Question 37

[bcapper22053]

I need help understanding the logic of this question. I understand from the passage that as the temperature decreases, the resistance will increase. With a greater resistance it makes sense that the current will decrease, but only assuming that the voltage stays the same. (V=IR). Why does the voltage drop as well?

[jellywing_2058]

As the temperature of the water decreases, the resistance of the thermistor increases; therefore, the resistance of the circuit increases.
The resistance of the circuit (series) is:
R = R1 + RT
Where
R1= the fixed resistor and
RT= the thermistor resistance RT

Therefore, the current in the circuit decreases.

You are right, this is assuming that the voltage stays the same; however which voltage?
The supply voltage that is 2 V. But this 2 V is divided amongst R1 and RT in a different proportion, if the RT changes.
With RT increasing and R1 remaining the same, the voltage across RT will increase (VT = I x RT). This will lead V1 to decrease as V1=2V – VT and an increase in VT will lead to a lesser voltage now available for R1.


More mathematically,
I = U x R
I = 2/(R1 + RT)

V1 = I x R1 Voltage drop across R1 as shown by the Voltmeter reading
VT = I x RT Voltage drop across RT (the Thermistor)
Replacing the I:
V1 = (2/(R1 + RT)) x R1
VT = (2/(R1 + RT)) x RT

And
V1 + VT = 2 volts

Since VT = (2/(R1 + RT)) x RT,
As RT increases, VT increases

And since V1 = 2 volts – VT,
As VT increases, V1 decreases.


So, answer choice D: “decreases because the current I decreases” is correct.
As I decreases, the voltage across R1 (V1 = I x R1) will decrease as well.