3.5.1.5

Potentiometer

Circuits & Potential Dividers — AQA A-Level Physics

Key Definition
Potentiometer — A variable potential divider using a single resistor with a sliding contact. Output p.d. varies continuously from 0 to V_in.
PotentiometerA potential divider with a sliding contact that allows continuous variation of the output voltage from zero to the supply voltage. circuit
A single resistor with three terminals: two ends connected across V_in, a sliding contact providing V_out.
Key Definition
Temperature sensor circuit — A potential divider using a thermistorA component whose resistance decreases significantly as temperature increases (negative temperature coefficient). and a fixed resistor. As temperature rises, thermistor resistance falls, so V_out across the thermistor falls.
Key Definition
Light sensor circuit — A potential divider using an LDRLight-dependent resistor. A component whose resistance decreases as light intensity increases. and a fixed resistor. As light intensity increases, LDR resistance falls, so V_out across the LDR falls.
ThermistorA component whose resistance decreases significantly as temperature increases (negative temperature coefficient). and LDRLight-dependent resistor. A component whose resistance decreases as light intensity increases. sensor circuits
Two potential dividerA circuit that uses two or more resistors in series to produce a fraction of the source voltage across one of the resistors. circuits side by side. Left: thermistor as R_2 with V_out across it. Right: LDRLight-dependent resistor. A component whose resistance decreases as light intensity increases. as R_2 with V_out across it.
Worked Example
A 9.0 V supply is connected to a 1.0 k-ohm fixed resistor (R_1) and a thermistor (R_2). At 20 degrees C, the thermistor has a resistance of 2.0 k-ohm. Calculate V_out across the thermistor. What happens to V_out if the temperature increases to 50 degrees C and the thermistor resistance drops to 500 ohm?
Show Solution
1
At 20 degrees C

$V_out = R_2 / (R_1 + R_2) \times\;\text{V}_in = 2000 / (1000 + 2000) \times 9.0 = 6.0\;\text{V}.$

2
At 50 degrees C

$V_out = 500 / (1000 + 500) \times 9.0 = 500/1500 \times 9.0 = 3.0\;\text{V}.$

3

V_out halves as the thermistor resistance falls.

Answer
6.0 V at 20 degrees C; 3.0 V at 50 degrees C. V_out decreases as temperature increases.
Common Mistake MEDIUM
Students often: Stating V_out always decreases when temperature increases, without specifying which component V_out is measured across.
Instead: V_out across the thermistor decreases. V_out across the fixed resistor increases. Always state which component.
Examiner Tips and Tricks
  • AQA commonly asks 'explain how V_out changes when temperature increases.' State the sequence: temperature up, thermistor resistance down, share of p.d. across thermistor decreases, so V_out across thermistor decreases.
  • Give all three steps.
Circuits & Potential Dividers Overview