Log for current charging

Discussion in 'Homework Help' started by user123456, Mar 13, 2014.

  1. user123456

    Thread Starter New Member

    Jan 26, 2014
    12
    0
    Hello everyone,

    [​IMG]

    I have figured out the amount of current it takes, it functions like this:

    = 12/2(1 - e^-4/3.5)

    =6( 1 - e^-1.1428)

    4.0866 = 6 ( 0.6811)

    Cant figure out the seconds because i'm not sure how to find -t of logs:

    So setup the second equation we have

    5 = 12/2( 1 - e^-t/3.5)

    Thanks
     
  2. ericgibbs

    Senior Member

    Jan 29, 2010
    2,499
    380
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