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    (Original post by tanyapotter)
    Do you remember what you got for the very last transformer question, about why high voltages are needed? It's definitely so that current is lower, right?
    No, why would you want current to be lower? That increases power loss due to resistance. I put because industry often needs high voltages (B I think?).
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    (Original post by tanyapotter)
    Do people remember which way the coil is turned for max/min flux?

    I put the very bottom box: pi/3 anticlockwise and pi/6 clockwise?
    Yes, I put that!

    How did you find the paper overall? I thought it had some easier bits and some harder bits.
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    Did people get C for the one about 48V and the variable capacitor? I think C was 96V and charge value that was in all the other rows except for one?
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    My MC

    D B B D D C B A B C B B B A B A B A C A A D C C A

    Can someone please check these if they were confident on their answers.
    I know that the last two are definitely wrong.
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    (Original post by tanyapotter)
    Do people remember which way the coil is turned for max/min flux?

    I put the very bottom box: pi/3 anticlockwise and pi/6 clockwise?
    Yeah I got that one
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    (Original post by Cadherin)
    No, why would you want current to be lower? That increases power loss due to resistance. I put because industry often needs high voltages (B I think?).
    Current causes heating. Decreased current decreases heating power loss
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    (Original post by jess.olo)
    this is probs what I got or lower I'm hoping for 50-58 tbh. I have no idea why this would be !? anyone ??

    Will be around 51 for A and 57 for A*
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    (Original post by C0balt)
    Current causes heating. Decreased current decreases heating power loss
    So did you put D then?
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    (Original post by Cadherin)
    So did you put D then?
    D is correct
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    (Original post by Cadherin)
    So did you put D then?
    Yes
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    (Original post by Cadherin)
    Yes, I put that!

    How did you find the paper overall? I thought it had some easier bits and some harder bits.
    I've dropped loads of silly mistake marks so I'm not too thrilled, but I'm glad my performance wasn't further excasterbated by a difficult paper. It was decent.

    Need an A for university so hopefully I've done enough!!
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    (Original post by Jay1421)
    If I recall correctly the answer was 3.6*10^7, I don't think it was A, sorry
    I can't remember my figures but did you remember to subtract the earth radius? Because if so then i got the same
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    (Original post by notanooblol)
    D is correct
    Damn... Yet another silly mistake I made. Argh, think I dropped 10 sodding marks... :/
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    (Original post by tanyapotter)
    I've dropped loads of silly mistake marks so I'm not too thrilled, but I'm glad my performance wasn't further excasterbated by a difficult paper. It was decent.

    Need an A for university so hopefully I've done enough!!
    Well, I hope you get it. I need an A*... Which I doubt I'll get after my terrible ISA performance and the silly mistakes I made on this. Oh well...
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    displacement and acceleration 180 out of phase for q5 multiple choice?
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    (Original post by particlestudent)
    A field line and equipotential isn't the same. The charge was moving away from the point charge I think
    Ah yes I remember now, potential was increasing. as you move away from the charge, potential becomes less negative as max value is 0
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    (Original post by Cadherin)
    Damn... Yet another silly mistake I made. Argh, think I dropped 10 sodding marks... :/

    Losing 10 marks should be 120 ums, as long as you haven't lost too many more, then dont worry, I think I've lost about the same
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    (Original post by lucabrasi98)
    I can't remember my figures but did you remember to subtract the earth radius? Because if so then i got the same
    Yeah, the final answer using the total radius was 4.2*10^7, and it gave Earth's radius as 6.4*10^6 if I recall correctly
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    (Original post by Cadherin)
    No, why would you want current to be lower? That increases power loss due to resistance. I put because industry often needs high voltages (B I think?).
    Current lower = less power loss in the cables as P (Power Loss) = Current^2 x Resistance
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    (Original post by notanooblol)
    Did people get C for the one about 48V and the variable capacitor? I think C was 96V and charge value that was in all the other rows except for one?
    yeah I got that
 
 
 
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