9478 / 2027
Circuits overview

Lesson 6 of 9

Potential dividers and sensors

  • I can solve one-source series/parallel circuits, including potential dividers using NTC thermistors and LDRs.Syllabus

    Syllabus 9478, 16(j). Solve one-source series/parallel circuits, including potential dividers using NTC thermistors and LDRs.

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:

I = Vs/(Rupper + Rlower)
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

A wiper selects a fraction of a uniform trackA twenty-kilohm uniform potentiometer spans the eight-volt source from A at vertical coordinate sixty to C at two hundred and sixty. The wiper contacts the track at coordinate one hundred and ninety, leaving thirty-five percent of the track below it. The lower resistance is seven kilohms and upper resistance thirteen kilohms. An ideal voltmeter connects B, the wiper terminal, to C and reads 2.80 volts.+-VABC13 kΩ7 kΩ20 kΩwhole track8.0 V2.80 VOutput is between B and C.

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

NTC below the output junctionThe upper fixed resistance is four kilohms; the lower NTC resistance falls from twelve to four kilohms as its supplied state warms. Output B/C falls from six to four volts. The source maintains eight volts across A/C, and the voltmeter is connected to B and C with negligible loading. These are supplied sensor states, not a universal calibration.+-VABC8.0 VOutput4 kΩNTC12 kΩ to 4 kΩ

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

LDR above the output junctionThe upper LDR resistance falls from fourteen to two kilohms as its supplied state gets brighter; the lower fixed resistance is two kilohms. Output B/C rises from one to four volts. The source maintains eight volts across A/C, and the voltmeter is connected to B and C with negligible loading. These are supplied sensor states, not a universal calibration.+-VABC8.0 VOutputLDR2 kΩ14 kΩ to 2 kΩ

Brighter LDR: upper resistance falls from 14 to 2 kilohm, so output rises from 1.0 to 4.0 V.

Each circuit has its output between B and C. The potentiometer's wiper position divides a uniform track; the NTC and LDR examples place their sensors in different arms, so their stated changes produce different output responses.

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:

RBC = 0.35 × 20 = 7.0 kΩ
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:

Initially: I = 8.0/16000 = 0.500 mA
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:

Initially: I = 8.0/16000 = 0.500 mA
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 load is another B/C branchAn eight-volt source drives a four-kilohm upper resistor from A to B. Both the four-kilohm lower divider resistor and a separate four-kilohm load join B to C in parallel. Their equivalent is two kilohms. Source current is four-thirds milliampere and divides equally between the lower branches. The output is eight-thirds volt, not four volts.+-ABC8.0 V4 kΩ4 kΩLoad4 kΩ1.333 mASame B/C p.d. across both branches

The parallel load changes the lower equivalent resistance. Output is 2.67 V; each lower branch carries about 0.667 mA.

The load shares B and C with the lower divider resistor. The source current passes through the upper resistor and then splits between the two equal lower branches.
Rlower, effective = (4.0 × 4.0)/(4.0 + 4.0)
= 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.

Try it step by step

Work it out in steps: Total resistance R1 + R2 in ohm, then Output across R2: V = Vs x R2 / (R1 + R2) in V.

  1. Problem 1. A 9.0 V supply is connected across R1 = 1.5 kilo-ohm in series with R2 = 4.5 kilo-ohm. The output is taken across R2. Find the output p.d.
  2. Problem 2. A 12 V supply is connected across R1 = 2.2 kilo-ohm in series with R2 = 1.0 kilo-ohm. The output is taken across R2. Find the output p.d.
  3. Problem 3. A 5.0 V supply is connected across R1 = 470 ohm in series with R2 = 330 ohm. The output is taken across R2. Find the output p.d.
Show the answer

Problem 1.

Total resistance R1 + R2 in ohm: R1 + R2 = 1500 + 4500 = 6000 ohm

Output across R2: V = Vs x R2 / (R1 + R2) in V: V = 9.0 x 4500 / 6000 = 6.75 V (3 sf)

Problem 2.

Total resistance R1 + R2 in ohm: R1 + R2 = 2200 + 1000 = 3200 ohm

Output across R2: V = Vs x R2 / (R1 + R2) in V: V = 12 x 1000 / 3200 = 3.75 V (3 sf)

Problem 3.

Total resistance R1 + R2 in ohm: R1 + R2 = 470 + 330 = 800 ohm

Output across R2: V = Vs x R2 / (R1 + R2) in V: V = 5.0 x 330 / 800 = 2.06 V (3 sf)

Check your understanding

Two 4.0 kilohm divider arms connect across an ideal fixed 8.0 V supply. A 4.0 kilohm load is then connected across the lower arm B/C. What is V_B - V_C?

  1. 5.33 V
  2. 4.0 V
  3. 2.67 V
Show the answer

2.67 V

The lower arm and load are parallel, giving 2.0 kilohm. The new output is 8.0 x 2.0/(4.0 + 2.0) = 8/3 V.