Did the answers make a wrong assumption?

[kenhung123]

Ok from Q10c, the answers got 10 degrees. They got this by using the value of R in question b where it is 625ohms at 20 degrees.

The current passing through the thermistor must be equal to the current passing through both the LED and resistor.

So I(Thermistor)=I(R)+I(LED) where I(LED)=4.8mA. Which is same as I(Thermistor)-I(R)=4.8mA

Now, using I=V/R we substitute the known voltages
(10-2)/R1-2/R2=4.8mA
Well we only have 1 equation with 2 variables which we cannot solve...However if I subbed in R2=625ohms as the question used for R when at 20 degrees I get the same answer...

So does a resistor vary resistance in response to temperature or thats only thermistors?

[Stojad]

a) 0.5 kΩ

b)

The resistance of the thermistor is 500 Ω. I can only presume that the voltage drop across R is 2.5 V like it is across the LED because they are parallel, so the the voltage drop across the thermistor is 7.5 V.

Hence the current across the thermistor is 7.5V/500Ω, 15 mA. The sum of the current through R + 11 mA is equal to 15 mA (the total current of the circuit as well as the thermistor), so the current through R then must be 4 mA. R's resistance is therefore 2.5V/0.004A or 625 Ω.

c)

I'm making the assumption that R remains at 625 Ω. The voltage across it is now 2 V, so the current through it is 2V/625Ω or 3.2 mA. The question says that the current though the LED is now 4.8 mA, so the total current of the parallel path, equal to the current through the thermistor, is 8 mA.

The current through the thermistor is 8 mA and the voltage drop across it is 8V (since 2 V of the total 10 V is lost across the fixed resistor). Therefore the resistance of the thermistor must be 8V/0.008A, 1kΩ. From the graph, 10°C gives a necessary resistance of 1kΩ, and any temperature greater than that will provide a resistance small enough for the LED to work.

The question does not make the assumption that the resistance of R changes with temperature. You should assume that only the thermistor's resistance changes with temperature.