[Mr Nonsense]

So this is the I-V characteristic for a filament lamp:



So what is the best answer for the question where you have to "explain the shape of the I-V characteristic for a filament lamp", typically 4 - 5 marks??

This is what I have:

Initially - with positive potential differences - the current is directly proportional to the p.d. However, as the current through the filament increases, the heating effect caused in the lamp also increases and so the temperature of the filament rises. This increase in the filament's temperature also increases the resistance of the filament. As a result the rate of increase of the current decreases and a greater change in the potential difference is required to cause a change in the current. This can be seen on the curve as the gradient becomes more shallow (greater resistance). This same pattern is repeated when a negative potential difference is applied across the filament.

So how can I improve this answer? Is it correct? Is there anything that i have missed?

Thanks in advance!

[teachercol]

Looks fine generally - I'd refer to the gradient rather "the rate of increase of current".

[Faith01]

sounds like you know your stuff, exam is tomorrow (AQA), Ill need to do really good in that before i flop unit4

[Ezraeil]

Yeah...that'd get you all the marks im pretty sure...try and shorten it a bit maybe to save time in the exam...You doing AQA Physics A?

[Mr Nonsense]

(Original post by SyedT)
Yeah...that'd get you all the marks im pretty sure...try and shorten it a bit maybe to save time in the exam...You doing AQA Physics A?

yeah - i find the questions so vague in the exams generally (not this question though).


Which bits do you think i could chop then to make it shorter?

(Original post by teachercol)
Looks fine generally - I'd refer to the gradient rather "the rate of increase of current".

ok, so the "rradient decreases indicating that a larger change in the p.d. is required to cause a change in the current (i.e. greater resistance)"??

thanks everyone!

[Ezraeil]

Well...my answer would be something like:

Initially I is proportional to V. As the current is increased,the temperature of the filament increases,which leads to an increase in its resistance.This then leads to the current and voltage not increasing proportionally. The same pattern is repeated when a negative P.D is applied.

So in your answer,just say the temperature rises,rather than saying stuff about "heating effect" etc.