Free Pearson Edexcel International A Level Physics lesson
Potential dividers and sensor decisions
Turn a resistance change into a useful voltage signal.
What you’ll learn
- Derive the unloaded divider equation.
- Predict how sensor position reverses output response.
- Calculate output loading and choose a useful resistance range.
A voltage fraction from a resistance fraction
Connect two resistors in series between an ideal supply and its return. Name the resistor from supply to junction Rtop and the resistor from junction to return Rbottom. If a voltmeter or control input draws negligible current from the junction, both resistors carry I = Vs/(Rtop + Rbottom). The output measured from junction to return is IRbottom, giving the familiar divider ratio.
The numerator belongs to the resistor across which the output is measured. Reversing the measurement points changes the sign; taking the complementary voltage across the upper resistor changes the numerator. Avoid memorising a diagram without labels. When both resistances are multiplied by the same factor, an unloaded output ratio stays fixed, although supply current and total dissipated power change.
The ideal equation assumes negligible source resistance and negligible output loading. It also assumes each resistance remains at its stated operating value, so large self-heating or sensor changes during measurement can invalidate a prediction. A divider is useful for creating a signal level but is usually an inefficient way to supply substantial power at a reduced voltage. A circuit that requires stable load power needs its source and load behaviour considered explicitly.
Vout = Vs Rbottom/(Rtop + Rbottom), for negligible output current
Symbols and units
- Vs is supply voltage in V; Rtop and Rbottom are series resistances in Ω; Vout is voltage across the bottom resistor.
How the equation explains the idea
- With negligible output current, the same I = Vs/(Rtop + Rbottom) flows through both resistors.
- Multiply by Rbottom: Vout = Vs Rbottom/(Rtop + Rbottom). With 6 V, 2 kΩ on top and 1 kΩ below, the output is 2 V.
Worked example
A labelled two-arm divider
- A 9.0 V supply drives Rtop = 3.0 kΩ and Rbottom = 6.0 kΩ.
- Series current = 9.0/9000 = 1.0 mA.
- Output across the lower arm = 1.0 mA × 6.0 kΩ = 6.0 V.
The lower resistor belongs in the numerator because the output is measured across it.
Keep this idea. Label the output terminals and put their resistor in the numerator of the unloaded divider ratio.
Sensor position decides the output direction
An NTC thermistor’s resistance falls as temperature rises, while an LDR’s resistance usually falls as illumination increases. In a divider, that change does not specify the output direction until the sensor’s position and output terminals are known. If the sensor is the lower arm and output is taken across it, falling sensor resistance reduces its share of the supply voltage and output falls.
Put the same sensor in the upper arm with a fixed lower resistor and the result reverses. Falling upper resistance increases total current and the voltage across the fixed lower arm rises. A clear explanation links the environmental input to sensor resistance, then to the divider ratio. It is incomplete to say that heat makes voltage increase without describing the circuit.
Choose the fixed resistor to place the output in a useful range over the intended temperatures or light levels. A sensor that varies from 1 kΩ to 10 kΩ paired with a 1 MΩ resistor may produce a very small or nearly saturated output over much of its range. Use measured sensor data rather than assuming resistance changes linearly with temperature or illumination. Account for self-heating, ambient conditions and the input resistance of the device reading the output.
Sensor below: Vout = Vs Rsensor/(Rfixed + Rsensor). Sensor above: Vout = Vs Rfixed/(Rsensor + Rfixed).
Symbols and units
- Vs and Vout are supply and output voltages in V; Rsensor and Rfixed are positive resistances in Ω.
How the equation explains the idea
- With a sensor below the output junction, increasing Rsensor increases its fraction of the total and raises Vout.
- With the sensor above, Vout = Vs Rfixed/(Rsensor + Rfixed), so increasing sensor resistance lowers output. For an NTC thermistor, warming lowers sensor resistance; first identify its position, then reverse the resistance trend.
Conditions. Ideal unloaded divider supplied by 6 V with a fixed 1 kΩ resistor. Output is measured from the junction to the bottom of the divider. The zero-resistance endpoint is a mathematical limit.
Sensor resistance / kΩ; Output voltage / V.
Putting the sensor below the junction makes output rise as its resistance rises. Putting it above makes output fall. Both outputs are 3 V at 1 kΩ sensor resistance; warming an NTC sensor moves left along these curves.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
A bright-light signal
- A 6.0 V divider has an LDR above a fixed 2.0 kΩ lower resistor.
- In dim light the LDR is 10 kΩ, giving output 6 × 2/12 = 1.0 V.
- In brighter light the LDR falls to 1.0 kΩ, giving output 6 × 2/3 = 4.0 V.
This layout produces a rising output as illumination increases.
Keep this idea. Trace input → sensor resistance → divider ratio, and specify the output terminals.
The measuring device becomes part of the circuit
The simple divider equation assumes the output device draws negligible current. A real load connected from the junction to return is in parallel with the lower resistor. Their equivalent resistance is smaller than either positive resistance alone. Replace the lower arm by this parallel equivalent, then use the divider equation with the upper resistor to calculate the actual output.
Loading lowers the output for this common layout because it reduces the effective lower resistance. The supply current rises, but it now divides between the lower resistor and the load. It is incorrect to keep the original series current and simply subtract a load current afterward. The entire steady circuit has changed, so solve its new equivalent network first and then recover the branch currents.
A high-input-resistance voltmeter or electronic control input causes less loading than a low-resistance load. Scaling both divider resistors upward saves current when unloaded but makes the same finite load more influential. This creates a real design trade-off between power consumption and signal stiffness. Check the required output tolerance, sensor heating and load input rather than assuming that a high-resistance divider is always an improvement.
Rlower,effective = (1/Rbottom + 1/Rload)⁻¹; Vout = Vs Rlower,effective/(Rtop + Rlower,effective)
Symbols and units
- Rbottom is lower-divider resistance in Ω; Rload is output load resistance in Ω; Reff is their parallel equivalent.
- Vs and Vout are supply and output voltages in V.
How the equation explains the idea
- The load draws current from the output, so replace the lower resistor by Reff = Rbottom Rload/(Rbottom + Rload).
- Reff is smaller than Rbottom for any finite positive load. Substituting Reff into the divider formula therefore lowers Vout relative to the unloaded value.
- For equal 1 kΩ divider resistors across 6 V, unloaded output is 3 V. A 1 kΩ load makes Reff = 500 Ω and output 2 V.
Worked example
A load changes a half-supply output
- An unloaded 10 V divider has two 2.0 kΩ resistors and gives 5.0 V.
- A 2.0 kΩ load across the lower resistor makes the lower equivalent 1.0 kΩ.
- The loaded output is 10 × 1/(2 + 1) = 3.33 V.
The output is no longer half the supply because the lower arm has changed.
Keep this idea. Model the output load as a parallel branch before recalculating the divider.
Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.
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