Current/Voltage relationship for a filament lamp?

[Diamond Dog]

Hi there, First post here guys!
Recently i have been assigned my coursework on measuring the resistance of a filament lamp whereby an experiment was carried out by increasing the voltage through the filament lamp in intervals of 0.5 volts and measuring the current.
I have since made a graph of current against voltage (current on Y axis and Voltage on X axis) which looks like this:
http://www.gcse.com/IV_lamp.htm
However, I am now analysing this graph and struggling. Why does the graph look like this? Why does the graph curve and why does the beginning of the line lok almost straight.
Your help will be much appreciated
Thanks!

 

[cristo]

Well, what does the straight line tell you about the initial relationship between voltage and current?

Do you know an equation relating voltage and current?

What property of the lamp changes as the current increases?

(NB: in future please post such questions in the homework forums)

 

[Diamond Dog]

Hi, Cristo, sorry i didnt know there was a homeworky forum, will remember for next time, i thinki i might be on here quite frequently now!

1) Well the straight line at the beginning tell me that they are proportional, but what happens when it starts to curve?

2) yes, R=V/I which means the gradient would be resistance

3) and i believe as current increases, temperature increases?

I think i understand the basis, it's just trying to explain what is happeing, on the graph i can't quite say it.

Sorry again and thanks!

 

[NoTime]

3) and i believe as current increases, temperature increases?

What happens to most materials as temperature increases?

 

[cristo]

Hi, Cristo, sorry i didnt know there was a homeworky forum, will remember for next time, i thinki i might be on here quite frequently now!

That's ok; welcome to the forums, by the way

1) Well the straight line at the beginning tell me that they are proportional, but what happens when it starts to curve?

Correct, they are directly proportional

2) yes, R=V/I which means the gradient would be resistance

Careful, the gradient of the curve is 1/R (since Δy/Δx=ΔI/ΔV=1/R) Edit: note the corrected typo in this equation[/color]

3) and i believe as current increases, temperature increases?

Correct, and what other property of the wire inside in the lamp increases as the temperature of the wire increases?

 

[Diamond Dog]

They begin to melt?
Does this mean that the grash is curving because its getting hotter and hotter and will eventually melt?

 

[Diamond Dog]

That's ok; welcome to the forums, by the way

Correct, they are directly proportional


Careful, the gradient of the curve is 1/R (since Δy/Δx=ΔI/ΔV=R)


Correct, and what other property of the wire inside in the lamp increases as the temperature of the wire increases?

Would that be resistance?

 

[cristo]

Would that be resistance?

Correct. Now look at my equation above ΔI/ΔV=1/R (noting the typo) Let's write this equation as V=IR. Since as the current increases, the resistance increases, what can you say about the increase of the voltage? i.e. compare this to a point where R does not depends on I; a small increase in I will result in an increase in V. Now, if we increase I by the same small amount, but at a point where the resistance is now bigger, how big is the increase in V? ( is it the same as the point with fixed resistance, greater or smaller?)

Hope this makes sense!

 

[Diamond Dog]

Correct. Now look at my equation above ΔI/ΔV=1/R (noting the typo) Let's write this equation as V=IR. Since as the current increases, the resistance increases, what can you say about the increase of the voltage? i.e. compare this to a point where R does not depends on I; a small increase in I will result in an increase in V. Now, if we increase I by the same small amount, but at a point where the resistance is now bigger, how big is the increase in V? ( is it the same as the point with fixed resistance, greater or smaller?)

Hope this makes sense!

Ahhh, i see now ! Thanks Cristo you have been a lot of help. I like the way you asked questions in order for me to figure it out by myself, helps me understand more rather than you spoon feeding me the answer. Think i will be posting more on here for further in this coursework!
Thans again, very much appreciated!

 

[cristo]

Ahhh, i see now ! Thanks Cristo you have been a lot of help.

You're very welcome!

I like the way you asked questions in order for me to figure it out by myself, helps me understand more rather than you spoon feeding me the answer.

That's the way these forums should work. It's a pleasure to help a student that has clearly thought about the question before he posts, and is willing to listen to the hints and answer the question himself, rather than simply demanding we give him the answer!

You'll fit in very well here! Good luck with the coursework.

 

[Darryl]

i agree, that was a pleasure to read, i hope you are a teacher by trade cristo.
if not, you can "talk the talk".