# Investigation on a Light Dependant Resistor

 P: 4 united fan- i think measuring the current in the circuit is right but i dont think the lamp should be part of the circuit. it should be seperate. hope that helps
 P: 4 oh and btw thanks megegg!
P: 13
 Quote by sorrel united fan- i think measuring the current in the circuit is right but i dont think the lamp should be part of the circuit. it should be seperate. hope that helps
Oh, ok. cheers mate.

So do you think i should have one circuit with the LDR and the ammetre and then another one with the lamp and a resistor so that the current doesn't get too high?

Thanks for the help
 P: 5 i too have this assignment, and reading through this thread i'm confused by the how you all seem to be overcomplicating the intensity ^^ the formula is simple also, you'll want to use diffraction grating to calculate the wavelength of light passing through each colour filter, as you'll have to account for the fact that the method is imperfect, so just getting wavelength figures off the net will create potential error in the data.
P: 5
 Quote by sorrel hey im confused on part b...how the wavelength of the light falling on the LDR is determined...as if u r using a diffraction grating or prism how do u know?? sorrel
there are several formulae for diffraction gratings that can be found on the net.

unfortunately i can't really help you, as i ignored them all and devised my own formula - it works fine but not many people would use it, so it could be seen by the examining body as collaboration.

try wiki?
 P: 5 Intensity = power/area :P simple really
 P: 5 one query - part d) the range and precision of any instruments that would be used: what is the average range and precision of a standard digital ammeter? or voltmeter?
P: 13
 Quote by animecrazy one query - part d) the range and precision of any instruments that would be used: what is the average range and precision of a standard digital ammeter? or voltmeter?
Sorry no idea mate. So with the intensity thing are you going to calculate the power around the LDR using an ammetre and voltmetre when different wavelegths shine on the LDR?
 P: 5 yes. the filament lamp would give off radiation in all directions, so the area would be a sphere, using the distance from the lamp to the LDR as the radius to calculate area. and it's ok, i found some average values
P: 13
 Quote by animecrazy yes. the filament lamp would give off radiation in all directions, so the area would be a sphere, using the distance from the lamp to the LDR as the radius to calculate area. and it's ok, i found some average values

alright cheers mate, i'll work on that. I think you can get coloured filters which come with wavelength specifications which would be a lot simpler then diffraction grating i think.
 P: 13 So what have you guys decided to do for intensity?
P: 3
 Quote by animecrazy yes. the filament lamp would give off radiation in all directions, so the area would be a sphere, using the distance from the lamp to the LDR as the radius to calculate area. and it's ok, i found some average values
I'm not sure what you mean by that. I think I'm just going to relate a change in voltage to the intensity.
P: 13
 Quote by Mr. Kipling I'm not sure what you mean by that. I think I'm just going to relate a change in voltage to the intensity.

Yeah i was thinking of doing something like that. i will probably measure current and do a graph of wavelegth against current( as current should increase as intensity increases) or wavelegth against resistance.
 P: 6 Hi, I am also doing this scenario and investigation. I was just reading the reast of the threads and noticed that alot of people have suggested using the current for intensity, however, I do belive that one of the graphs we need to present/example is Intensity against resistance, or in our case resistance against current. Will that turn out as a bodge job or be a legitimate representation? I haven't tried it yet as only recieved it today ad was doing some research on how to measure intensity, so what does everyone think? Cheers
P: 13
 Quote by Bob Spoongfield Hi, I am also doing this scenario and investigation. I was just reading the reast of the threads and noticed that alot of people have suggested using the current for intensity, however, I do belive that one of the graphs we need to present/example is Intensity against resistance, or in our case resistance against current. Will that turn out as a bodge job or be a legitimate representation? I haven't tried it yet as only recieved it today ad was doing some research on how to measure intensity, so what does everyone think? Cheers

Well i was working under the idea that if intensity increases the resistance will decrease and so the current will rise or in other words if intensity rises current will rise and so using current as a substitute for intensity seems plausible.i hope it is at the least as thats what i'm planning to do.
P: 6
 Quote by United fan Well i was working under the idea that if intensity increases the resistance will decrease and so the current will rise or in other words if intensity rises current will rise and so using current as a substitute for intensity seems plausible.i hope it is at the least as thats what i'm planning to do.
Yer, I'm hoping to use that as well. I will be checking with my teacher tomorrow! see what he says. I was looking at the formula for intensity as well and it seems much too complicated for this level so I guess there must be some simpler way. will post back tomorrow with more info. lte's hope it is plausible eh !
 P: 13 yep hope so too mate. if your teacher does say its alright to use current, let us know mate otherwise i've got a bit of a problem as the exams on thursday
P: 6
 Quote by United fan yep hope so too mate. if your teacher does say its alright to use current, let us know mate otherwise i've got a bit of a problem as the exams on thursday
unlucky. Mine's A-level plan and is in for the 15th so I have a bit of time to get it right, but I will definitely post back to you all asap.

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