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    Hello,

    How did the mark scheme arrive to this solution?

    24) A lamp rated at 12 V 60 W is connected to the secondary coil of a step-down transformer and is at full brightness. The primary coil is connected to a supply of 230 V. The transformer is 75% efficient.
    What is the current in the primary coil?

    Answer: A 0.25A

    Thank you!

    -- My working got the answer B for some reason so it must be wrong?

    ( (60/12)x12 ) / (75%x230) = 0.35A
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    You might be looking at the wrong markscheme - that question is in the 2015 paper
    http://filestore.aqa.org.uk/subjects...1-QP-JUN15.PDF
    fwiw
    power input = (4x60)/3
    =80 W

    80/230 = 0.347 A
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    (Original post by Joinedup)
    You might be looking at the wrong markscheme - that question is in the 2015 paper
    http://filestore.aqa.org.uk/subjects...1-QP-JUN15.PDF
    fwiw
    power input = (4x60)/3
    =80 W

    80/230 = 0.347 A
    You're right I was looking at the 2014 one but doing the 2015 paper! Thanks a tonne for the solution as well !
 
 
 
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