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

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Original post by `Mo

Grade boundary predictions?

I reckon they'll be quite high this year on the pure basis that the paper was the easiest of all the papers (In spite of me messing up the entire paper) and that they haven't had a paper with high grade boundaries yet.

A - 77-80 B - 70 C- 64-67 D < 60

Reply 1181

username916760

I predict a very harsh boundary. Dont want to let you off guys, but it was one of the easiest papers ever. I wont be surprised if it is:
100%: 95%
a: 75%
b and below: I dont have any knowledge of law boundaries.

Reply 1182

username916760

Original post by 4ever_drifting

i was pretty sure in the exam that the diode had a constant resistance because the line was straight... Now i'm a bit confused!

If we assume the line to be straight (i know some people are saying it wasn't - i personally didn't check), we can model it using y=mx+c:
I=mv+c => v=(i-c)/m, so v/i=r=(i-c)/mi=1/m-c/mi

1/m-c/mi will eventually be constant as i increases, since c/mi will tend to 0 for large i... Obviously our graph didn't go up that high, so the resistance of our graph probably wasn't constant, though i am unsure what the markscheme will say.

Given that c was negative, let c=-k. Then r=1/m+k/mi, which will decrease as i gets bigger, so the resistance was probably decreasing.

don't be. The people here are different sorts of students. When you come out of the exam you wont ask the clueless guys for the answers. I would wait for "teachercol". He has been teaching for such a long time. He will hopefully put the answers up.

Reply 1183

HarryW95

Original post by Junaid96

It kinda does. A straight line has constant resistance.

No, it doesn't! If R=V/I and V is reasonably constant (which it was, hence the steep gradient after 1.8V), and current is rapidly increasing, R decreases. Just substitute values into the equation and check for yourself.

Example from past paper, from what I remember this graph was very similar to ours. January 2010 if anybody wants to check.

http://gyazo.com/eafb7b433afb3946ae23333e57aa99f6

Answer to question: [markscheme]

no current/no light/does not conduct until V is greater than 1.5 V
brightness/intensity of LED increases with current/voltage above 1.5 V
above 1.8 V current rises almost linearly with increase in p.d./AW
the LED does not obey Ohm’s law
as I is not proportional to V/AW
below 1.5 V, LED acts as an infinite R/ very high R/acts as open switch
above 1.5 V, LED resistance decreases (with increasing current/voltage)

(edited 11 years ago)

Reply 1184

Emissionspectra

Original post by Malabarista

Check out the Jan 12 mark scheme, (I think Jan 12), it is very similar and the LED is identified as having a constant resistance after a certain voltage.

I dont even do OCR just helping out :P

Reply 1185

Malabarista

Original post by Emissionspectra

I dont even do OCR just helping out :P

Nah, sorry man, you were right, I'm really annoyed by that now, 3 marks gone. Makes it all that much harder to concentrate on next exams...

Reply 1186

HarryW95

Original post by Malabarista

Nah, sorry man, you were right, I'm really annoyed by that now, 3 marks gone. Makes it all that much harder to concentrate on next exams...

It's only three marks out of 100. Don't worry about it. :redface:

Reply 1187

Doglamlico

Original post by HarryW95

Well yeah you can see it decreasing as it curves off, but then when it's straight (around >1.8V) then it's a constant R. R = dV/dI, which remains constant after 1.8V.

Reply 1188

Malabarista

Original post by HarryW95

It's only three marks out of 100. Don't worry about it. :redface:

Thanks, also, please tell me for Q6 wave pulse one anyone here got something like this:

Reply 1189

Emissionspectra

Original post by Malabarista

Nah, sorry man, you were right, I'm really annoyed by that now, 3 marks gone. Makes it all that much harder to concentrate on next exams...

Don't worry, it's only 3 marks, I'm sure you did fine.

Reply 1190

HarryW95

Original post by Junaid96

Well yeah you can see it decreasing as it curves off, but then when it's straight (around >1.8V) then it's a constant R. R = dV/dI, which remains constant after 1.8V.

... The markscheme states that:

above 1.5 V, LED resistance decreases (with increasing current/voltage)

It doesn't say resistance decreases up to 1.8V and then remains constant. Resistance continues to decrease from that point (1.5V)

above 1.8 V current rises almost linearly with increase in p.d./AW

If current is rising linearly with an increase in current, resistance is decreases (R=V/I)

Reply 1191

HarryW95

Original post by Malabarista

Thanks, also, please tell me for Q6 wave pulse one anyone here got something like this:

Yes mine looked like that except I didn't include the straight line of zero displacement at the beginning. Guess I didn't notice that. Oh well, hopefully I can still get 3/4 marks for that.

Reply 1192

Malabarista

Original post by HarryW95

Yes mine looked like that except I didn't include the straight line of zero displacement at the beginning. Guess I didn't notice that. Oh well, hopefully I can still get 3/4 marks for that.

I'm glad I'm getting some similar answers then. I was really disappointed by the lack of and 'explain the photoelectric effect' in that exam.

Reply 1193

Doglamlico

Original post by HarryW95

... The markscheme states that:

It doesn't say resistance decreases up to 1.8V and then remains constant. Resistance continues to decrease from that point (1.5V)

If current is rising linearly with an increase in current, resistance is decreases (R=V/I)

If the current is rising linearly with voltage, R = V/I and so 2V/2I, 3V/3I are all going to give the same resistance.

Reply 1194

HarryW95

Original post by Junaid96

If the current is rising linearly with voltage, R = V/I and so 2V/2I, 3V/3I are all going to give the same resistance.

Current was rising linearly where as voltage was just increasing.

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.

Reply 1195

Doglamlico

Original post by HarryW95

Hmm. Anyone got the paper yet?

Reply 1196

HarryW95

Original post by Malabarista

I'm glad I'm getting some similar answers then. I was really disappointed by the lack of and 'explain the photoelectric effect' in that exam.

I was glad that didn't come up tbh! Think that question came up in Jan 11 if I remember correctly..

Reply 1197

Nick_

Original post by Junaid96

If the current is rising linearly with voltage, R = V/I and so 2V/2I, 3V/3I are all going to give the same resistance.

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

(edited 11 years ago)

Reply 1198

Ionus7

Original post by Z REFAN Z

When voltage was 1.4V what is the resistance from the reading of the graph.

Which the answer is infinite or high resistance

I put 0... Darn it. Now I know why that makes sense. D'OH!

Reply 1199

KLW

I thought it was a really good paper, but some of the questions were a bit odd, I think boundaries will approx be 75/80% - A, 65/70% - B and then in rough 10 marks intervals per grade