Predict sensor divider outputs and loading changes
Explain why NTC and LDR output trends depend on their position, then calculate a supplied loaded example.
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
Before you begin
- State NTC and LDR resistance trends.
- Calculate divider output and two-resistor parallel resistance.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does increased illumination do to an LDR resistance under unchanged conditions?
What you will learn
- Predict sensor divider trends for both placements.
- Recalculate an output with a supplied parallel load.
Place the sensor before predicting the trend
Define output from the divider junction to the lower reference end. In an unloaded divider, reducing the lower resistance reduces its fraction of the total and lowers output. Reducing the upper resistance increases the lower fraction and raises output. The same sensor change can therefore give opposite outputs in two placements.
Warming an NTC lowers its resistance: output falls if it is below the junction and rises if it is above. Increasing LDR illumination likewise lowers its resistance and gives those placement-dependent trends. Keep the other resistor and source p.d. fixed; no universal warmer-means-lower or brighter-means-higher output rule exists.
Account for the connected load
A load connected from junction to lower end is in parallel with the lower resistor. Their equivalent is lower than either finite positive value. Use that effective lower resistance in the divider expression because the upper current now splits into lower-resistor and load currents.
For 12 V, upper 2 kΩ and lower 4 kΩ, the unloaded output is 8 V. Attaching a 4 kΩ load gives a lower equivalent of 2 kΩ, so output becomes 6 V. This resistive example demonstrates loading; a real sensor interface requires its specified input impedance and operating data. Do not retain Rupper/Rlower = Vupper/Vlower for the original pair when their currents differ.
Sources for this lesson
Pause and explain
A 12 V divider has upper 2 kΩ, lower 4 kΩ and a 4 kΩ load across the lower resistor. What is output?
Worked example
A 12 V divider has upper 2 kΩ and lower 4 kΩ. A 4 kΩ load is connected across the lower resistor. Find output.
Show the worked solution
- The lower resistor and load give 4 × 4/(4 + 4) = 2 kΩ.
- The effective divider is 2 kΩ above and 2 kΩ below.
- Vout = 12 × 2/(2 + 2) = 6 V.
Answer 6 V loaded, compared with 8 V unloaded.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Predict lower-reference output when an NTC warms, first below then above the divider junction, unloaded with source and other resistor fixed.
Give me a hint
Its resistance falls in both placements; inspect the lower fraction.
Compare my reasoning
- NTC below: lower resistance fraction falls, so output falls.
- NTC above: upper resistance falls, so lower fraction rises.
- The outputs change in opposite directions.
Lower NTC: output falls; upper NTC: output rises.
Look for these in your work
- State both placements.
- Keep the same output reference.
A 12 V divider has upper 2 kΩ and lower 4 kΩ. Find output before and after a 4 kΩ load is attached across the lower resistor.
Give me a hint
Replace the loaded lower pair by its parallel equivalent.
Compare my reasoning
- Unloaded output is 12 × 4/6 = 8 V.
- Loaded lower equivalent is 4 × 4/8 = 2 kΩ.
- Loaded output is 12 × 2/4 = 6 V.
8 V unloaded; 6 V loaded.
Look for these in your work
- Calculate the parallel equivalent first.
- Use the defined output pair.
An LDR below the junction brightens and its output is claimed to rise; a load is then attached but the original resistor voltage ratio is retained. Correct both steps.
Give me a hint
Check the sensing fraction and the new branch currents.
Compare my reasoning
- Brighter lower LDR means lower resistance and lower unloaded output.
- The load is parallel with that lower resistance.
- Recalculate its effective resistance; original upper and lower resistor currents are no longer equal.
The lower-reference unloaded output falls; with appreciable loading use the lower parallel equivalent.
Look for these in your work
- Correct the placement trend.
- Identify the changed equal-current assumption.
Common mistakes
- Name the sensor placement and reference end.
- A load adds a current branch.
- The original pair no longer shares current when loading is appreciable.
What can you explain now?
For lower-reference output, what happens when an upper NTC warms with other conditions fixed?
Compare with the explanation
Output rises in the unloaded model.
The upper resistance falls, increasing the fixed lower share of the total.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Use sensor placement and the output reference to predict trends, and replace a loaded lower section by its parallel equivalent.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.