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    (Original post by particlestudent)
    Could you take a picture of your working? And post .

    I've attached my working. Hopefully you can read my messy handwritting! Thanks
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    (Original post by philo-jitsu)
    Am I right in thinking that the q=q0e^(-t/RC) equation is only applicable when the resistance is constant?

    case in point the june 14 paper section B Q4Aii, I just could not get a resistance out, then I realised the voltage I was finding was for the capacitor hence the pd across the resistor was the max pd - capacitors pd as they're on the same row.....unit 1 electronics coming back to ruin me.

    https://60abffc9b401b1c0936e01291c15...%20Physics.pdf
    Yeah in the decay equation for a capacitor R is constant, with this you've got to use unit 1 knowledge.... total pd - pd across capacitor to find pd across resistor then go from there... just trying to catch people out i think
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    (Original post by liverpool2044)
    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

    can someone explain to me 5b onward?

    i dont really get emf and its relationship with linkage, why is emf greatest at these values
    It is greatest when the flux linkage is zero because at this point the rate of change of flux is greatest - and by Faraday's law, this will be when the EMF is greatest
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    (Original post by kingaaran)
    It is greatest when the flux linkage is zero because at this point the rate of change of flux is greatest - and by Faraday's law, this will be when the EMF is greatest
    ahh thanks
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    Name:  IMG_20160618_145235.jpg
Views: 90
Size:  450.1 KBCould someone please explain this goddamn mark scheme for me?

    I don't understand the relative voltages between the the capacitor and the resistor while the capacitor discharges. Firstly why is the E (which is the source pd apparently not electric field strength) negative initially across the capacitor when it is fully charged (I thought maybe it could be due to direction of charge flow when discharging but then remembered voltage is a scalar).

    Also it says that the resistor pd discharges from E to zero but isn't the resistor pd zero to begin with when fully charged? So why would it discharge to zero?

    Im so baffled by capacitors in general lol, it's the only topic in unit 4 that drives me nuts. Im so annoyed at how this is in the mark scheme when no where does it mention any of this in the textbook ffs 😾

    So if anyone could help me, i'll love you forever.
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    Why is it that the magnetic force is always perpendicular to the electron's velocity. Ik it's a centripetal force but what's the reason why it has this property. Since ofc an electric force acting perpendicular to the direction of the initial velocity of a charged particle has the same direction throughout, what makes the magnetic force constantly change direction like this?
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    (Original post by treeporn)
    Name:  IMG_20160618_145235.jpg
Views: 90
Size:  450.1 KBCould someone please explain this goddamn mark scheme for me?

    I don't understand the relative voltages between the the capacitor and the resistor while the capacitor discharges. Firstly why is the E (which is the source pd apparently not electric field strength) negative initially across the capacitor when it is fully charged (I thought maybe it could be due to direction of charge flow when discharging but then remembered voltage is a scalar).

    Also it says that the resistor pd discharges from E to zero but isn't the resistor pd zero to begin with when fully charged? So why would it discharge to zero?

    Im so baffled by capacitors in general lol, it's the only topic in unit 4 that drives me nuts. Im so annoyed at how this is in the mark scheme when no where does it mention any of this in the textbook ffs 😾

    So if anyone could help me, i'll love you forever.
    Have you learnt Kirchoff's (dunno how to spell cba) laws? It's all to do with conservation of charge
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    (Original post by Nikhilm)
    Why is it that the magnetic force is always perpendicular to the electron's velocity. Ik it's a centripetal force but what's the reason why it has this property.
    I could be wrong, but I think no one actually knows why it happens, only how to describe it i.e. in terms of vector calculus, which is something I guess we'll learn at University
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    (Original post by kother2015)
    I've attached my working. Hopefully you can read my messy handwritting! Thanks
    Ft=mv-(-mu)
    Ft=mv+mu
    Ft-mu=mv
    v=(Ft-mu)/m

    Ft doesn't change as it says it receives the same impulse so if u increases, then the top part of the fraction decreases and so v will decrease .
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    (Original post by particlestudent)
    Ft=mv-(-mu)
    Ft=mv+mu
    Ft-mu=mv
    v=(Ft-mu)/m

    Ft doesn't change as it says it receives the same impulse so if u increases, then the top part of the fraction decreases and so v will decrease .
    Oh my god
    You are a star
    Thanks so much
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    (Original post by MintyMilk)
    I could be wrong, but I think no one actually knows why it happens, only how to describe it i.e. in terms of vector calculus, which is something I guess we'll learn at University
    Oh damn, and lol not me i'm doing law
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    (Original post by Nikhilm)
    Have you learnt Kirchoff's (dunno how to spell cba) laws? It's all to do with conservation of charge
    No I haven't I don't think its part of the syllabus? But thanks for mentioning it, i'll be sure to research it!
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    I quite like PHYA4, bit this will be the first AQA exam I've done this year, apparently the other AQA science exams have been rubbish
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    (Original post by MintyMilk)
    I could be wrong, but I think no one actually knows why it happens, only how to describe it i.e. in terms of vector calculus, which is something I guess we'll learn at University
    Couldnt you essentially have perpetual motion if it wasnt perpendicular? I think i was sort kf explained this when we were doing lenz law....too close to exams to go down that rabbit hole though
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    I'm doing June 2014 section B and the capacitance question says that a capicitor is charged at a constant current for 60 s and the p.d becomes 4.4v, the question is to find the resistance after 30s. In the mark scheme is says the p.d after 30s is 2.2v but i thought that capicitors didnt charge uniformly. Is it only uniform becuase the current is constant?
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    (Original post by rich1334)
    I quite like PHYA4, bit this will be the first AQA exam I've done this year, apparently the other AQA science exams have been rubbish
    Normally one of the units are hard and the other is easy/okay. So if 4 is hard expect 5 to be okay and vice versa
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    can anyone explain how to work out question 19 of the multiple choice on june 2014?
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    Can someone explain why the time is 40. Please
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    (Original post by Imtheish)
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    Can someone explain why the time is 40. Please
    Maybe I am being stupid but I don't understand the question
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    (Original post by cmrol)
    Maybe I am being stupid but I don't understand the question
    It says calculate the revolution, and to do that you need to find the time first from he graph. The mark scheme says time is 40ms, and I'm asking why it is. It's the June 2013 paper
 
 
 
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