Topic 6 of 9
Potential dividers and sensors
A divider output depends on both resistances and the selected endpoints. Identify the output arm before predicting whether its voltage rises or falls.
Let Rupper connect A to B and Rlower connect B to C. Apply a terminal supply Vs across A/C and take the output VBC. With negligible output current, both resistors carry the same current:
VBC = IRlower
= VsRlower/(Rupper + Rlower)
The numerator contains the resistance across the chosen output, not automatically the sensor. If source internal resistance matters, calculate the actual terminal Vs first. E.m.f. is not necessarily the p.d. across A/C.
A wiper selects a fraction of a uniform track
The lower 35% of the track gives 35% of 8.0 V: 2.80 V across B/C. The voltmeter draws negligible current.
NTC below the output junction
Warmer NTC: lower resistance falls from 12 to 4 kilohm, so output falls from 6.0 to 4.0 V.
LDR above the output junction
Brighter LDR: upper resistance falls from 14 to 2 kilohm, so output rises from 1.0 to 4.0 V.
Uniform potentiometer: convert position to resistance
A 20 kΩ track is connected across 8.0 V. Its wiper is 35% of the track length from C. With a uniform track and an unloaded output:
RAB = 13.0 kΩ
VBC = 8.0 × 7.0/20.0 = 2.80 V
I = 8.0/20000 = 0.400 mA
Here 1 kΩ = 103 Ω and 1 mA = 10-3 A. Moving the wiper changes the selected fraction, while the unloaded end-to-end resistance stays 20 kΩ. A nonuniform track would require its resistance distribution rather than position alone.
NTC in the lower arm
Keep the supply at 8.0 V and upper resistor at 4.0 kΩ. The lower NTC changes from 12 kΩ to 4.0 kΩ when warmed:
VBC = 8.0 × 12/16 = 6.0 V
Warmer: I = 8.0/8000 = 1.00 mA
VBC = 8.0 × 4/8 = 4.0 V
Warming lowers the NTC resistance and total resistance, so total current rises. Nevertheless, the NTC's share of the supply falls and VBC decreases. The complementary upper-arm p.d. rises from 2.0 to 4.0 V.
LDR in the upper arm
Now use an upper LDR and a fixed lower 2.0 kΩ resistor, again at 8.0 V. Increasing illumination changes the supplied LDR resistance from 14 kΩ to 2.0 kΩ:
VBC = 8.0 × 2/16 = 1.0 V
Brighter: I = 8.0/4000 = 2.00 mA
VBC = 8.0 × 2/4 = 4.0 V
Lower LDR resistance increases the current through the unchanged lower resistor, so its output rises. Swapping which arm contains the sensor changes this conclusion. Neither example supplies a universal resistance-temperature or resistance-illumination calibration.
In a real sensor investigation, record the actual supply and output, control the other environmental conditions, and allow thermal response to settle. Keep electrical self-heating small when measuring ambient temperature. A threshold for a connected device requires that device's specified input behaviour as well as the divider calculation.
A connected load changes the divider
Two 4.0 kΩ resistors across 8.0 V give an unloaded 4.0 V output across the lower resistor. Connect a 4.0 kΩ load across B/C: it is parallel with the lower resistor, not in series with the divider.
The load is another B/C branch
The parallel load changes the lower equivalent resistance. Output is 2.67 V; each lower branch carries about 0.667 mA.
= 2.0 kΩ
VBC = 8.0 × 2.0/(4.0 + 2.0)
= 2.67 V
The source current is 8.0/6000 = 1.333 mA, splitting equally into about 0.667 mA per lower branch. The output is no longer 4.0 V. A voltmeter is also an output load; calling it negligible means its resistance is sufficiently large for the required accuracy.