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AQA Physics Unit 1 PHYA1 20th May 2013

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Could anyone please explain question 6c on the June 2011 paper? I'm confused as to why the mark schemes putting the lines where it is


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For filament bulbs, if the pd is decreased, does the resistance decrease and the power increase? Also what happens to the current in this case and the temperature?
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Bixel
5
What will we be asked about oscilloscopes?
Original post by kashagupta
Could anyone please explain question 6c on the June 2011 paper? I'm confused as to why the mark schemes putting the lines where it is


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6. c i). It had double the internal resistance so the gradient will be doubled (and pass through the same y-intercept as the EMF is unchanged).

6. c ii). It has no internal resistance so the gradient is 0. Hence, it's just a horizontal line. :smile:
Original post by rainerised
For filament bulbs, if the pd is decreased, does the resistance decrease and the power increase? Also what happens to the current in this case and the temperature?


Think about it. V=IR. what happens to r when you decrease pd? P=V^2/R, what happens to power r is decreased?
Original post by Paul Thornton
I just feel like its gonna be another one on circuits, something on a filament lamp hasn't come up before


could do although I'm thinking they will do something different they have used almost all the components
Original post by Bixel
What will we be asked about oscilloscopes?


I can't really think about anything other than last years..

http://www.thestudentroom.co.uk/attachment.php?attachmentid=217144&d=1368903799

check out the top one
Original post by StalkeR47
Think about it. V=IR. what happens to r when you decrease pd? P=V^2/R, what happens to power r is decreased?


I think that resistance does decreases as pd decreases, but not sure what happens to current, does that also decrease? and if so does temperature decrease?
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ccashman
Does any one have any idea what the 6/8 marker question will be on? I always lose my marks there
Original post by usycool1
6. c i). It had double the internal resistance so the gradient will be doubled (and pass through the same y-intercept as the EMF is unchanged).

6. c ii). It has no internal resistance so the gradient is 0. Hence, it's just a horizontal line. :smile:


Thanks so much! I'm just finding annoying little questions which I should know the answer to and just don't! It's really frustrating! On the same paper, how would you go about working out 7bi?


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Original post by rainerised
I think that resistance does decreases as pd decreases, but not sure what happens to current, does that also decrease? and if so does temperature decrease?


I=V/R so decreasing the pd decreases the current. But the resistance increases.
This might sound really stupid but in experiments where you have to change temperature, how would you do this, because wouldn't the components get wet? On one of them it said to heat the thermistor in a beaker of water, so would you heat it first then take it out and do the experiment? I dont really get how it would work.
Original post by kashagupta
Thanks so much! I'm just finding annoying little questions which I should know the answer to and just don't! It's really frustrating! On the same paper, how would you go about working out 7bi?


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work out the voltage of z which is 6v. 10-6 is 4v. so, 4/3 is the current at y. EG, I=V/R, I=4/3 = 1.33A.
Original post by rainerised
This might sound really stupid but in experiments where you have to change temperature, how would you do this, because wouldn't the components get wet? On one of them it said to heat the thermistor in a beaker of water, so would you heat it first then take it out and do the experiment? I dont really get how it would work.


just say you'll use a water bath, look at the mark schemes to help you on certain questions :smile:
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Bixel
5
Original post by BenChard
I can't really think about anything other than last years..

http://www.thestudentroom.co.uk/attachment.php?attachmentid=217144&d=1368903799

check out the top one


Brilliant mate, thank you!
Reply 535
Avatar for Clarky747
Clarky747
Original post by rainerised
This might sound really stupid but in experiments where you have to change temperature, how would you do this, because wouldn't the components get wet? On one of them it said to heat the thermistor in a beaker of water, so would you heat it first then take it out and do the experiment? I dont really get how it would work.

Use a water bath WITH THERMOMETER! Make sure you add something like stir the water before taking measurement to ensure temperature is equal throughout water. Ignore the water-electricity=bad things.
Original post by StalkeR47
I=V/R so decreasing the pd decreases the current. But the resistance increases.


Ok that makes sense. Thanks :smile:
Original post by Clarky747
Use a water bath WITH THERMOMETER! Make sure you add something like stir the water before taking measurement to ensure temperature is equal throughout water. Ignore the water-electricity=bad things.



Ok thanks, I think I'm just making things more complicated :smile:
Original post by rainerised
Ok that makes sense. Thanks :smile:


no problem:cool:
Reply 539
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Clarky747
Original post by Nav_Mallhi
Do you have the mark scheme for this paper, by any chance? :smile:

:smile:
AQA-PHYA1-W-MS-JAN13.pdf129.8KB

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