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    here's the ms and the question, so i need to know how to draw the graph for the last part, i.e. shape and axis intercepts
    thanks
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    (Original post by gorilla_baby)
    here's the ms and the question, so i need to know how to draw the graph for the last part, i.e. shape and axis intercepts
    thanks
    I don't see them...:confused:
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    oops, sorry
    here they are, in the same doc
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  1. File Type: doc A circuit is set up as shown in the diagram.doc (36.0 KB, 111 views)
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    (Original post by Muppety_Kid)
    I don't see them...:confused:
    i posted them above
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    (Original post by gorilla_baby)
    here's the ms and the question, so i need to know how to draw the graph for the last part, i.e. shape and axis intercepts
    thanks
    I think you should probably post this in the physics section here

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    You need to find the "half life" of the voltage , i.e the time taken for the voltage to halve , and use the formula t(1/2) = ln2.T to find time constant T.

    The second part is straightforward but you have to first find the resistance given the initial amperage using Ohm's Law.
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    ^^ For that part, can't you find T by taking 1/e of original value? i.e. (9V * 1/e) = time constant, cause it is an exponential graph. The NAS book does that for I = Io e^ (-t / RC) ... but since V is directly proportional to I, we can take time constant using a V -t graph too, right? I think the MS did include it.

    My method is basically as I described above - find (1/e) * 9V, find the time taken for the voltage to fall from 9V to that value. That's the time constant.

    Next part : When Io = 0.19 * 10^-3 A. Io = (Vbattery)/Resistance, so you can find the resistance. And since t = RC (time constant = resistance * capacitance), C = t/R.

    3rd part : As the potential difference across the resistor is decreasing, the potential difference across the capacitor is increasing. So its increasing but in a curve-way and it gets less steep (as the MS says). Not sure why it has to end at 7.5V / 15 s though!
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    (Original post by rohitkhannak)

    3rd part : As the potential difference across the resistor is decreasing, the potential difference across the capacitor is increasing. So its increasing but in a curve-way and it gets less steep (as the MS says). Not sure why it has to end at 7.5V / 15 s though!
    yea, that's what i thought, and i also dont get the 7.5V/15s thing??!!
 
 
 

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