Resistance relation with temp [THERMISTORS] helpWatch
So I know that for thermistors the lower the temperature, the greater the resistance.
Since V=IR, shouldn't a greater R mean a greater V?
The mark scheme says that the voltmeter reading decreases
The circuit diagram is basically an input V of 6V with a variable resistor and a 1000ohm resistor in series. But the resistor parallel with a voltmeter (if anyone's wondering)
The temperature of the thermistor subsequently increases
The resistance of the thermistor decreases due to the increase in temperature
The effective resistance of the thermistor and parallel resistor increases (given by Σ(r1 +r2)-1)
The potential difference therefore increases in the thermistor and parallel resistor part
The potential difference therefore decreases for the other resistor
You understand that the resistance of the thermistor and the resistor in parallel to it is due to the sum of the negative reciprocals of the resistances? So if the resistance of the thermistor decreases, the effective resistance is greater due to the law of parallel resistances.
If Rt is the resistance of the thermistor, and Rn is the resistance of the resistor in parallel:
If the resistance of the thermistor decreases, you must be able to see that the effective resistance will increase.