Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the total charge from the graph

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

    image.jpg

    I didn't understand how it became like this

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

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding the total charge from the graph
  1. Graph interpretation (Replies: 2)

Loading...