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    This question is in the sample paper for edexcel unit 4 physics:

    (a) A 2200 μF capacitor is charged to a potential difference of 12 V and then discharged through an electric motor. The motor lifts a 50 g mass through a height of 24 cm.
    (b) The capacitor is charged to 12 V again and then discharged through a 16 ohm resistor.
    (i) Show that the time constant for this discharge is approximately 35 ms.

    done

    (ii) Sketch a graph of current against time for this discharge on the grid below. You should indicate the current at t = 0 and t = 35 ms.


    I can't seem to do this! I know the shape, but the values they give us in the m/s aren't the same as what I get, why?
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    What values do you get?
    Initial current at start of graph = V/R (V is initial pd on capacitor=12V)
    Current after 35ms is V/R where V is pd after 35ms. This V is 0.37 times the start value.
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    (Original post by Stonebridge)
    What values do you get?
    Initial current at start of graph = V/R (V is initial pd on capacitor=12V)
    Current after 35ms is V/R where V is pd after 35ms. This V is 0.37 times the start value.
    I get 750ms for the initial current. The markscheme uses 1.6V for some reason instead of 12V - any idea why?
    How do you know that the V is 0.37 times the start value?
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    I recognise this question. I spoke to my physics teacher about it, and it turns out the markscheme is wrong.

    V = IR, I = V/R which means 12V/16 = 0.75 A

    The initial value is 750mA and the current at the time constant is around 275mA. you know the graph shape
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    (Original post by DeyTa)
    I recognise this question. I spoke to my physics teacher about it, and it turns out the markscheme is wrong.

    V = IR, I = V/R which means 12V/16 = 0.75 A

    The initial value is 750mA and the current at the time constant is around 275mA. you know the graph shape
    That would make so much sense!
    Thank you!

    Although I don't get what you mean by the time constant being around 275mA? don't you mean that the current at 35ms is around 275mA?
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    (Original post by unamed)
    How do you know that the V is 0.37 times the start value?
    Because the graph is a reverse exponential graph, and according to the equation It = Io x e^(- time/time constant), at the time where t = time constant, it becomes It = Io x e^-1, which is the same as multiplying by 1/e, which is approximately 0.368.

    Io is the initial current, which we now know to be 750mA,

    So current at the time constant = 750 x 1/e = 750 x 0.368 = 276mA

    I'll see if i can upload a graph
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    (Original post by DeyTa)
    Because the graph is a reverse exponential graph, and according to the equation It = Io x e^(- time/time constant), at the time where t = time constant, it becomes It = Io x e^-1, which is the same as multiplying by 1/e, which is approximately 0.368.

    Io is the initial current, which we now know to be 750mA,

    So current at the time constant = 750 x 1/e = 750 x 0.368 = 276mA

    I'll see if i can upload a graph
    Oh, I understand now! thank you.
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    just in case :P as you can see my MS Paint skills are exquisite

    attachment below (i hope)
    Attached Images
     
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    (Original post by DeyTa)
    just in case :P as you can see my MS Paint skills are exquisite

    attachment below (i hope)
    That's actually I good graph - considering your software! :p:
    Are you ready for the exam tomorrow? :eek:
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    (Original post by unamed)
    Are you ready for the exam tomorrow? :eek:
    I guess I'm pretty ready, although I still get a bit confused about electromagnetic induction sometimes As it turns out, the explanation for electromagnetic induction is just a model! So we might actually be learning wrong stuff hehe

    good luck!
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    (Original post by DeyTa)
    I guess I'm pretty ready, although I still get a bit confused about electromagnetic induction sometimes As it turns out, the explanation for electromagnetic induction is just a model! So we might actually be learning wrong stuff hehe

    good luck!
    I hope electromagnetism doesn't come up at all - I hate it with a passion. I'd be much happier with particle physics - after all, for all we know, it might be wrong! :p:

    you too!
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    It's the same as charging mate, for current anyway. (exponential decline)
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    Hi guys tomorrow i am having both phy4 and phy5
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    for this type of question, would you estimate the value of I when time is 35ms using the graph or do an actual working and multiply 750ms by 1/e?
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    (Original post by Kameo)
    for this type of question, would you estimate the value of I when time is 35ms using the graph or do an actual working and multiply 750ms by 1/e?
    well, considering we need the value to make the graph - you work it out. If they give us the graph, then it makes sense to use the graph.
 
 
 
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