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Jan 03 phy6 question help? watch

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    (Original post by S1M)
    hmmm i see your point... maybe its got to do with the resistances before it is charged or whilst it is charging... that resistance would be higher which will screw up the results through heat loss... iirc there is a resistor there too... wouldn't exposing it to high voltages heat it up resulting in energy loss whilst the capacitor is charging?
    it is charged without resistance
    but discharged through the FINE insulated wire which I think has resistance
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    (Original post by keisiuho)
    it is charged without resistance
    but discharged through the FINE insulated wire which I think has resistance
    well erm..... clutching at straws here... wouldn't the rest of the circuits (erm... the other wires) lose heat energy too?
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    Official markscheme says for one mark each:

    1) Quality of written communication
    2) Wc is very small
    3) Even at 50V...it is only 6J
    4) Any change in T is difficult to measure so something like a thermocouple is needed
    5) Wire might melt/fuse
    6) Heat/energy loss to surroundings/air....[not to connecting wires]

    Hope this helps

    G
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    so it is supposed to adopt a caloric measurement to determine the energy dissipated
 
 
 
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