Q#34 (Past Exam Paper – November 2010 Paper 22 Q6)
The variation with temperature of the resistance RT of a thermistor is shown in Fig. 6.1.
Fig. 6.1
The thermistor is connected into the circuit of Fig. 6.2.
Fig. 6.2
The battery has e.m.f. 9.0 V and negligible internal resistance. The voltmeter has infinite
resistance.
(a) For the thermistor at 22.5 °C, calculate
(i) the total resistance between points A and B on Fig. 6.2, [2]
(ii) the reading on the voltmeter. [2]
(b) The temperature of the thermistor is changed. The voltmeter now reads 4.0 V.
Determine
(i) the total resistance between points A and B on Fig. 6.2, [2]
(ii) the temperature of the thermistor. [2]
(c) A student suggests that the voltmeter, reading up to 10 V, could be calibrated to measure temperature.
Suggest two disadvantages of using the circuit of Fig. 6.2 with this voltmeter for the
measurement of temperature in the range 0 °C to 25 °C. [2]
Solution:
(a)
(i)
At temperature = 22.5 oC, RT = 1600 Ω or 1.6 kΩ
Total resistance = (1/1600 + 1/1600)-1 = 800 Ω
(ii)
EITHER use potential divider formula OR Current = 9 / 2000 (= 4.5mA)
{RT = 800 Ω = 0.8 kΩ}
V = 0.8 / (0.8+1.2) × 9 = 3.6 V V = IR = (9/2000) × 800 = 3.6 V
(Total resistance = 0.8 + 1.2 = 2.0 kΩ)
(b)
(i)
{Voltmeter reading = 4.0V.
So, p.d. across the 1.2 kΩ is 9 – 4 = 5.0V
Ohm’s law: V = IR. (that is, p.d. V is proportional to resistance R}
A p.d. of 5.0V corresponds to 1.2k Ω. So, 4.0V would correspond to (4/5) × 1200)
Total resistance between A and B = (4/5) × 1200 = 960 Ω
(ii)
For parallel combination, 1/960 = (1/1600 + 1/RT)
RT = 2400Ω / 2.4 kΩ
{From graph,}
Temperature = 11 oC
(c)
Any 2 sensible suggestions:
e.g.: Only a small part of the scale is used / small sensitivity
Non-linear
{A linear graph is a straight line graph. If they do no depend linearly on each other, a similar change in voltage would correspond to different ranges of temperature at the different regions of temperature. The separation of the scale would small in one region and large at another region.}
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