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AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

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anyone know how to do q8?IMG_20150517_173346.jpg
Original post by wat a wizard
anyone know how to do q8?IMG_20150517_173346.jpg


C. (CV^2)/2t
@Ubisoft
Why over 2t?
P=IV, I=Q/t so P=QV/t but Q=CV so P=CV^2/t
Reply 1563
Guys, I don't get it if it's me with the lack of knowledge or it's just exam technique or something else. I did the June 14 Past paper for Turning points and the first question tripped me up. ( it was about gas discharging tube). I have never come across such question in my book or notes. I have done a lot of past papers so exam technique shouldn't be a problem. What do you people recommend ?
Original post by Somniare
@Ubisoft
Why over 2t?
P=IV, I=Q/t so P=QV/t but Q=CV so P=CV^2/t


i think this is correct

ubisoft
.
(edited 8 years ago)
Reply 1565
Original post by 0151
Anyone gathered together the grade boundaries for the past papers to see which are the most difficult?


Again, grade boundaries do not reflect difficulty too much. Last year's PHYA4 paper was rather hard but the boundaries were consistent with past years.


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Reply 1566
Original post by Mehrdad jafari
No, that's 100% correct. I chose the wrong one even though i was thinking right. But C2 was being charged by C1 so when V2 was maximum the electrons were being drawn from the positive plate of the capacitor to terminal Y to the negative plate through R2


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Original post by Mehrdad jafari
Sorry, i was talking about the conventional current and not electron flow. So in that case the current was flowing from Y to C2 even though the electrons were being drawn from that plate.ImageUploadedByStudent Room1431869605.091468.jpg


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I see. So which one was correct?

I can't recall whether the question mentioned conventional current or electron flow.

I chose C2 to Y because I thought the current would reverse and discharge through R2, producing that dip, but I'm not so sure now.


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Reply 1567
Original post by Mehrdad jafari
Here are the answers but to be honest i couldn't make sense of the first two
ImageUploadedByStudent Room1431874642.382549.jpg


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That doesn't make sense why it's 2.0mA to me. I understand why the current is initially 3.0mA but why it's changed I'm not sure.


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Original post by sykik
Guys, I don't get it if it's me with the lack of knowledge or it's just exam technique or something else. I did the June 14 Past paper for Turning points and the first question tripped me up. ( it was about gas discharging tube). I have never come across such question in my book or notes. I have done a lot of past papers so exam technique shouldn't be a problem. What do you people recommend ?


I can understand you. In such a case the answer is already suggested to you, you only need to explain that. Which part of the first question did you have difficulty with?


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CD223,

I believe that no current flowed. Because at that instant the 'push' from the electrostatic charge on C2 was equal to the 'push' from the pd on the other capacitor.
Is there a thread for Unit 5 anywhere?!


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Original post by CD223
That doesn't make sense why it's 2.0mA to me. I understand why the current is initially 3.0mA but why it's changed I'm not sure.


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We don't need to worry about that. In OCR there is so much on capacitors and they learn capacitors in series and in parallel, that's why our predictions differ.


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Reply 1572
Original post by Somniare
@Ubisoft
Why over 2t?
P=IV, I=Q/t so P=QV/t but Q=CV so P=CV^2/t


[br]E=12CV2[br][br]E =\dfrac{1}{2} CV^2[br]

As the capacitor stores half the energy supplied by the battery.

[br]P=ΔEΔt[br][br]P =\dfrac{\Delta E}{\Delta t}[br]

[br]P=CV22Δt[br][br]\Rightarrow P=\dfrac{CV^2}{2 \Delta t}[br]


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(edited 8 years ago)
Original post by Fred Cantoni
Is there a thread for Unit 5 anywhere?!


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It's linked in the first post as well, but here
Original post by Somniare
CD223,

I believe that no current flowed. Because at that instant the 'push' from the electrostatic charge on C2 was equal to the 'push' from the pd on the other capacitor.


But C2 was discharged. I read the whole question again in the exam


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Reply 1575
Original post by Somniare
CD223,

I believe that no current flowed. Because at that instant the 'push' from the electrostatic charge on C2 was equal to the 'push' from the pd on the other capacitor.


All three options make sense to me. I don't dismiss any of them because all arguments have convincingly "correct" supporting evidence.

I wasn't sure if it meant the instant that V2 was at a max or just after.


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Original post by Somniare
@Ubisoft
Why over 2t?
P=IV, I=Q/t so P=QV/t but Q=CV so P=CV^2/t


Doesn't that assume current and voltage are constant? I don't know, do you have the mark scheme answer op?
Reply 1577
Original post by Mehrdad jafari
But C2 was discharged. I read the whole question again in the exam


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I see. So you think it was C2 to Y or Y to C2?


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Reply 1578
Original post by ubisoft
Doesn't that assume current and voltage are constant? I don't know, do you have the mark scheme answer op?


http://www.thestudentroom.co.uk/showthread.php?p=55896435#post55896435

It does say "mean" power though, suggesting it varies with the current and voltage.


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Original post by CD223
I see. So you think it was C2 to Y or Y to C2?


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I think the current flowed from Y to C2 even though the electron flow was in the reverse direction.


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