Question 34
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]
Reference: Past Exam Paper – November 2010 Paper 22 Q6
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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