thermistors

please could someone explain how the voltage decreases in a thermistor as the temperature increases?
thanks!
You are going to need to include the circuit. If the circuit is just a battery and a thermistor in series, then voltage does not decrease as temperature increases (what do you think happens?)

I'm assuming you mean you have a resistor of constant resistance and a thermistor in series with a battery. One way is to think about the total resistance of the circuit, which is $R_{total} = R_{resistor} + R_{thermistor}$. You know $V_{total}=IR_{total} =V_{resistor} + V_{thermistor}$. As temperature increases, the resistance of the thermistor decreases. Therefore, the total resistance decreases, and hence the current increases. The pd across the resistor increases, so to keep the total pd constant the pd across the thermistor decreases.
If the circuit is just a battery and thermistor in series, as the temperature increases, the current will decrease and the voltage will stay the same?
I have attached how the circuit looks below and it is what you have described it to be. But how would the voltage across the resistor increase and why should the voltage be constant?
Kirchoff's Second law (K2) states that the sum of potential differences in a closed loop is 0. This means that the potential differences across every component (as the circuit is in series) is equal to the potential difference across the battery, which is constant.

I'm not too sure how to explain it further without reiterating my points, but I'll try.
From K2, we have $V_{total} = V_{battery}$, and as the voltage of the battery is constant we have $V_{total}$ is constant. Now, when the resistance of the thermistor decreases, the total resistance decreases. As we have $V_{total}=IR_{total}$ and $R_{total}$ decreases, to keep $V_{total}$ constant $I$ must increase. Therefore as $V_{resistor} = IR_{resistor}$ then $V_{resistor}$ increases. As $V_{total} =V_{resistor} + V_{thermistor}$, then if V_resistor increases to keep V_total constant V_thermistor must decrease.

You are right that for the first example the potential difference does not change and current decreases as temperature increases, although you don't necessarily need to consider the changing resistance, current, etc. to find how voltage changes. By K2 you have $V_{battery} = V_{total}$ and $V_{total} = V_{thermistor}$, so therefore the p.d. remains the same.
thanks for the reply! isn't the voltage in a series circuit shared between the components?
Yes, this is the $V_{total} =V_{resistor} + V_{thermistor}$ part of my explanation.
and how did you get this line : Therefore as V_{resistor} = IR_{resistor} then V_{resistor} increases. Why should the voltage of the resistor increase?
Original post by MouldyVinegar
Yes, this is the $V_{total} =V_{resistor} + V_{thermistor}$ part of my explanation.

ahh, yes sorry that makes sense
Original post by Ashirs
and how did you get this line : Therefore as V_{resistor} = IR_{resistor} then V_{resistor} increases. Why should the voltage of the resistor increase?

As we have $V_{total}=IR_{total}$ and $R_{total}$ decreases, to keep $V_{total}$ constant $I$ must increase. This $I$ is the same in $V_{resistor} = IR_{resistor}$, and as it increases so does $V_{resistor}$
Original post by MouldyVinegar
As we have $V_{total}=IR_{total}$ and $R_{total}$ decreases, to keep $V_{total}$ constant $I$ must increase. This $I$ is the same in $V_{resistor} = IR_{resistor}$, and as it increases so does $V_{resistor}$

ahh yes thank you so much your explanation was very clear!!!
Original post by Ashirs
If the circuit is just a battery and thermistor in series, as the temperature increases, the current will decrease and the voltage will stay the same?
I have attached how the circuit looks below and it is what you have described it to be. But how would the voltage across the resistor increase and why should the voltage be constant?

A side note, the question can be solved using potential divider rule without invoking Kirchoff's Second law.