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Edexcel Physics Unit 2 "Physics at work" June 2013

Guys do we think the final answer was a decrease or an increase in resistance? I firmly believe it is a decrease. I don't think you can use ohm's law because it was a series circuit so V is not the same across each lamp. They also said the bulbs didn't light up which was a clue. A MASSIVE clue was they asked us for the effect of low current ON the resistance, they didn't ask us to explain an increase/decrease in resistance USING the low current. Do you guys kinda get what im trying to say?

Therefore I believe it was the lattice vibrations argument. It doesn't make sense to have 4 marks on an equation derivation question either imo.

I firmly believe the opposite haha! But if the current decreases and so does the resistance won't that mean nothing has changed? and so the light bulb should remain on?

I HATE THIS QUESTION.

Reply 841

GCSE-help

6

Original post by Superhans34

I firmly believe the opposite haha! But if the current decreases and so does the resistance won't that mean nothing has changed? and so the light bulb should remain on?

I HATE THIS QUESTION.

It's difficult to explain and I hate it too, but the more and more I look at it, the more confident I'm getting it is the vibration/electron collision argument.

But like you said, a lot of people firmly believe the opposite, the majority in fact, so you may be right!

Reply 842

flou_fboco

I think I going to retake nxt year

Reply 843

a10

19

Original post by study beats

was this exam hard or easy guys?? i dont do this...but damn i did the physics unit 1 exam and it ruined my life..

it wasn't that difficult it was same difficulty as jan 13 (unit 2). Unit 1 was horrible and I feel your pain bro :cry2:

Reply 844

asneivethan

does anyone know whether the bulbs are connected in series or parallel for the last question

Reply 845

x0x

OP

7

Original post by asneivethan

does anyone know whether the bulbs are connected in series or parallel for the last question

Series

Reply 846

krisshP

14

Original post by a10

it wasn't that difficult it was same difficulty as jan 13 (unit 2). Unit 1 was horrible and I feel your pain bro :cry2:

I hated it so much as well and I worked my ass off :frown:

Reply 847

Phteven

15

Original post by Superhans34

Yeah during the photoelectric effect but when the electron just jumps between an energy level, the energy lost is emitted as a photon. Not a photo-electron/electron because that doesn't make sense.

Ahh sorry, I misread :colondollar:

. Yes, it's a photon then.

Reply 848

jollygood

8

Original post by awesomesoccerfan

I referred to the equation R=(rho x l)/A and said that R is inversely proportional to A and directly proportional to l.As l increases and A decreases so R would be higher.Am i right? :s-smilie:

yes i referred to equation as well and said same thing. its right.

Reply 849

jollygood

8

Original post by Freddy-Francis

guys what is a critical angle ? :O

Angle of incidence in denser medium for which the angle of refraction is 90 degree in rare medium.

Reply 850

jollygood

8

Original post by ALevel96

If you double the length the number of charge carriers doubles too! There's no two ways about it

if no. of charge carriers is doubled the current should become doubled as well but the mcq said the current is constant.

Reply 851

ambbs

12

What did people think about the difficulty compared to Jan 13, where it was 56 for an A? I found Jan a bit easier personally, but maybe that's because I only sat it as a mock and there wasn't the pressure you get in a real exam - I didn't like how this paper favoured electricity explanations rather than calculations though!

Reply 852

Jaydude

11

So what are your opinions on the grade boundaries?

I think 54. We'll wait and see. :biggrin:

Reply 853

JoshThomas

Original post by jollygood

if no. of charge carriers is doubled the current should become doubled as well but the mcq said the current is constant.

Yaa by referring to i=naqv, current is number of charge carriers per m^3, if the length is doubled the number of charge carriers per m^3 doesnt change, nowhere in i=naqv does length affect current.....

Reply 854

hamzeh h

Original post by JoshThomas

Yaa by referring to i=naqv, current is number of charge carriers per m^3, if the length is doubled the number of charge carriers per m^3 doesnt change, nowhere in i=naqv does length affect current.....

yup i've been trying to tell him that he wont accept it

Reply 855

jollygood

8

Original post by Alexsk

I wrote that 3 is C. Is that wrong?

for third one i wrote C as well because if u look at two waves one is 90 degrees ahead of other. the other one starts at zero. i think they just extended the Y one to trick us.

Reply 856

x0x

OP

7

Original post by jollygood

for third one i wrote C as well because if u look at two waves one is 90 degrees ahead of other. the other one starts at zero. i think they just extended the Y one to trick us.

Trace Y and move it backwards a quarter of a cycle
You will find that it goes over X
So it is half a radian ahead of it
so i think its D
nevertheless you may be right :P

Reply 857

a10

19

for the area one (the I =nvqa) what did everyone put? I put 2v

Reply 858

B-Stacks

13

Original post by a10

for the area one (the I =nvqa) what did everyone put? I put 2v

That's what I put. But apparently its wrong and most people put v/2.

Reply 859

x0x

OP

7

Original post by B-Stacks

That's what I put. But apparently its wrong and most people put v/2.

From the resistivity equation l and A both are doubled so the current stays the same
But from I=nAvq A is doubled and since I doesn't change it has to be v/2
that is my reasoning .. i may be wrong so please if you can provide an alternative/better explanation please do so

Edexcel Physics Unit 2 "Physics at work" June 2013

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