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Edexcel Physics Unit 2 "Physics at work" June 2013 Watch

  • View Poll Results: The last question - Does resistance increase or decrease?
    It increases ( using V=IR or some other method)
    70.73%
    It decreases using the 'lattice vibrations' theory
    29.27%

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    (Original post by CoolRunner)
    Edexcels gonna **** us over again. Gosh, how i hate edexcel physics.
    I know. They may tear us apart.
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    (Original post by muonz)
    Hm? P is a fixed resistor, when the lamp is in parallel with the fixed resistor, the resistance of the parallel combination is smaller than the resistance of P. Therefore now that the lamp is removed you have a potential divider, the ratio of the resistance between P and R changes so that P has a higher proportion of the voltage (as the resistance of P > the resistance of P and the lamp in parallel), hence the voltmeter reading increases.
    But but...
    I thought like this...where did I go wrong?

    The 12 V is shared between P (and all its parallels) and Q(Q nah it was? Can't see the pic in this app!).
    So v of P= 12 - v of Q

    Now the total resistance of the circuit increases. So the current decreases. Resistance of Q is constant so its V decreases. So V of P increases!!
    Sorry for the mixing up edits but as I posted firstly it increases

    I don't seem to follow you. :confused:

    EDIT: take Q as the other resistor not the bulb. The pic does not show up in the app. So Q in my above explanation is not the bulb
    EDIT: yes I get it increases. Lol I just got it now and I assumed I typed in it increases. Not a typo a genuine mistake...lol now I'm not confused.

    Thanks for pointing out

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    Guess this paper can't be as hard as the Unit 1 paper.
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    (Original post by StUdEnTIGCSE)
    Guess this paper can't be as hard as the Unit 1 paper.
    Who knows. Edexcel may throw us a real bomb as a surprise
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    Its possible that it could be really hard. In one of the past papers i saw it was 45/80 for an A and 55/80 for 120 UMS. LOL Its kind of good having a hard paper cuz you can loose like 30 marks and get an A or B
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    (Original post by popo)
    Its possible that it could be really hard. In one of the past papers i saw it was 45/80 for an A and 55/80 for 120 UMS. LOL Its kind of good having a hard paper cuz you can loose like 30 marks and get an A or B
    Ah yes...
    But a hard paper always gives that suspension feeling that lasts untill results day.

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    (Original post by CoolRunner)
    Edexcels gonna **** us over again. Gosh, how i hate edexcel physics.
    how much u got in jan paper?
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    did anyone do unit 1 in june 13? and if not does anyone know what 120 ums was in jan? both papers were hard so I'm thinking it was quite low
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    (Original post by lucilleJR)
    did anyone do unit 1 in june 13? and if not does anyone know what 120 ums was in jan? both papers were hard so I'm thinking it was quite low
    52=A

    64= 120ums (or 62 marks, cant remember!)
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    (Original post by Daniel Atieh)
    how much u got in jan paper?
    a D lol.
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    Has anyone got markscheme notes for physics unit 2? similar to biology?
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    Just wanted to confirm why the answer to this was C and not D
    Attached Images
     
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    (Original post by Zoeyyy)
    Just wanted to confirm why the answer to this was C and not D
    Top part resistance=8 ohms
    Bottom part resistance=8 ohms
    1/R=1/8+1/8
    1/R=1/4
    R=4 ohms
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    (Original post by Zoeyyy)
    Just wanted to confirm why the answer to this was C and not D
    Well the total resistance of resistors depend on whether they are in parallel or series.
    In series the total resistance is the sum of each individual resistances.
    In parallel it follows this

     \dfrac{1}{R}= \dfrac{1}{r_{1}} + \dfrac{1}{r_{2}} + \dfrac{1}{r_{3}} + \dfrac{1}{r_{3}} .....

    Where R is the total resistance and r is the individual resistances.

    Now before you solve for parallel you should find the resistance for each branch. The two 4ohm resistors are in series in each branch so adding it each branch has resistance of 8ohm. Now adding as in parallel gives 4ohms.

    An easier way to find the resistance of two branches in parallel is

     R= \dfrac{r_{1} \times r_{2}}{r_{1} + r_{2}}

    And when there are n branches each with the same r resistance its easier. The total resistance us simple r/n

    EDIT: I corrected one thing...typo



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    Does anyone have access to the 2013 jan paper? Thanks!
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    (Original post by Jaydude)
    Does anyone have access to the 2013 jan paper? Thanks!
    http://www.thestudentroom.co.uk/show....php?t=2262586
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    Expectations for our Wednesday exam, guys?



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    (Original post by krisshP)
    Top part resistance=8 ohms
    Bottom part resistance=8 ohms
    1/R=1/8+1/8
    1/R=1/4
    R=4 ohms
    Ahh, thanks a bunch!
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    (Original post by StUdEnTIGCSE)
    Well the total resistance of resistors depend on whether they are in parallel or series.
    In series the total resistance is the sum of each individual resistances.
    In parallel it follows this

     \dfrac{1}{R}= \dfrac{1}{r_{1}} + \dfrac{1}{r_{2}} + \dfrac{1}{r_{3}} + \dfrac{1}{r_{3}} .....

    Where R is the total resistance and r is the individual resistances.

    Now before you solve for parallel you should find the resistance for each branch. The two 4ohm resistors are in series in each branch so adding it each branch has resistance of 8ohm. Now adding as in parallel gives 4ohms.

    An easier way to find the resistance of two branches in parallel is

     R= \dfrac{r_{1} \times r_{2}}{r_{1} + r_{2}}

    And when there are n branches each with the same r resistance its easier. The total resistance us simple r/n

    EDIT: I corrected one thing...typo



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    Super! Thanks
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    hey, does anyone on here know whether we need to learn about the evidence which support the wave/particle theory or whether we only need to know how to explain the evidence?
 
 
 
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