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    Hay guys, as I see this thread is pretty active right now. So maaaaybe somebody could explain me 6th question (b) in the 2013 January paper , please ? It's just that in the mark scheme there are just values given, no solutions..

    p.s. GOOD LUCK TOMORROW EVERYONE, we can do it !!!!!
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    (Original post by PickwickianGeek)
    anyone care to explain 5 b) on this paper? http://filestore.aqa.org.uk/subjects...1-W-SQP-07.PDF
    You have a total pd of 12V. The resistors are in parallel with a diode in parallel in it's forward bias. (Hint: What is the voltage at 6ohm resistor) ?
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    (Original post by StalkeR47)
    YEP that is true! But I can help you if you need any. Today is the last day hahahaa:cool:
    Oh don't remind me! I want to study physics at uni but damn electricity just annoys me I do have one question though, in the context of Q6C Jan 2013 (and others similar) how do you go about talking about proportion?

    Up until a week ago I always got it right but all of a sudden I lost it :/
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    (Original post by pushkin_)
    Hay guys, as I see this thread is pretty active right now. So maaaaybe somebody could explain me 6th question (b) in the 2013 January paper , please ? It's just that in the mark scheme there are just values given, no solutions..

    p.s. GOOD LUCK TOMORROW EVERYONE, we can do it !!!!!
    Hi! I will explain this to you...
    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)
    anyone care to explain 5 b) on this paper? http://filestore.aqa.org.uk/subjects...1-W-SQP-07.PDF
    As it's a diode, you know that the pd across it is 0.6v, therefore the pd across the resistor is 12-0.6= 11.4v. 11.4/4 = 2.85A. Add this to 12/6 = 2A, and your answer should be 4.85A Is this correct?
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    (Original post by Abbseh)
    Oh don't remind me! I want to study physics at uni but damn electricity just annoys me I do have one question though, in the context of Q6C Jan 2013 (and others similar) how do you go about talking about proportion?

    Up until a week ago I always got it right but all of a sudden I lost it :/
    Can you give more details on the question you have just asked?
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    (Original post by yajman)
    As it's a diode, you know that the pd across it is 0.6v, therefore the pd across the resistor is 12-0.6= 11.4v. 11.4/4 = 2.85A. Add this to 12/6 = 2A, and your answer should be 4.85A Is this correct?
    yeah that's correct, cheers for the explanation. So current is split in a parallel circuit but is the same throughout a series circuit, yes?
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    (Original post by yajman)
    As it's a diode, you know that the pd across it is 0.6v, therefore the pd across the resistor is 12-0.6= 11.4v. 11.4/4 = 2.85A. Add this to 12/6 = 2A, and your answer should be 4.85A Is this correct?
    yajman! you are not allowed to give straight answers! check the guidelines for the student room. that is why I have him a hint.
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    (Original post by westarmy)
    Would you like to balance my pivot by placing weights on either side? :ahee:
    That made me giggle and squirm- and I'm not even sure what you meant by that!... I'm still grinning. Like an idiot.




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    (Original post by StalkeR47)
    Hi! I will explain this to you...
    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..
    Ohh, I see now ! I was doing it hard way and messed up with current ratio .. Thanks a lot !!!
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    (Original post by StalkeR47)
    yajman! you are not allowed to give straight answers! check the guidelines for the student room. that is why I have him a hint.
    Sorry there! As you can see, I'm fairly new to TSR, so I'm extremely sorry for that.
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    (Original post by PickwickianGeek)
    yeah that's correct, cheers for the explanation. So current is split in a parallel circuit but is the same throughout a series circuit, yes?
    That's correct, yes. Apparently, I'm not supposed to give straight answers, so I'll try and be discrete next time. Good luck if you're sitting the exam tomorrow
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    (Original post by yajman)
    Sorry there! As you can see, I'm fairly new to TSR, so I'm extremely sorry for that.
    It's ok!
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    (Original post by pushkin_)
    Ohh, I see now ! I was doing it hard way and messed up with current ratio .. Thanks a lot !!!
    No probs! Ratio always confuses me and therefore I tend to use calculations. The reason why I replied all that so quickly is because I just explained all that to someone else before, so I copied my explanations from page 23 of this post. HHAA:cool:
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    (Original post by yajman)
    That's correct, yes. Apparently, I'm not supposed to give straight answers, so I'll try and be discrete next time. Good luck if you're sitting the exam tomorrow
    Ah okay, thanks! Oh, are you not? It's not like it's a practical exam though, surely we are allowed to give away correct calculations on old papers? Thanks, you too if you're sitting it!
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    (Original post by PickwickianGeek)
    Ah okay, thanks! Oh, are you not? It's not like it's a practical exam though, surely we are allowed to give away correct calculations on old papers? Thanks, you too if you're sitting it!
    Yeah, I'm not sure why. I'll be sitting the exam tomorrow too, so thanks for the wishes. Praying that it will be a good paper with not too many difficult questions, particularly the 6-marker. That will make-or-break my chances of getting a good result
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    Does anyone have any idea what the big six marker question could be on?! The one last year was good, hope we have one like that or something on the photoelectric effect
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    (Original post by reemaisthinking)
    That made me giggle and squirm- and I'm not even sure what you meant by that!... I'm still grinning. Like an idiot.




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    :P well I'm pretty much doing the same! And you know what I meant
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    For the 6 marker, I would suggest to read the question 3-4 times. Then write the answer.
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    (Original post by chrishy2012)
    Does anyone have any idea what the big six marker question could be on?! The one last year was good, hope we have one like that or something on the photoelectric effect
    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.
 
 
 
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