Brighter light lowers the resistance of a light-dependent resistor, or LDR. In a series potential divider with a fixed supply, the voltage across the fixed resistor then rises, while the voltage across the LDR falls. Read the voltmeter connections before deciding which way the output moves.
The 2027 SEC G3 Physics K323 syllabus includes simple potential-divider problems involving LDRs. This example follows the same circuit in dim and bright conditions and compares the two possible output measurements.
- Identify the component across which the output is measured.
- Recall that brighter light lowers LDR resistance.
- Find the series current from the supply voltage and total resistance.
- Use the current and the selected component’s resistance to find its voltage.
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Mark the two output terminals
An LDR and a fixed resistor are connected in series across a constant supply. The same current flows through both. Their potential differences add to the supply voltage.
The output is the potential difference between the two selected terminals. If those terminals are across the fixed resistor, use its resistance in V = IR. If the terminals are across the LDR, use the LDR resistance for that light level.
The worked example assumes an ideal fixed supply, negligible connecting-wire resistance and a voltmeter that draws negligible current.
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Worked example | A 6 V divider in dim and bright light
A 2 kΩ fixed resistor is in series with an LDR across 6 V. In dim light the LDR is 4 kΩ. In bright light it is 1 kΩ. Find the voltages across both components.
| Light | Total resistance | Series current | Fixed resistor voltage | LDR voltage |
|---|---|---|---|---|
| Dim | 6 kΩ | 1 mA | 2 V | 4 V |
| Bright | 3 kΩ | 2 mA | 4 V | 2 V |
In dim light, 6 V ÷ 6 kΩ gives 1 mA. Across the fixed resistor, 1 mA × 2 kΩ gives 2 V. The remaining 4 V is across the LDR.
In bright light, 6 V ÷ 3 kΩ gives 2 mA. Across the fixed resistor, 2 mA × 2 kΩ gives 4 V. The LDR now has the remaining 2 V.
The fixed resistor has the same resistance in both cases. Its voltage rises because the series current rises. The LDR voltage falls because the two voltages must still add to 6 V.
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Practise with a different supply and resistor
An ideal 9 V supply feeds a 3 kΩ fixed resistor in series with an LDR. At one light level the LDR is also 3 kΩ. The two resistances are equal, so each has 4.5 V across it.
In brighter light the LDR drops to 1.5 kΩ. The total is 4.5 kΩ and the current is 2 mA. The fixed resistor has 6 V across it, and the LDR has 3 V.
If the question moves the output terminals from the fixed resistor to the LDR, report the change from 4.5 V to 3 V. The supply and components are the same, but the measured pair of terminals has changed.
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Frequently asked questions
How can I check the two voltages quickly?
Add them and compare with the supply voltage. In series, the larger resistance should also have the larger voltage drop.
Why can I multiply milliamps by kilo-ohms directly?
The factors of one thousand cancel, leaving volts. You can also convert to amps and ohms before multiplying.
Does a larger current make both voltages rise?
The fixed supply sets their sum. In this divider, the fixed resistor gains a larger share of the supply as LDR resistance falls.
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