# URGENT A level physics question

A potential divider circuit is shown below.
V
9.0 V
The battery has electromotive force (e.m.f.) 9.0 V and negligible internal resistance.
At room temperature the potential difference (p.d.) across the thermistor is 4.5 V.
The temperature of the thermistor is increased and its resistance decreases by 20% from its
previous value.
What is the p.d. across the thermistor now?
A 3.6 V
B 4.0 V
C 5.0 V
D 5.4 V

first you have to set up the potential divider equation
so the pd across thermistor = 9 * R_t/(R_t+R_r)
where R_t = resistance of thermistor and R_r = fixed resistance of resistor

then R_t decreases by 20% so
now, the pd across thermistor = 9 * 0.8*R_t/(0.8*R_t+R_r)

Solving equation 1), you get R_t = R_r
then subbing that into equation 2) you can work out the pd = 4V

hope that helps and feel free to ask more qs if youre stuck
Original post by mosaurlodon
first you have to set up the potential divider equation
so the pd across thermistor = 9 * R_t/(R_t+R_r)
where R_t = resistance of thermistor and R_r = fixed resistance of resistor
then R_t decreases by 20% so
now, the pd across thermistor = 9 * 0.8*R_t/(0.8*R_t+R_r)
Solving equation 1), you get R_t = R_r
then subbing that into equation 2) you can work out the pd = 4V
hope that helps and feel free to ask more qs if youre stuck

thanks
Original post by mosaurlodon
first you have to set up the potential divider equation
so the pd across thermistor = 9 * R_t/(R_t+R_r)
where R_t = resistance of thermistor and R_r = fixed resistance of resistor
then R_t decreases by 20% so
now, the pd across thermistor = 9 * 0.8*R_t/(0.8*R_t+R_r)
Solving equation 1), you get R_t = R_r
then subbing that into equation 2) you can work out the pd = 4V
hope that helps and feel free to ask more qs if youre stuck

why does it not work if you use V1/V2=R1/R2
Original post by tpugh1265
why does it not work if you use V1/V2=R1/R2

It works. I would say using this ratio to solve the problem is easier.