistent (all in ohms and volts). Note $R_1 + R_2 = 6000 + 3000 = 9000\ \Omega$.
3. $\frac{3000}{9000} = \frac{1}{3}$ — The fraction tells you $V_{out}$ is one-third of the supply. This is a useful sanity check: $R_2$ is half of $R_1$, so it claims one-third of the total voltage.
4. $V_{out} = \frac{1}{3} \times 9 = 3\ \text{V}$ — [A1] Final answer with unit. Sanity check: 3 V is less than 9 V, which is physically correct for a potential divider output.

Answer: $V_{out} = 3\ \text{V}$

Now try the reverse: finding a resistance given the desired output voltage. This is the type of question that separates confident students from those who only memorise the formula in one direction.

Worked example: Finding a Resistance for a Required Output Voltage

A potential divider has $R_1 = 4000\ \Omega$ and a 12 V supply. Calculate the value of $R_2$ needed to produce an output voltage of 4 V across $R_2$. [3 marks]

1. $V_{out} = \frac{R_2}{R_1 + R_2} \times V_{in}$ — [M1] Starting from the standard equation avoids sign errors when rearranging.
2. $4 = \frac{R_2}{4000 + R_2} \times 12$ — Known: $V_{out} = 4\ \text{V}$, $V_{in} = 12\ \text{V}$, $R_1 = 4000\ \Omega$. Unknown: $R_2$.
3. $4(4000 + R_2) = 12 R_2$ $16000 + 4R_2 = 12R_2$ $16000 = 8R_2$ — [M1] Multiply both sides by $(4000 + R_2)$, then expand and collect $R_2$ terms.
4. $R_2 = \frac{16000}{8} = 2000\ \Omega$ — [A1] Sanity check: $\frac{2000}{4000+2000} \times 12 = \frac{2000}{6000} \times 12 = \frac{1}{3} \times 12 = 4\ \text{V}$. Confirmed correct.

Answer: $R_2 = 2000\ \Omega$

4. Variable Resistors in Potential Dividers

A fixed potential divider gives a fixed output — useful, but limited. Replace one of the resistors with a variable resistor (rheostat) and the output voltage becomes adjustable. This is how volume controls, dimmer switches, and many calibration circuits work.

Suppose $R_2$ is a variable resistor. As you increase $R_2$, the fraction $\frac{R_2}{R_1+R_2}$ increases, so $V_{out}$ increases. Decrease $R_2$ toward zero and $V_{out}$ approaches zero. The variable resistor gives you continuous control over the output voltage between 0 V and a maximum just below $V_{in}$.

This is the principle behind a potential divider supplying a variable potential difference from a fixed power supply — one of the core requirements of the AQA specification. The output is not fixed; it tracks the resistance setting of the variable component.

5. Sensor-Based Potential Dividers: Thermistors and LDRs

The real power of the potential divider comes when one resistor is a sensor — a component whose resistance changes in response to a physical quantity. The circuit then converts a physical change (temperature, light level) into a changing voltage that can trigger a switch or feed a microcontroller.

Two sensors appear explicitly in the AQA specification: the thermistor and the light-dependent resistor (LDR). Both are used in the same circuit topology, but they respond to different physical quantities. Understanding how each one shifts the output voltage is the key skill the examiner tests.

5a. Thermistor in a Potential Divider

An NTC thermistor has a resistance that decreases as temperature rises. Place the thermistor as $R_2$ (the lower resistor, across which $V_{out}$ is measured) with a fixed resistor $R_1$ above it. At low temperature, the thermistor resistance is high, so it claims a large share of $V_{in}$ — $V_{out}$ is high. As temperature rises, thermistor resistance falls, its share of the voltage falls, and $V_{out}$ decreases.

Swap the positions — put the thermistor as $R_1$ and the fixed resistor as $R_2$ — and the relationship inverts: rising temperature now produces a rising $V_{out}$. Choosing which position the sensor occupies lets you design the circuit response to suit the application.

5b. LDR in a Potential Divider

An LDR behaves similarly but responds to light intensity rather than temperature. Its resistance is high in darkness and decreases as light intensity increases. The analysis follows exactly the same logic as for the thermistor — only the physical trigger changes.

Consider an LDR placed as $R_1$ (upper resistor) with a fixed resistor $R_2$ below it. In bright light, the LDR resistance is low, so the fraction $R_2/(R_1+R_2)$ is large and $V_{out}$ is high. In darkness, the LDR resistance is high, the fraction is small, and $V_{out}$ is low. This configuration could drive an automatic light switch: when $V_{out}$ drops below a threshold voltage, a comparator circuit switches the lights on.

Notice the pattern: whichever resistor is the sensor, when its resistance decreases, the voltage across it decreases. The voltage across the other resistor increases by the same amount (since $V_1 + V_2 = V_{in}$ always). This conservation of voltage across a series circuit is the underlying principle that makes all sensor-based potential dividers work.

6. Worked Sensor Circuit Analysis

Exam questions on sensor potential dividers often give you a resistance value for the sensor at a specific condition and ask you to calculate $V_{out}$. The method is identical to the fixed-resistor case — you substitute the sensor's resistance at that condition into the standard equation.

Worked example: Thermistor Potential Divider: Calculating Output at Two Temperatures

A thermistor is connected as $R_2$ in a potential divider with a fixed resistor $R_1 = 5000\ \Omega$ and a supply voltage $V_{in} = 10\ \text{V}$. At 20°C the thermistor resistance is $15000\ \Omega$. At 80°C the thermistor resistance is $1000\ \Omega$. Calculate $V_{out}$ at each temperature and describe the trend. [4 marks]

1. $V_{out} = \frac{15000}{5000 + 15000} \times 10 = \frac{15000}{20000} \times 10$ — [M1] Correct substitution of thermistor resistance at 20°C.
2. $\frac{15000}{20000} = 0.75$, so $V_{out} = 0.75 \times 10 = 7.5\ \text{V}$ — [A1] At low temperature, the thermistor has high resistance and claims most of the supply voltage.
3. $V_{out} = \frac{1000}{5000 + 1000} \times 10 = \frac{1000}{6000} \times 10$ — [M1] Correct substitution of thermistor resistance at 80°C.
4. $\frac{1000}{6000} = \frac{1}{6} \approx 0.167$, so $V_{out} = 0.167 \times 10 \approx 1.67\ \text{V}$ — [A1] At high temperature, the thermistor resistance is low, so it claims a small fraction of the voltage. $V_{out}$ has fallen from 7.5 V to 1.67 V as temperature increased — confirming the inverse relationship when the thermistor is the lower resistor.
5. As temperature increases from 20°C to 80°C, $V_{out}$ decreases from 7.5 V to 1.67 V. — This is because the thermistor resistance decreases with temperature, reducing its share of $V_{in}$. A describe question requires you to state what happens in sequence — no 'because' needed, but the direction and the quantities strengthen the answer.

Answer: $V_{out} = 7.5\ \text{V}$ at 20°C; $V_{out} \approx 1.67\ \text{V}$ at 80°C. Output voltage decreases as temperature increases.

7. Exam Technique and Common Mistakes

Potential divider questions are reliable mark-earners if you follow a consistent method. The mistakes that cost marks are predictable — and avoidable once you know what to look for.

Potential dividers are a compact but versatile topic. The core equation, the sensor behaviour rules, and the position-of-sensor logic together cover every question type the examiner can set. Practise identifying the output resistor, substituting confidently, and checking your answer against $V_{in}$ — and this topic becomes a reliable source of marks.