Potential divider — AQA A-Level Physics
Teaching note checked against the official specification.
A potential divider is a circuit that uses two or more resistors in series to split a source voltage into smaller output voltages. The output voltage across a resistor is proportional to its resistance relative to the total resistance, given by V_out = V_in x (R2 / (R1 + R2)).
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Potential divider explained
Potential divider covers 2 specification points, including The potential divider used to supply constant… and Examples should include the use of variable….
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Open a point below for its explanation and revision guidance.
The potential divider used to supply constant or variable potential difference from a power supply.
A potential divider consists of two or more components in series across a power supply. The supply p.d. is shared between the components in proportion to their resistances, so the output p.d. across one component can be set to a required value. For two resistors, V_out = V_in × R₂ / (R₁ + R₂). If one resistor is replaced by a variable resistor or a potentiometer, the output p.d. can be varied continuously. A thermistor or light-dependent resistor can also form part of a potential divider so that the output p.d. changes with temperature or light intensity. This makes the potential divider a key circuit for supplying a constant or variable p.d. to another component or device.
Examples should include the use of variable resistors, thermistors, and light dependent resistors (LDR) in the potential divider.
This statement requires you to apply the potential divider to three sensing components. A variable resistor changes its resistance by hand, so the output p.d. can be set or adjusted. A thermistor's resistance falls as temperature rises (negative temperature coefficient), so in a divider its output p.d. changes with temperature. An LDR's resistance falls as light intensity rises, so its output p.d. changes with illumination. In each case use V_out = V_in × R₂ / (R₁ + R₂), where R₂ is the resistor across which the output is taken. For example, with a 6.0 V supply, a fixed 2.0 kΩ resistor and a thermistor of resistance 4.0 kΩ, the p.d. across the thermistor is 6.0 × 4000 / 6000 = 4.0 V. Decide which component is R₂, then substitute.
Your focus
- Describe the structure and purpose of a potential divider circuit.
- Calculate the output potential difference of a potential divider using the resistance ratio.
- Explain how a variable resistor or sensor can be used in a potential divider to produce a changing output p.d.
Show all 6 objectives
- Describe how a variable resistor, thermistor and LDR each change resistance in response to a physical quantity.
- Calculate output p.d. in potential divider circuits containing these components.
- Explain how a potential divider converts a resistance change into a usable output p.d. change.
Which command word is this?
You must use the correct formula, substitute the given values, and show your working. The final answer must include the correct unit and be given to an appropriate number of significant figures (usually 2 or 3).
Potential divider exam tips
Topic Overview
The potential divider is a fundamental circuit configuration in A-level Physics that allows a source voltage to be split into smaller, usable voltages. It consists of two or more resistors in series, and the output voltage is taken across one of them. This principle is essential for understanding sensor circuits, volume controls, and biasing in electronic devices.
In the AQA A-Level specification, potential dividers are studied under 'Electricity' and are often combined with thermistors and light-dependent resistors (LDRs) to create sensing circuits. Mastery of this topic is crucial because it links directly to practical applications and provides a foundation for more advanced circuit analysis, such as internal resistance and electromotive force.
Key Concepts
- →The output voltage of a potential divider is given by V_out = V_in x (R2 / (R1 + R2)), where R2 is the resistor across which the output is measured.
- →The current through each resistor in a series potential divider is the same, and the sum of the potential differences across the resistors equals the input voltage (Kirchhoff's second law).
- →The ratio of the resistances determines the ratio of the voltages: V1/V2 = R1/R2.
- →When a load resistor is connected in parallel with the output resistor, the equivalent resistance of that section decreases, which reduces the output voltage.
- →Thermistors and LDRs can be used in potential dividers to produce a voltage that varies with temperature or light intensity, enabling sensing circuits.
Marking Points
- Describe a potential divider as two or more components connected in series across a power supply.
- State that the potential difference across each component is proportional to its resistance when the current is the same.
- Use the potential divider equation V_out = V_in × R₂ / (R₁ + R₂) to calculate an output p.d.
- Explain how replacing a fixed resistor with a variable resistor or using a potentiometer allows the output p.d. to be varied.
- Explain how a thermistor or light-dependent resistor in a potential divider produces an output p.d. that changes with temperature or light intensity.
- Design or interpret a potential divider circuit to supply a required constant or variable p.d. to a component.
- Applies V_out = V_in × R₂ / (R₁ + R₂) to circuits containing a variable resistor, thermistor or LDR.
- Explains that a variable resistor allows the output p.d. to be adjusted manually.
- Explains that a thermistor's resistance decreases as temperature increases, changing the output p.d.
- Explains that an LDR's resistance decreases as light intensity increases, changing the output p.d.
- Identifies which component the output p.d. is measured across before calculating.
Examiner Tips
- 💡Sketch the circuit clearly and label the input p.d., output p.d. and component resistances before calculating.
- 💡When using the potential divider equation, identify which resistor is R₂ by checking where the output p.d. is measured.
- 💡For variable output questions, state how the resistance change affects the share of the supply p.d. and hence the output.
- 💡Name the sensing component and state how its resistance changes before doing the calculation.
- 💡Write the divider equation, identify R₂, then substitute values with units.
- 💡For explain questions, link the resistance change to the change in share of the supply p.d. across each component.
- 💡Always write down the potential divider formula before substituting numbers. This ensures you use the correct resistor in the numerator and avoids careless errors.
- 💡When explaining changes in output voltage due to a sensor (e.g., thermistor), refer to the change in resistance and how it affects the proportion of the input voltage across each component. Use phrases like 'the resistance of the thermistor decreases, so the potential difference across it decreases'.
- 💡For calculation questions, check that your final answer is sensible: the output voltage must be less than the input voltage and should be proportional to the resistance ratio.
Common Mistakes
- Assuming the output p.d. is always half the supply p.d. Correction: the output p.d. depends on the ratio of the resistances, so it is half only when the two resistances are equal.
- Using the resistance of the component across which the output is not taken in the numerator of the potential divider equation. Correction: the numerator is the resistance across which the output p.d. is measured.
- Ignoring the effect of connecting a load across the output. Correction: a load in parallel with the output resistor changes the effective resistance and therefore changes the output p.d.
- Reversing the thermistor or LDR resistance trend; correct this by remembering that resistance falls as temperature or light intensity rises.
- Taking the output across the wrong component; correct this by stating which resistor is R₂ before substituting into the equation.
- Mixing resistance units; correct this by converting all resistances to the same unit, such as ohms or kilohms, before calculating.
- Students often think that the output voltage is always across the larger resistor. In fact, it depends on which resistor the output is taken across; the larger resistor will have the larger voltage drop, but the output could be across either.
- Many students assume that adding a load resistor in parallel does not affect the potential divider. In reality, the load draws current and changes the equivalent resistance, altering the output voltage.
- Some students confuse potential difference with current, thinking that the output voltage is determined by the current through the resistors. The output voltage is determined by the resistance ratio, not the absolute current.
Revision Plan
- 1Day 1-2: Revise the potential divider formula and practice simple calculations with two fixed resistors. Ensure you can rearrange the formula to find any variable.
- 2Day 3-4: Learn how thermistors and LDRs behave (resistance changes with temperature/light) and how they are used in potential dividers. Practice explaining the effect on output voltage.
- 3Day 5-6: Study the effect of connecting a load resistor in parallel with the output resistor. Practice calculations involving parallel combinations and explain why the output voltage changes.
- 4Day 7-8: Attempt past paper questions on potential dividers, focusing on both calculations and explanations. Review mark schemes to understand the level of detail required.
- 5Day 9-10: Create a summary sheet with key formulas, graphs, and example circuits. Test yourself with active recall questions and teach the topic to a peer.
Exam Question Types
- 📋Calculation questions: These ask you to calculate the output voltage given resistor values and input voltage, or to find a resistor value given the desired output voltage. Always show your working and include units.
- 📋Explanation questions: These require you to describe and explain how the output voltage changes when a sensor's resistance changes (e.g., temperature increase for a thermistor). Use precise language linking resistance change to voltage division.
- 📋Circuit analysis with load: These questions involve a load resistor connected in parallel with the output resistor. You must calculate the new equivalent resistance and then the new output voltage, and often explain why the output voltage decreases.
- 📋Practical application questions: These may ask you to design a potential divider to provide a specific voltage or to explain how a potential divider can be used in a sensing circuit (e.g., a light-sensing circuit with an LDR).
Command Word Expectations (AQA)
You must use the correct formula, substitute the given values, and show your working. The final answer must include the correct unit and be given to an appropriate number of significant figures (usually 2 or 3).
You must provide a logical chain of reasoning that links cause and effect. For example, 'As temperature increases, the resistance of the thermistor decreases, so the potential difference across it decreases, hence the output voltage decreases.' Simply stating the change without reasoning will not gain full marks.
You must state what happens without necessarily giving a reason. For example, 'The output voltage decreases as temperature increases.' However, in A-level, 'describe' often requires a brief explanation to secure full marks, so always add a reason if you can.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A potential divider consists of a 12 V battery connected to two resistors in series: R1 = 300 Ω and R2 = 600 Ω. Calculate the output voltage across R2.
- 1.Step 1: Identify given facts: V_in = 12 V, R1 = 300 Ω, R2 = 600 Ω, output across R2.
- 2.Step 2: Apply the potential divider formula: V_out = V_in x (R2 / (R1 + R2)).
- 3.Step 3: Substitute values: V_out = 12 x (600 / (300 + 600)) = 12 x (600 / 900) = 12 x 0.6667 = 8 V.
- 4.Step 4: State final answer with units: V_out = 8 V.
Question: A thermistor is used in a potential divider with a fixed resistor R = 1.0 kΩ. At 20 °C, the thermistor resistance is 2.0 kΩ. The input voltage is 9.0 V. Calculate the output voltage across the thermistor. Describe and explain how the output voltage changes as the temperature increases.
- 1.Step 1: Identify given facts: V_in = 9.0 V, R_fixed = 1.0 kΩ, R_thermistor = 2.0 kΩ at 20 °C, output across thermistor.
- 2.Step 2: Apply potential divider formula: V_out = V_in x (R_thermistor / (R_fixed + R_thermistor)).
- 3.Step 3: Substitute values: V_out = 9.0 x (2.0 / (1.0 + 2.0)) = 9.0 x (2.0 / 3.0) = 6.0 V.
- 4.Step 4: Explain temperature effect: As temperature increases, thermistor resistance decreases (negative temperature coefficient). This reduces the proportion of the input voltage across the thermistor, so V_out decreases.
- 5.Step 5: State final conclusion: At 20 °C, V_out = 6.0 V; as temperature increases, V_out decreases.