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    (Original post by Clarky747)
    Do you happen to have the January 2013 paper? My teachers aren't able to log on to the teacher section of the AQA website.

    I don't see anything wrong with doing so as for the people doing AS next year the June 2013 paper will be the unknown one that they can't access.

    Don't worry if you can't.
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    (Original post by henpen)
    Do you happen to have the January 2013 paper? My teachers aren't able to log on to the teacher section of the AQA website.

    I don't see anything wrong with doing so as for the people doing AS next year the June 2013 paper will be the unknown one that they can't access.

    Don't worry if you can't.
    Yeah
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    how is the voltmeter measuring 'lost volts'? is it not measuring pd across the variable resistor?
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    (Original post by BenChard)
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    how is the voltmeter measuring 'lost volts'? is it not measuring pd across the variable resistor?
    Lost volt is the voltage lost inside the battery.
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    (Original post by BenChard)
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    how is the voltmeter measuring 'lost volts'? is it not measuring pd across the variable resistor?
    I don't understand what your getting at?
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    (Original post by StalkeR47)
    Lost volt is the voltage lost inside the battery.
    the text says about as current increase, volts lots increase and pd decreases. I just need this explained please
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    (Original post by Clarky747)
    I don't understand what your getting at?
    voltmeter is not measuring lost volt. it is measuring the voltage at the resistor. however, as the current increases, the lost volt inside the battery also increases so the pd across battery terminal decreases. and therefore, the voltmeter reads lower value of voltage at the resistance. Again, this is because more voltage is being used up in the battery (increase in lost volt). Did you get this?
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    (Original post by BenChard)
    the text says about as current increase, volts lots increase and pd decreases. I just need this explained please
    As current increases. Heat dissipated in the battery increases and so more voltage is lost in the internal resistance of the battery as increased temperature increases voltage needed. Therefore as more volts are lost in the battery. The voltage shown on the voltmeter will have decreased.
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    (Original post by StalkeR47)
    voltmeter is not measuring lost volt. it is measuring the voltage at the resistor. however, as the current increases, the lost volt inside the battery also increases so the pd across battery terminal decreases. and therefore, the voltmeter reads lower value of voltage at the resistance. Again, this is because more voltage is being used up in the battery (increase in lost volt). Did you get this?
    thanks that's cleared things up
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    (Original post by Clarky747)
    As current increases. Heat dissipated in the battery increases and so more voltage is lost in the internal resistance of the battery as increased temperature increases voltage needed. Therefore as more volts are lost in the battery. The voltage shown on the voltmeter will have decreased.
    Yes! that is what I said.
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    (Original post by BenChard)
    thanks that's cleared things up
    Your welcome!
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    (Original post by StalkeR47)
    Yes! that is what I said.
    Sorry, you must of posted it while I was typing.
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    (Original post by Clarky747)
    As current increases. Heat dissipated in the battery increases and so more voltage is lost in the internal resistance of the battery as increased temperature increases voltage needed. Therefore as more volts are lost in the battery. The voltage shown on the voltmeter will have decreased.
    this also helped, thanks:rolleyes:
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    (Original post by Clarky747)
    Sorry, you must of posted it while I was typing.
    Oh! It is ok.
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    (Original post by BenChard)
    the text says about as current increase, volts lots increase and pd decreases. I just need this explained please
    The volts are lost to increased heating due to current.... it becomes much lesser than the EMF value (which can be considered as like the true pd).

    This is why you can only find EMF when the current is zero... because there's no heating in the circuit therefore no resistance.... therefore no loss of volts so you get the EMF, this of course practically never happens... there's always some kind of resistance So that's why you use the graph from experimental data to predict the EMF value
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    Hi does anyone have a list of all the experiments that could come up for 6 marker for unit 1?
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    Guys, what are the hardest electricity questions you have ever come across in the past papers?
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    (Original post by Clarky747)
    Yeah
    Excellent, you are by far the most useful person I know wrt the January 2013 papers.
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    (Original post by Jatt.B)
    Hi does anyone have a list of all the experiments that could come up for 6 marker for unit 1?
    http://www.thestudentroom.co.uk/atta...4&d=1368903799
    that's what has come up, I think it will be:

    - graph of emf and internal resistance
    - graph of threshold frequency and work function
    - filament lamp circuit
    - osiclloscope
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    (Original post by BenChard)
    http://www.thestudentroom.co.uk/atta...4&d=1368903799
    that's what has come up, I think it will be:

    - graph of emf and internal resistance
    - graph of threshold frequency and work function
    - filament lamp circuit
    - osiclloscope
    Thank you! Also, is tommorow likely to be a harder paper? since most people doing it will be retaking?
 
 
 
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