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

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    anyone know how to do q8?Name:  IMG_20150517_173346.jpg
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    (Original post by wat a wizard)
    anyone know how to do q8?Name:  IMG_20150517_173346.jpg
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    C. (CV^2)/2t
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    @Ubisoft
    Why over 2t?
    P=IV, I=Q/t so P=QV/t but Q=CV so P=CV^2/t
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    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 ?
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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
    i think this is correct

    (Original post by ubisoft)
    .
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    (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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    (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.Attachment 400511


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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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    (Original post by Mehrdad jafari)
    Here are the answers but to be honest i couldn't make sense of the first two
    Attachment 400593


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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.
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    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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    (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
    

E =\dfrac{1}{2} CV^2

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

    

P =\dfrac{\Delta E}{\Delta t}

    

\Rightarrow P=\dfrac{CV^2}{2 \Delta t}


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    (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
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    (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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    (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?
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    (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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    (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/show...5#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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