Book answer wrong (quick answer check)

Discussion in 'Math' started by ihaveaquestion, Jul 1, 2009.

  1. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
    314
    0
    Hi everyone,

    Rather simple question:

    The current entering the positive terminal of a device is i(t) = 3e^(-2t) A and the voltage across the device is v(t) = 5di/dt V.

    Find the charge delivered to the device between t = 0 and t = 2 s.

    since i = dq/dt, q = integral of i with respect to t

    I'll have my integral sign be [

    So q = [ i dt = [ 3e^(-2t) dt = (-3/2)e^(-2t) evaluated from 0 to 2:

    -3/2e^(-4) - (-3/2) = 1.47 C

    Answer in the back of the book = 1.297 C

    Who's wrong?

    Thanks in advance.
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,136
    1,786
    I think your answer is correct. If you differentiate the result of the integration; do you get back the original integrand? Check! There is no problem with the numerical evaluation.

    FWIW the book's answer just happens to be what you get if you compute
    Code ( (Unknown Language)):
    1.  
    2. -3/2 * [ e^(-2) - 1] = 1.297 Coulombs
    3.  
    This would also be correct if t was equal to 1 in the original problem.
    Have a bit more confidence next time. I think you've earned it.
     
  3. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    ihaveaquestion,

    I get the same answer as you do.

    Ratch
     
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