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    nI was given a graph of q/t which was an exponential decay I think.
    The question said what was the current at
    (i) t=0s
    and
    (ii) t=15s


    for (i) I just put 0 as I have no idea how to start

    for (ii) I worked out time of half-life(20s) and then subbed into the equation given to get 1.41x10-5A
    the charge at 15s on the graph is 240x10-6 C not sure how to use it
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    (Original post by bruh2132)
    nI was given a graph of q/t which was an exponential decay I think.
    The question said what was the current at
    (i) t=0s
    and
    (ii) t=15s


    for (i) I just put 0 as I have no idea how to start

    for (ii) I worked out time of half-life(20s) and then subbed into the equation given to get 1.41x10-5A
    the charge at 15s on the graph is 240x10-6 C not sure how to use it
    The clue is in the question:

    q/t = instantaneous current. i.e. remember that current is defined as the rate at which charge flows.

    The current at any point on the curve is therefore given by the gradient of q/t.

    For i), draw a tangent to the curve at t=0 and continue the line, noting where it crosses the time axis. The current at t=0 is then simply:

    i_{(t=0)} = \frac{q_{(t=0)}}{t_{(q=0)}}

    Do the same for t=15 seconds.

    At t=o, the point at where the initial discharge gradient crosses the time axis, is known as the time-constant of the discharge as long as it's an exponential decay.
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    (Original post by uberteknik)
    The clue is in the question:

    q/t = instantaneous current. i.e. remember that current is defined as the rate at which charge flows.

    The current at any point on the curve is therefore given by the gradient of q/t.

    For i), draw a tangent to the curve at t=0 and continue the line, noting where it crosses the time axis. The current at t=0 is then simply:

    i_{(t=0)} = \frac{q_{(t=0)}}{t_{(q=0)}}

    Do the same for t=15 seconds.

    At t=o, the point at where the initial discharge gradient crosses the time axis, is known as the time-constant of the discharge as long as it's an exponential decay.
    Thanks a lot, I wasn't expecting to have to draw stuff, was expecting something more theoretical, thanks a lot though!
 
 
 
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