# REALLY stuck on resistance temperature questions

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(These questions are taken from the AQA Physics A AS book, page 55)

1) A filament bulb is labelled 3.0V, 0.75W

a) Calculate it's current and it's resistance at 3.0V

2) A certain thermistor has a resistance of 50 000 ohms at 20C and has a resistance of 4000 ohms at 60C.

It is connected in series with an ammeter and a 1.5V cell. Calculate the ammeter reading when the thermistor is

(a) at 20C

(b) at 60C

4) The resistance of a certain metal wire increased from 25.3 ohms at 0C to 35.5 ohms at 100C. Assuming the resistance over this range varies linearly with temperature, calculate

(a) the resistance at 50C

(b) the temperature when the resistance is 30.0 ohms

**Even looking at the answers page, I still can't udnerstand these temperature and resistance questions. I don't even remember my teacher going over stuff like this, I would**__GREATLY__appreciate it if somebody explained to me how they got the answers for the following questions, as my understanding on this is really bad1) A filament bulb is labelled 3.0V, 0.75W

a) Calculate it's current and it's resistance at 3.0V

2) A certain thermistor has a resistance of 50 000 ohms at 20C and has a resistance of 4000 ohms at 60C.

It is connected in series with an ammeter and a 1.5V cell. Calculate the ammeter reading when the thermistor is

(a) at 20C

(b) at 60C

4) The resistance of a certain metal wire increased from 25.3 ohms at 0C to 35.5 ohms at 100C. Assuming the resistance over this range varies linearly with temperature, calculate

(a) the resistance at 50C

(b) the temperature when the resistance is 30.0 ohms

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#2

1)

Have you done the formula for electrical power P=VI

Just plug in then use Ohm's Law for the resistance.

2)

Just use Ohm's Law on the given resistance and pd. to find the current.

3)

50 degs is

Given the resistance at 0 and 100, what value would you expect it to have at the half way point if it varies uniformly?

Have you done the formula for electrical power P=VI

Just plug in then use Ohm's Law for the resistance.

2)

Just use Ohm's Law on the given resistance and pd. to find the current.

3)

50 degs is

**half way between**0 and 100Given the resistance at 0 and 100, what value would you expect it to have at the half way point if it varies uniformly?

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(Original post by

1)

Have you done the formula for electrical power P=VI

Just plug in then use Ohm's Law for the resistance.

2)

Just use Ohm's Law on the given resistance and pd. to find the current.

3)

50 degs is

Given the resistance at 0 and 100, what value would you expect it to have at the half way point if it varies uniformly?

**Stonebridge**)1)

Have you done the formula for electrical power P=VI

Just plug in then use Ohm's Law for the resistance.

2)

Just use Ohm's Law on the given resistance and pd. to find the current.

3)

50 degs is

**half way between**0 and 100Given the resistance at 0 and 100, what value would you expect it to have at the half way point if it varies uniformly?

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#4

(Original post by

Thank you! I got the right answers Your gonna have to bear with my stupidity when it comes to electricity, but I thought that filament lamps and thermistors don't obey ohms law?? I dont understand why the answers are right though, how can we apply the ohms law equation to a device when the temperature doesn't remain constant? x

**dreams150**)Thank you! I got the right answers Your gonna have to bear with my stupidity when it comes to electricity, but I thought that filament lamps and thermistors don't obey ohms law?? I dont understand why the answers are right though, how can we apply the ohms law equation to a device when the temperature doesn't remain constant? x

Strictly speaking I should have said "use the formula V=IR" to calculate... etc

The formula R=V/I defines what we mean by resistance. The resistance of anything that carries a current I when a pd V is applied to it is

*by definition*V/I

Ohm's Law is saying that for some materials (notably metals) their resistance only depends on physical circumstances such as temperature and mechanical strain. So if you keep temperature and other physical conditions constant, the resistance is constant. Or current is proportional to applied pd.

You

**can**always use the formula R=V/I to calculate the resistance of a material. It's the definition of resistance.

You

**can't**assume that V/I will always give the same value for a material. I.E. That it's resistance will stay constant.

In question 2 you are just applying the formula R=V/I to the material with a given value of R and V.

In question 1 you are given the conditions under which the bulb operates normally. A bulb will work normally at a given applied pd. It then works at a known power rating. (given). It gets hot and has a particular resistance.

These values of V, I and R are the bulbs "operating" values. The operating value of R for the bulb will still be = V/I by definition. Power = VI by definition.

What is also true is that this value of R for the bulb will not be constant if you change the applied pd and the bulb gets hotter or cooler.

Note the question asks to calculate the values

*at an applied pd of 3V*.

The precise operating conditions have been specified.

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(Original post by

THis "Ohm's Law" thing can be a bit confusing. It's common to say (I said it!) "use Ohm's Law" to calculate...etc

Strictly speaking I should have said "use the formula V=IR" to calculate... etc

The formula R=V/I defines what we mean by resistance. The resistance of anything that carries a current I when a pd V is applied to it is

Ohm's Law is saying that for some materials (notably metals) their resistance only depends on physical circumstances such as temperature and mechanical strain. So if you keep temperature and other physical conditions constant, the resistance is constant. Or current is proportional to applied pd.

You

You

In question 2 you are just applying the formula R=V/I to the material with a given value of R and V.

In question 1 you are given the conditions under which the bulb operates normally. A bulb will work normally at a given applied pd. It then works at a known power rating. (given). It gets hot and has a particular resistance.

These values of V, I and R are the bulbs "operating" values. The operating value of R for the bulb will still be = V/I by definition. Power = VI by definition.

What is also true is that this value of R for the bulb will not be constant if you change the applied pd and the bulb gets hotter or cooler.

Note the question asks to calculate the values

The precise operating conditions have been specified.

**Stonebridge**)THis "Ohm's Law" thing can be a bit confusing. It's common to say (I said it!) "use Ohm's Law" to calculate...etc

Strictly speaking I should have said "use the formula V=IR" to calculate... etc

The formula R=V/I defines what we mean by resistance. The resistance of anything that carries a current I when a pd V is applied to it is

*by definition*V/IOhm's Law is saying that for some materials (notably metals) their resistance only depends on physical circumstances such as temperature and mechanical strain. So if you keep temperature and other physical conditions constant, the resistance is constant. Or current is proportional to applied pd.

You

**can**always use the formula R=V/I to calculate the resistance of a material. It's the definition of resistance.You

**can't**assume that V/I will always give the same value for a material. I.E. That it's resistance will stay constant.In question 2 you are just applying the formula R=V/I to the material with a given value of R and V.

In question 1 you are given the conditions under which the bulb operates normally. A bulb will work normally at a given applied pd. It then works at a known power rating. (given). It gets hot and has a particular resistance.

These values of V, I and R are the bulbs "operating" values. The operating value of R for the bulb will still be = V/I by definition. Power = VI by definition.

What is also true is that this value of R for the bulb will not be constant if you change the applied pd and the bulb gets hotter or cooler.

Note the question asks to calculate the values

*at an applied pd of 3V*.The precise operating conditions have been specified.

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#6

(Original post by

Seriously you have no idea how grateful I am for your help Have you got the AQA PHYA1 Physics exam coming up? x

**dreams150**)Seriously you have no idea how grateful I am for your help Have you got the AQA PHYA1 Physics exam coming up? x

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#7

Hi

i don't know if this thread is still alive but

do you know how to do part b fir question 4?

Thanks guys

i don't know if this thread is still alive but

do you know how to do part b fir question 4?

Thanks guys

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#8

(Original post by

Hi

i don't know if this thread is still alive but

do you know how to do part b fir question 4?

Thanks guys

**HikariUchiha**)Hi

i don't know if this thread is still alive but

do you know how to do part b fir question 4?

Thanks guys

Having said that, the question tells you the resistance at 0 and 100 and asks you to find it at 50, which is exactly half way between the two. It also tells you that the resistance varies uniformly. So it's just a case of simple proportion.

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#9

if the graph show the resistance start from 0 to 3 and the temperature from also 0 to 90 degrees so what is the actual temperature on this .

why the temperature is too high and what it should be ?

need help guys on this question ......

why the temperature is too high and what it should be ?

need help guys on this question ......

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