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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 ambbs)
    I put C too. I think we're right, because I remembered seeing this in the textbook Attachment 223879Attachment 223880


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    I think it is D
    If it was C then Y should start from the origin and go below the X axis
    i traced the wave to check :P
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    Can anyone upload the Doppler? What was the answer?
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    (Original post by vinhvu95)
    Did anyone see through the trick question in question 17? It wasn't a negative gradient graph, because that's when the voltmeter is measuring the load resistance, but in this question the voltmeter was measuring the Pd at the battery, so it would've been a positive graph


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    Are you 100% sure?
    I saw a few books with exactly the same circuit as the one we were shown in the exam and they used the same method?
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    Upload the books!!!!!
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    (Original post by vinhvu95)
    Did anyone see through the trick question in question 17? It wasn't a negative gradient graph, because that's when the voltmeter is measuring the load resistance, but in this question the voltmeter was measuring the Pd at the battery, so it would've been a positive graph


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    it was definitely a negative gradient graph with EMF as the intercept and gradient r
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    So what did the graph look like?


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    (Original post by x0x)
    I think it is D
    If it was C then Y should start from the origin and go below the X axis
    i traced the wave to check :P
    I'm still convinced it's C :dontknow:
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    (Original post by Alexsk)
    Upload the books!!!!!
    Dude calm down! I am not saying you're wrong :P
    I borrowed that book and i don't have it anymore
    and i remember doing a paper 3 question about internal resistance. I drew the same arrangement and my teacher told me that it would work as there is no difference. Did your physics teacher tell you that your reasoning is right? If so then you're likely to be right
    there were 5 marks for that part .. how did you find the internal resistance and the e.m.f?
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    (Original post by a10)
    it was definitely a negative gradient graph with EMF as the intercept and gradient r
    I put positive gradient, but with a non zero start that curved at the end...am I completely wrong? Lol. I agree with how to work out emf and r though.


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    (Original post by a10)
    it was definitely a negative gradient graph with EMF as the intercept and gradient r
    Yes, that's what I did

    But I stopped my graph when it intercepted the horizontal axis and I stopped it when it intercepted the vertical axis, is that right?
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    (Original post by a10)
    it was definitely a negative gradient graph with EMF as the intercept and gradient r
    I'm not 100% sure but if you imagine that the load resistor had infinite resistance, Pd at the battery would be very low and overall current would be very small, if the load resistor had a very low resistance, then pd at the battery would be very high and overall current would be high because there is less resistance

    Edit: So my graph looked like a positive gradient graph where the maximum y value was emf and gradient is r rather than -r

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    What's the answer everyone put for 17b?
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    (Original post by ambbs)
    I'm still convinced it's C :dontknow:
    Can you explain why you think its C?
    half a radian is equal to 90 degrees
    90 degrees is a quarter of a full single cycle
    if you move Y backwards 90 degrees its new position is over X hence that is why i think its D but again i might be wrong :P
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    (Original post by krisshP)
    Yes, that's what I did

    But I stopped my graph when it intercepted the horizontal axis and I stopped it when it intercepted the vertical axis, is that right?
    yeah ... did the same
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    (Original post by Jaydude)
    I put positive gradient, but with a non zero start that curved at the end...am I completely wrong? Lol. I agree with how to work out emf and r though.


    Posted from TSR Mobile
    it should be negative gradient because from the equation, but I think u can get away with drawing a positive graph but mentioning r as negative

    E = V + Ir

    in the form y= c + m.x

    V=E - Ir

    so the graph was V against I hence gradient should be negative, and the value is r
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    (Original post by x0x)
    Dude calm down! I am not saying you're wrong :P
    I borrowed that book and i don't have it anymore
    and i remember doing a paper 3 question about internal resistance. I drew the same arrangement and my teacher told me that it would work as there is no difference. Did your physics teacher tell you that your reasoning is right? If so then you're likely to be right
    there were 5 marks for that part .. how did you find the internal resistance and the e.m.f?
    HAHA I am calm and I agree with you for this question!
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    (Original post by x0x)
    Can you explain why you think its C?
    half a radian is equal to 90 degrees
    90 degrees is a quarter of a full single cycle
    if you move Y backwards 90 degrees its new position is over X hence that is why i think its D but again i might be wrong :P
    If you look at the pictures I had in my original post you'll see why I think it's C I saw a very similar question in the textbook
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    (Original post by vinhvu95)
    I'm not 100% sure but if you imagine that the load resistor had infinite resistance, Pd at the battery would be very low and overall current would be very small, if the load resistor had a very low resistance, then pd at the battery would be very high and overall current would be high because there is less resistance

    Edit: So my graph looked like a positive gradient graph where the maximum y value was emf and gradient is r rather than -r

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    that should still be correct by the sounds of it
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    (Original post by x0x)
    yeah ... did the same
    E=V+Ir
    V=-rI+E

    Gradient is negative, but I still don't get how we are supposed to know that the graph stops at I=0 and at V=0.
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    (Original post by a10)
    it should be negative gradient because from the equation, but I think u can get away with drawing a positive graph but mentioning r as negative

    E = V + Ir

    in the form y= c + m.x

    V=E - Ir

    so the graph was V against I hence gradient should be negative, and the value is r
    It wasn't Big V, it was small v that the voltmeter was measuring. Because the voltmeter was across the battery


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