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what was the exact question for q1b phy 5 - capacitors watch

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    hi does anyone remember the question of the capacitor

    was it to calculate the charge across the 4µF capacitor?
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    (Original post by andyj72)
    hi does anyone remember the question of the capacitor

    was it to calculate the charge across the 4µF capacitor?
    Yep
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    48 micro C??
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    ...i did the energy stored......just luvly!!
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    woopsie!!
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    yeah

    i got that wrong, i did 24 x the capacitance, forgot to divide the voltage by 2, d'oh.
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    oh s**** did u have 2 divide it by 2????? i just multiplied voltage by capacitance Q=CV...
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    (Original post by Steve8052)
    oh s**** did u have 2 divide it by 2????? i just multiplied voltage by capacitance Q=CV...
    i done 12V * 4microF

    only 1 mark, not a big prob if u got that wrong
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    I just said 12V

    Anything electrical I always get wrong X(
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    well i jst did 24*2 yh 1 mrk but since the paper was short (wat was total mrks again) 1 mrk =quite large %
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    The paper is out of 40 I believe, but losing one mark isn't that bad!
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    (Original post by roochis)
    The paper is out of 40 I believe, but losing one mark isn't that bad!

    phew, i did it right then. thats good.

    it all adds up roochis, it can be the difference between an A and a B
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    (Original post by Steve8052)
    oh s**** did u have 2 divide it by 2????? i just multiplied voltage by capacitance Q=CV...
    yep u had to divide by 2 cos according to Kirchoffs 2nd law the voltages will be split in proportion to their resistances(around a closed loop the emf = sum of p.ds). the resisntances being the capacitors. and since they were the same capacitance u divide the voltage by 2 .

    i just stated it to just in case.

    u might get pity marks :hmmmm: :stupido:
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    i also never devided by 2...but i reckon i smacked the rest of the paper
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    (Original post by andyj72)
    yep u had to divide by 2 cos according to Kirchoffs 2nd law the voltages will be split in proportion to their resistances(around a closed loop the emf = sum of p.ds). the resisntances being the capacitors. and since they were the same capacitance u divide the voltage by 2 .

    i just stated it to just in case.

    u might get pity marks :hmmmm: :stupido:
    Lets settle this!

    If you used the value of total capacitance i.e.2µF then you can do Q=CV and you get 48µC - you dont split the voltage as you consider the whole circuit, and charge is split!

    Now if you considered the 4µF capacitor in series, individually, then tis correct to say voltage is split, and 12V passed through it; in that case you'd get Q=CV -> 12x4µF=48µC also

    so the answer is 48µC in both cases:cool: - even conferred with Hash, and he's going to oxford :rolleyes: hehe

    Pk
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    (Original post by Phil23)
    Lets settle this!

    If you used the value of total capacitance i.e.2µF then you can do Q=CV and you get 48µC - you dont split the voltage as you consider the whole circuit, and charge is split!

    Now if you considered the 4µF capacitor in series, individually, then tis correct to say voltage is split, and 12V passed through it; in that case you'd get Q=CV -> 12x4µF=48µC also

    so the answer is 48µC in both cases:cool: - even conferred with Hash, and he's going to oxford :rolleyes: hehe

    Pk
    You're going to oxford too :rolleyes: :p:
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    (Original post by sonja)
    You're going to oxford too :rolleyes: :p:
    that makes the two of us then :cool: hehe - everyone satisfied that its 48µC now??? :rolleyes:

    pk
 
 
 
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