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    Hi i can't find the mark scheme for these questions anywhere and have no idea whether or not I have answered them correctly so could anyone experienced in the physics field let me know if they are correct? Thanks

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    Btw I know part (iv) is wrong so any help will be appreciated


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    (Original post by jessyjellytot14)
    Hi i can't find the mark scheme for these questions anywhere and have no idea whether or not I have answered them correctly so could anyone experienced in the physics field let me know if they are correct? Thanks

    Name:  ImageUploadedByStudent Room1459172425.272063.jpg
Views: 65
Size:  115.5 KB

    Btw I know part (iv) is wrong so any help will be appreciated


    Posted from TSR Mobile
    You are on the right track with part iv. Just follow it through and make RT the subject.

    Part b) is incorrect.

    As the temperature increases, the thermistor resistance decreases.

    Which means the parallel combination resistance will also decrease.

    Since the 540 ohms series resistor remains constant, the total circuit resistance also falls and the circuit current increases causing the pd across the 540 ohms resistor to rise, whilst the pd across the parallel combination must fall.
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    (Original post by jessyjellytot14)
    Hi i can't find the mark scheme for these questions anywhere and have no idea whether or not I have answered them correctly so could anyone experienced in the physics field let me know if they are correct? Thanks

    Name:  ImageUploadedByStudent Room1459172425.272063.jpg
Views: 65
Size:  115.5 KB

    Btw I know part (iv) is wrong so any help will be appreciated


    Posted from TSR Mobile
    Part (iv)

    First, find the total resistance in the circuit using the EMF and current:
    V = IR
    V / I = R
    R = 15/(10x10^-3)
    R = 1500 Ohms

    Now, you can form an equation with the unknown resistance:
    (1/1200 + 1/X)^-1 + 540 = 1500

    Rearrange it:
    (1/1200 + 1/X)^-1= 960

    Get rid of the power of -1 on the left:
    (1/1200 + 1/X)= 1/960

    Rearrange now to get the unknown on its own:
    1/X = 1/960 - 1/1200
    1/X = 1/4800

    To get from 1/X to X just times by the power of -1:
    (1/X)^-1 = (1/4800)^-1
    X = 4800

    So the thermistor has a resistance of 4800 Ohms, to check just substitute this value into the original resistance calculation and you should get 1500 Ohms.

    My method is a little confusing, but I hope it helps
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    (Original post by derpz)
    Part (iv)

    First, find the total resistance in the circuit using the EMF and current:
    V = IR
    V / I = R
    R = 15/(10x10^-3)
    R = 1500 Ohms

    Now, you can form an equation with the unknown resistance:
    (1/1200 + 1/X)^-1 + 540 = 1500

    Rearrange it:
    (1/1200 + 1/X)^-1= 960

    Get rid of the power of -1 on the left:
    (1/1200 + 1/X)= 1/960

    Rearrange now to get the unknown on its own:
    1/X = 1/960 - 1/1200
    1/X = 1/4800

    To get from 1/X to X just times by the power of -1:
    (1/X)^-1 = (1/4800)^-1
    X = 4800

    So the thermistor has a resistance of 4800 Ohms, to check just substitute this value into the original resistance calculation and you should get 1500 Ohms.

    My method is a little confusing, but I hope it helps
    Thanks for your reply.

    It's worth remembering that the best learning is when the student reaches the answer for themselves which is why it's best to guide rather than give out full solutions which can then be copied verbatim.

    We had a saying at uni' when some lecturers wrote solutions onto the whiteboard: it's the most efficient method for getting what is in the lecturers handwriting into our own handwriting without anything passing through either brain!
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    (Original post by uberteknik)
    Thanks for your reply.

    It's worth remembering that the best learning is when the student reaches the answer for themselves which is why it's best to guide rather than give out full solutions which can then be copied verbatim.

    We had a saying at uni' when some lecturers wrote solutions onto the whiteboard: it's the most efficient method for getting what is in the lecturers handwriting into our own handwriting without anything passing through either brain!

    Just thought I would show each step so they can see how to do it, since they got the first bit right. But you have a good point and I will bear that in mind
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    uberteknik derpz Thank you for your help! I think it was just the re-arranging part I got stuck on but it makes sense now
 
 
 
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