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    Can I get a bit of help on June 10 Q6c)ii) ?

    It's about a filament bulb, I know increasing the current through a filament will increase it's resistance but in the question, the pd is reduced which causes power to increase. How does that happen?
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    (Original post by NedStark)
    Can I get a bit of help on June 10 Q6c)ii) ?

    It's about a filament bulb, I know increasing the current through a filament will increase it's resistance but in the question, the pd is reduced which causes power to increase. How does that happen?
    Current is lower therefore temperature is lower therefore resistance is lower therefore current is higher than predicted and so the power rating is therefore higher than expected for the lower pd.


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    (Original post by PickwickianGeek)
    Most are speculating it'll be on oscilloscopes. I too, hope it's on the photoelectric effect. So much can be said to get 6 marks. I personally hope it's not on semi-conductors or diodes, those are the worst.
    It's unlikely to be diodes as that was what one of the ISA's was on this year, but it could still come up (although there's a very low chance).
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    http://filestore.aqa.org.uk/subjects...1-W-SMS-07.PDF

    Someone help me on the very last question please
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    (Original post by dunkdeboffin)
    Ahah! (Thank you )
    Are there any differences between the following:
    weak nuclear force, weak force, weak interaction, weak nuclear interaction?
    Are some of them nonsense?
    Hey! I will give you the definitions: 1) Strong interaction- Interaction between 2 hadrons. 2) Strong nuclear force- attractive force that holds the nucleons in the nucleus. -----------Both of these definitions are worth 2 marks. Hope this clears the confusion. EDIT______------ And yeah.... Weak interaction- interaction between 2 leptons (I.e. electron and a neutrino). Weak nuclear force- Force responsible for beta decay. This is all you need to know for the exam on interactions.
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    (Original post by Goods)
    Current is lower therefore temperature is lower therefore resistance is lower therefore current is higher than predicted and so the power rating is therefore higher than expected for the lower pd.


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    How do we know the current is lower?
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    (Original post by StalkeR47)
    what? from past paper?

    its from jan 2013!
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    (Original post by NedStark)
    How do we know the current is lower?
    We know current is lower when resistance in bigger. This is because increase in resistance makes it harder for electrons to pass through.
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    (Original post by Micheal123456)
    its from jan 2013!
    What question?
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    i need help on a question 2b & 6b jan 2013 thanks!
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    (Original post by StalkeR47)
    What question?
    questin 2b & 6b
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    (Original post by StalkeR47)
    We know current is lower when resistance in bigger. This is because increase in resistance makes it harder for electrons to pass through.
    I'm confused on the link here, the question tells us the pd is reduced. So that's clearly the first thing we find out.

    How does the pd being reduced lead on to either the resistance or current being changed?
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    (Original post by Goods)
    Current is lower therefore temperature is lower therefore resistance is lower therefore current is higher than predicted and so the power rating is therefore higher than expected for the lower pd.


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    I agree with the lower resistance part.... however with lower resistance wouldn't there be higher current ? Therefore it will also heat up.

    I=V/R

    Power = IV ... therefore greater power if greater current ? :confused:

    So confused
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    (Original post by Micheal123456)
    questin 2b & 6b
    Q2b) for 1 electron to be created, 0.510999eV is needed. For a pair (2 of them) you need 0.510999eV times 2. Does that helps? Q6b)Ok.. we have 4 resistors in series. 2 in each series. Total voltage is 12 volts. (Remember voltage in parallel is the same all the way around and in series it adds up to). Since the resistors are in series, they must have voltage adds up to total depending on the amount of resistance. There is an easy way to handle this problem and a harder way to solve. Easy way----- Voltage in parallel----same. So, across ac and ce, they must share same amount of voltage. so it must be 6V for both ac and ce. This is total of 6 Voltage used up in A-E. So you have 6 left. Since the resistance of the resistor is twice the resistance of the thermistor. Resistor BD must get twice the voltage as DF. So, BD must be 4 V and thermistor which is DF or CD must get 2V. so our voltage adds up to total. 6+4+2=12V. Harder way... calculate the current through AE by using V=IR. I=12/4xx10^3 = 3x10^-4A... Use V=IR again to calculate voltage across AC. V=3x10^-4 times 20x10^3 = 6V. The rest can be solved by using V=IR. Hope this helps..
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    (Original post by PickwickianGeek)
    Most are speculating it'll be on oscilloscopes. I too, hope it's on the photoelectric effect. So much can be said to get 6 marks. I personally hope it's not on semi-conductors or diodes, those are the worst.
    That exactly what i was thinking but i really hope it isnt cause i have no idea what i would say. I just did the june 2012 paper and wrote nothing on that oscilloscpe question but im memorizing the answer right now from the markscheme.
    I wouldnt mind stuff on the diode, you'll end up get two marks just for saying repeats and connecting the circuit! Oh well lets just see what the paper holds for us!

    Good luck
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    (Original post by NedStark)
    I'm confused on the link here, the question tells us the pd is reduced. So that's clearly the first thing we find out.

    How does the pd being reduced lead on to either the resistance or current being changed?
    the temperature is lower (therefore the resistance is lower) hence if you calculate power using V^2/R since R is a smaller value the power rating will be larger?
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    (Original post by NedStark)
    I'm confused on the link here, the question tells us the pd is reduced. So that's clearly the first thing we find out.

    How does the pd being reduced lead on to either the resistance or current being changed?
    Can you tell me which question and past paper is it? it depends if the circuit have components in parallel or in series.
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    (Original post by StalkeR47)
    Q2b) for 1 electron to be created, 0.510999eV is needed. For a pair (2 of them) you need 0.510999eV times 2. Does that helps? Q6b)Ok.. we have 4 resistors in series. 2 in each series. Total voltage is 12 volts. (Remember voltage in parallel is the same all the way around and in series it adds up to). Since the resistors are in series, they must have voltage adds up to total depending on the amount of resistance. There is an easy way to handle this problem and a harder way to solve. Easy way----- Voltage in parallel----same. So, across ac and ce, they must share same amount of voltage. so it must be 6V for both ac and ce. This is total of 6 Voltage used up in A-E. So you have 6 left. Since the resistance of the resistor is twice the resistance of the thermistor. Resistor BD must get twice the voltage as DF. So, BD must be 4 V and thermistor which is DF or CD must get 2V. so our voltage adds up to total. 6+4+2=12V. Harder way... calculate the current through AE by using V=IR. I=12/4xx10^3 = 3x10^-4A... Use V=IR again to calculate voltage across AC. V=3x10^-4 times 20x10^3 = 6V. The rest can be solved by using V=IR. Hope this helps..
    were did you get the infomation that you need that about of energy for 1 electron to be created?
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    (Original post by StalkeR47)
    It will be harder than jan 2013.
    No way are you being serious wasn't the paper in jan hard enough


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    (Original post by StalkeR47)
    Can you tell me which question and past paper is it? it depends if the circuit have components in parallel or in series.
    June 10 6c)ii)

    Link: http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
 
 
 
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