Join TSR now and get all your revision questions answeredSign up now

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

    Offline

    3
    ReputationRep:
    (Original post by CD223)
    Thanks!

    Essentially you know it always takes 36ms for V to halve in value. This is because, for exponential decay the same decrease is observed over equal time intervals.

    As E is proportional to V squared, this means that when E becomes E/16, V must have halved again to become V/4.

    This means that the time for V to halve twice from its initial value and therefore become V/4 is:

    36ms + 36ms = 72 ms

    Does that help?


    Posted from TSR Mobile
    Thanks
    This would be the best way to approach that question because the change in voltage is an exact multiple of the voltage at 36 ms.
    I don't examiners will give inexact values so yeah, this is the easiest way to get the answer


    Posted from TSR Mobile
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Mehrdad jafari)
    Thanks
    This would be the best way to approach that question because the change in voltage is an exact multiple of the voltage at 36 ms.
    I don't examiners will give inexact values so yeah, this is the easiest way to get the answer


    Posted from TSR Mobile
    Thank my teacher lol.


    Posted from TSR Mobile
    Offline

    9
    ReputationRep:
    (Original post by Mehrdad jafari)
    Oh, sorry, i meant to write 72ms as i worked with ms in the question.
    And yeah, you're correct in saying that you cannot assume the proportionality of t with C as C is constant, but this proportionality has a different sense. t is proportional to c means that the greater the capacitance the longer the time the capacitor takes to discharge as it can contain more charge. I needed that proportionality to see how the voltage varies with time. If, for example, the greater the capacitance the less time it would take to discharge then proportionally would differ. It would become :
    Attachment 420273


    Posted from TSR Mobile
    Ah okay, I understand.

    Thank you
    Offline

    9
    ReputationRep:
    (Original post by getback339)
    yeah thanks for the reply, im not too sure i get the difference between A and B though
    Graph A shows that as frequency increases toward the natural frequency the amplitude decreases and Graph B shows that as frequency increases towards natural frequency the amplitude increases.

    Graph A's story is incorrect so it has to be Graph B.

    Does that help?
    Offline

    9
    ReputationRep:
    (Original post by Jed-Singh)
    Can anyone help me with the multiple choice questions that involve 'when potential or field strength etc' equals 0. I never get them correct
    Do you have an example question, it'd be easier to explain.
    Offline

    0
    ReputationRep:
    can someone help with q15 in mc section of the june 2014 paper?
    Offline

    3
    ReputationRep:
    (Original post by Disney0702)
    Do you have an example question, it'd be easier to explain.
    June 2013 Q13
    Offline

    2
    ReputationRep:
    (Original post by coqthepoliceman)
    can someone help with q15 in mc section of the june 2014 paper?
    Name:  ImageUploadedByStudent Room1433592670.383319.jpg
Views: 63
Size:  145.3 KB


    This is how I did it.


    Posted from TSR Mobile
    Offline

    9
    ReputationRep:
    (Original post by Jed-Singh)
    June 2013 Q13
    Name:  IMG_20150606_131841~01.jpg
Views: 94
Size:  390.1 KB

    Does that help?
    Offline

    3
    ReputationRep:
    (Original post by Disney0702)
    Name:  IMG_20150606_131841~01.jpg
Views: 94
Size:  390.1 KB

    Does that help?
    X=d-y! Youre a genius, i never thought of doing that. Thanks so much
    Offline

    9
    ReputationRep:
    (Original post by Jed-Singh)
    X=d-y! Youre a genius, i never thought of doing that. Thanks so much
    You're welcome
    Offline

    10
    ReputationRep:
    (Original post by Disney0702)
    Thank you so much Lauren!

    I've definitely never seen this before! Had you never uploaded this I would've probably freaked out in the exam hall.

    And the attachment you released is in good condition.
    Oh good, it worked then (stupid internet!). No problem, it's a little worrying how many teachers don't bother mentioning this at all - our teacher is virtually useless and even he managed to give us them!
    Offline

    2
    ReputationRep:
    (Original post by Disney0702)
    Graph A shows that as frequency increases toward the natural frequency the amplitude decreases and Graph B shows that as frequency increases towards natural frequency the amplitude increases.

    Graph A's story is incorrect so it has to be Graph B.

    Does that help?
    oh yeah haha, just confused there, thanks anyway
    Offline

    9
    ReputationRep:
    (Original post by getback339)
    oh yeah haha, just confused there, thanks anyway
    You're welcome
    Offline

    9
    ReputationRep:
    Can someone explain something to me please.

    Q8 Jan '08

    I understand that the answer is C but I do not see how D could be incorrect.
    Offline

    2
    ReputationRep:
    (Original post by Disney0702)
    Can someone explain something to me please.

    Q8 Jan '08

    I understand that the answer is C but I do not see how D could be incorrect.
    C=Q/V

    Which basically a graidient of this graphs.
    So X has greater gradient than Y.
    So Y has less gradient than X so less capacitance

    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by Disney0702)
    Can someone explain something to me please.

    Q8 Jan '08

    I understand that the answer is C but I do not see how D could be incorrect.
    C=Q/V

    For Y, The p.d is larger than the charge. Shown by a smaller gradient. So capacitance isn't increasing.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Disney0702)
    Can someone explain something to me please.

    Q8 Jan '08

    I understand that the answer is C but I do not see how D could be incorrect.
    The gradient of the graph is representative of the capacitance. As X has a greater gradient, its capacitance is greater.

    Both are straight lines - in other words the gradient, and hence the capacitance is constant for both.


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by CD223)
    Thank my teacher lol.


    Posted from TSR Mobile
    Hey man, hows it going?

    When we were discussing about the six markers a while back you said you thought it would be on SHM or circular motion. If the six marker is on SHM, what do you think they will ask? or like what would they be looking to see in your answers?
    Offline

    9
    ReputationRep:
    (Original post by a.a.k)
    C=Q/VWhich basically a graidient of this graphs.So X has greater gradient than Y.So Y has less gradient than X so less capacitancePosted from TSR Mobile
    (Original post by Adangu)
    C=Q/V

    For Y, The p.d is larger than the charge. Shown by a smaller gradient. So capacitance isn't increasing.
    Thank you very much both of you, that makes sense.
 
 
 
Poll
Which Fantasy Franchise is the best?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.