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Finding the total charge from the graph

  1. Oct 19, 2015 #1
    I hope this the right place to put my question


    I didn't understand how it became like this

    Thank you
  2. jcsd
  3. Oct 19, 2015 #2


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    A current of 1 A during one second means that 1 Coulomb has passed by.
    If the current is constant, all you have to do is multiply current and time to get charge.
    For example in the period 1-2 ms the current is constant, so the charge is 1 ms x 10 mA = 10 ##\mu##C
    Which is also the area in the graph under the red line from 1-2 ms.

    For 0-1 you have a non-constant current and the area under the red line is 10 x 1 /2 in mA x ms .


    Does this help ?
  4. Oct 19, 2015 #3
    Thank you very much
    I understood the first part
    But sorry I didn't get it the last part which is the 1/2 ,,,

    Thank you again
  5. Oct 19, 2015 #4


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    In the first msec the current increases from 0 to 10 mA. It isn't 10 mA all through this msec, so 10 x 1 would be too much by a factor of about 2 as you can see from geometry.
    In this case the average current during the first msec is 5 mA and the factor of 2 is exact.

    Could it be that you aren't all that familiar with integrals and that integrals represent an area ?

    Basically all that's done is that we add up all the little bits of charge = ##\ \ ## i x ##\Delta##t ##\ \ ## for the period 0 - 1 ms and do so for very small time steps. In the limit ##\Delta##t ##\downarrow ##0, this becomes the integral ##\int i \ dt##
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