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OCR Physics A G482, Electrons, Waves and Photons, 25th May 2012

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Reply 1240
Original post by mashmammad
what???? linear means constant instantaneous resistance. you may have a line that doesnt go through origin. but it doesnt mean the gradient is not constant.


You're making this so much more difficult than it needs to be.. Because you seem to be some what incapable to interpreting a graph, I've got the values for you.

http://gyazo.com/eafb7b433afb3946ae23333e57aa99f6

[R]
At 1.8 V = 1.8/9x10^-3 = 200 ohms
At 2.0 V = 2.0/38x10^-3 = 52.63 ohms

Hence the resistance is decreasing. Are we done here..?
no. what you are saying is just true if the graph was a line through origin. trust me mate, i dont have time to argue with you, but i asure you that this is wrong. but just promise if any of us was wrong let the other know through a message. on tsr of course.
Reply 1242
Original post by mashmammad
no. what you are saying is just true if the graph was a line through origin. trust me mate, i dont have time to argue with you, but i asure you that this is wrong. but just promise if any of us was wrong let the other know through a message. on tsr of course.


No, what YOU are trying to say is only valid if the line passes through the origin. It is not wrong and the markscheme clearly states that. The markscheme doesn't state that resistance is constant and there is one reason for that; because it isn't. My last post clearly shows resistance is decreasing after 1.8V and you can't argue with the figures.
The resistance of the LED decreases. This can be demonstrated by calculating two values from the graph. A constant gradient doesnt mean a constant resistance - the current was increasing at a greater rate than the p.d
Reply 1244
Original post by tooty_fruit
The resistance of the LED decreases. This can be demonstrated by calculating two values from the graph. A constant gradient doesnt mean a constant resistance - the current was increasing at a greater rate than the p.d


Thank you.. :rolleyes:
Reply 1245
Original post by Nick_
only if the line would go through the origin if you continued it which it did not, hence resistance was changing and in this case it was decreasing

this is because R does not equal dV/dI, it just equals V/I


This is genius.
if we all done that paper together, we would have aced it lol

can someoine do the mark scheme
Original post by Ionus7
I put 0... Darn it. Now I know why that makes sense. D'OH!


the answer is infinite resistance. this i because there was another similar question and was asked what the resistance is before that voltage. when looking at the mark scheme, it specifically said " infinite resistance do not accept 0 Ohm"

there u go
Original post by Malabarista
Thanks, also, please tell me for Q6 wave pulse one anyone here got something like this:

wave.png


bloody hell, if you drew something like that in the exam earlier on, you would get a zero for that one. youre trying to violate their work
Reply 1249
can someone please tell what the grade boundaries for an A was last year and jan 2012?? thanks:smile:
Am I the only one that felt that this paper was easier than previous papers?
Reply 1251
Original post by emah123
can someone please tell what the grade boundaries for an A was last year and jan 2012?? thanks:smile:

73/100
Original post by emah123
can someone please tell what the grade boundaries for an A was last year and jan 2012?? thanks:smile:


They're generally all between 67 and 73 in the past few years.
this was paper was not easier than jan 2012 end of
Original post by LewisMead
They're generally all between 67 and 73 in the past few years.


exactly so why should this paper be higher
Original post by Theafricanlegend
this was paper was not easier than jan 2012 end of


Yeah. Jan 2012 was very straight forward and plain.
Original post by Sarabande
Yeah. Jan 2012 was very straight forward and plain.


meaning that grade boundaries will be lower than 73.
my guess : 70-MAYBE69 for an a
Original post by Theafricanlegend
exactly so why should this paper be higher


I have not once said it would be.
Original post by LewisMead
I have not once said it would be.


i know i just used your comment to show people who think that the grade boundaries will be 80 that they are wrong
Original post by J.C
For the LED question on resistance after 1.8V, was it a straight line?!? I checked against my ruler and even calculated resistance at 1.9V and 2.0V and the resistance was clearly decreasing. Yet quite a lot of my friends said it was a straight line and so R is constant. However they did not check like I did.

Someone shed some light on this for me.


thats the same technique which i had used. i used r=v/I den put numbers in den found out resistance was decreasing. the only way the resistance could be constant is if the line was straight through the origin then you could have also used ohms law to explain it

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