A-LevelPhysicsD.C. circuits9702/41 · May/June 2021 · Q89 Marks
The variation with temperature of the resistance of a thermistor is shown in [Figure 8.1]. A student includes the thermistor and an ideal operational amplifier (op-amp) in the circuit of [Figure 8.2].
- (a)Calculate the potential V⁺ at the non-inverting input of the op-amp. V⁺ = .............................................................. V [2][2]
- (b)At 10 °C, the resistance of the thermistor is 2.5 kΩ. State and explain whether the light-emitting diode (LED) is emitting light.[2]
- (c)Explain why the student's circuit will not indicate any change in temperature above 0 °C.[2]
- (d)The resistor of resistance 5.0 kΩ is changed to a resistor of resistance R so that the LED switches on or off at a temperature of 20 °C. Determine R in kΩ. R = .............................................................. kΩ [3][3]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [1] (a) V+ = 3.0 × 3.0 / (2.5 + 3.0)
- [1] = 1.6 V
- [1] (b) V- is +2.0 V
- [1] V- > V+
- [1] output is negative so (LED) does not emit light
- [1] (c) at 0 °C, V- = 1.7 V
- [1] for all temperatures above 0 °C, resistance of thermistor < 4.2 kΩ
- [1] V- always greater than V+ (so no switching)
- [1] (d) (at 20 °C,) R_T = 1.8 kΩ
- [1] 2.5/3.0 = 1.8/R
- [1] [R/ (R + 1.8)] × 3.0 = 1.6
- [1] R = 2.2 kΩ
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →