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    (Original post by Xiomara)
    Really? Because I thought I got like 7xxxxx something, but I can't even remember :L Here's praying I got it right!
    The working was (V/R=I)0.75/5000= 1.5*10-4A for the curcuit current.
    The voltage supply was 6V so total resistance is (V/I=R)6/10*10-4=40000ohms.
    so total resistance is 40000ohms, take away the two 5000ohm resistors and you get 30000ohms for the variable resistor.
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    (Original post by ac3xx)
    I didn't think it was especially difficult, but not the easiest. I think it was more that it was a "different" paper compared to most of the past papers, slightly different style questions. I agree with the "midway" comment though.
    Spot on. They're changing the way they ask questions as opposed to increasing difficulty.
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    (Original post by Khenir)
    I had 16 since:

    8 vertical divisions,
    Peak to Peak = 128V
    so (128/8) = Y-sensitivity = 16.
    Would this not mean the waveform is touching the very top of the screen, is this ideal or? I think the mark scheme may allow anything sensible.
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    (Original post by Semanresu)

    Q6
    6.3V
    5.7V across 2 ohm resistor
    2.85A across 2 ohm resistor
    1.35A across unknown resistor
    4.2 ohm resistance
    2.86 ohms total resistance
    power: Internal resistor=26.5W, 2 ohm resistor=16.2W, other resistor=7.7W
    energy conserved: battery power-total power dissipated=0
    50.4-26.5-7.7-16.2=0
    shouldn't the voltage have been split between the resistors? making it 2.85V across the 2ohm resistor?
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    (Original post by benroxo)
    Would this not mean the waveform is touching the very top of the screen, is this ideal or? I think the mark scheme may allow anything sensible.
    it's probably the very bottom limit of what the mark scheme will probably allow but it's the most accurate I felt I could be.
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    (Original post by Khenir)
    shouldn't the voltage have been split between the resistors? making it 2.85V across the 2ohm resistor?
    I think the resistors were in parallel, so their pds were the same.
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    (Original post by Khenir)
    I had 16 since:

    8 vertical divisions,
    Peak to Peak = 128V
    so (128/8) = Y-sensitivity = 16.
    Put Y-sensitivity at 8, wasn't considering the negative direction.
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    (Original post by Semanresu)
    I think the resistors were in parallel, so their pds were the same.

    Well there goes my marks for that question.
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    (Original post by benroxo)
    The working was (V/R=I)0.75/5000= 1.5*10-4A for the curcuit current.
    The voltage supply was 6V so total resistance is (V/I=R)6/10*10-4=40000ohms.
    so total resistance is 40000ohms, take away the two 5000ohm resistors and you get 30000ohms for the variable resistor.
    Is there anyway you can work it out using the method we did? 6-1.5v....
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    (Original post by mahatma ghandi)
    Did you get these answers on the last page? you seem to know what you were doing.

    1.03x10^-4 A
    V=0.51
    Voltmeter reading will increase
    30,000 ohms. ?
    For the current it was 6.67x10^-5
    The voltage on the meter was 0.33V
    Last 2 are correct.
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    It looks like I got the electricity questions correct, however I left the answers for question 6 to one decimal place as the question left the numbers to one decimal place. Do you reckon I would get the marks still? Was I wrong to leave them at one decimal place?
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    (Original post by GeneralOJB)
    For the current it was 6.67x10^-5
    The voltage on the meter was 0.33V
    Last 2 are correct.
    for current, wouldn't it be 6.7x10^-5A since you were given values to 2sf? 6.7 is rounded to 2sf or it does not matter?
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    (Original post by ccashman)
    It looks like I got the electricity questions correct, however I left the answers for question 6 to one decimal place as the question left the numbers to one decimal place. Do you reckon I would get the marks still? Was I wrong to leave them at one decimal place?
    What value did you write for q6?
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    for Q.7 did anyone else get 30Kohms? well the last bit on questions!
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    (Original post by Xiomara)
    Is there anyway you can work it out using the method we did? 6-1.5v....
    I really don't know, sorry. Where is the 1.5V from?
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    (Original post by Xiomara)
    Is there anyway you can work it out using the method we did? 6-1.5v....
    My method was this:

    0.75 = 5000*I
    so I = (0.75/5000)

    I didn't think properly and had this:

    5.25 = I*(5000 + R)

    and then I realised that I*5000 was 0.75 (since I'd done it earlier DUH).

    so 4.5 = IR = (1.5*10^-4)*R

    so R= (4.5)/(1.5*10^-4) = 30,000.
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    (Original post by benroxo)
    I really don't know, sorry. Where is the 1.5V from?
    One of the resistors was know to have a voltage of 0.75 through it and 5000 ohms of resistance.

    The LDR was also known to have 5000ohms of resistance through it which means that it had the same voltage (0.75)

    so the 1.5v is simply the sum of those two.
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    (Original post by Khenir)
    One of the resistors was know to have a voltage of 0.75 through it and 5000 ohms of resistance.

    The LDR was also known to have 5000ohms of resistance through it which means that it had the same voltage (0.75)

    so the 1.5v is simply the sum of those two.
    ahh so you could either do 6V-1.5V/1.5*10-4 or 6/1.5*10-4 - (2*5000), surely both will get the marks?
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    (Original post by StalkeR47)
    for current, wouldn't it be 6.7x10^-5A since you were given values to 2sf? 6.7 is rounded to 2sf or it does not matter?
    Doesn't matter as they did not ask you to give your answer to the appropriate number of significant figures. Man you guys stress out about sig figs too much. I put em to whatever I feel like unless it tells me I'm being assessed on it. I really don't see the significance of them anyway, you're not losing accuracy by putting them to more than you are given. /rant
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    Im soo happy i absolutely destroyed this paper but i got the last last question wrong and i think that I messed up on a few theory questions but i reckon I got maybe around 62-64 for this paper


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