Results 1 to 12 of 12

Math Help - find charge, power, energy

  1. #1
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580

    Cool find charge, power, energy

    The current entering the positive terminal of a device is i(t) = 3e^{-2t}[A]
    and the voltage across the device is v(t) = 5\frac{di}{dt} [V]

    (a) Find the charge delivered to the device between t=0 and t=2s

    I would presume we use the intergal

    Q \triangleq\int_{t_0}^t{idt}

    but didn't know how to use the v(t) = 5\frac{di}{dt} V with this

    the other related question are. (b) Calculate the power absorbed and (c) Determin the energy absorbed in 3s.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Your idea with (a) is correct. Just carry that out. You don't need v(t) for part (a), but you do for part (b), and possibly (c). What do you get?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580

    Cool books ans was 1.297C

     <br />
Q = \int_{t=0}^2{3e^{-2t}}dt = 1.47253C<br />

    well somewhere theres something
    the books ans was 1.297C
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Hmm. Yeah, I get

    Q=\dfrac{3}{2}\left(1-e^{-4}\right)\approx 1.47253.

    I think the book is wrong on that one. Does the book have an analytical expression, or just the approximation?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580

    the same as you suggested

    the answer for that question in book is given as
    (a) 1.297C, (b) -90e^{-4t}, (c)-22.5J

    i will see if I can do (b) and (c)

    the example they had in the book was pretty much the same as you suggested.
    so maybe a typo in the answers.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    I agree with (b)'s answer. How is the book defining "energy absorbed"? Is it just the integral of the power over the desired time interval?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580

    Cool (b) calculate the power absorbed

    their definition of power is...."the rate of expending or absorbing energy, measured in Watts (W)"

    anyway this is how I did (b) assuming yours also...

    Since P=vi and

    v=5\frac{di}{dt} = -30e^{-2t}[V]

    i=3e^{-2t}[A]

    Then

    P=(-30e^{-2t})(3e^{-2t})=-90e^{-4t}[W]

    however in (c) "Determine the energy absorbed in 3s" I thot the setting  t=3 in -90e^{-4t} would give  P in Joules  [J]
    but

    -90e^{-12} =-0.0005529 [J] isn't -22.5 [J] which is the ans
    Last edited by bigwave; December 29th 2010 at 08:26 PM. Reason: added (c)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    I agree with your (b) procedure. For (c), I think you'll find that my hint in post # 6 will do the trick. Integrate from t = 0 to t = 3.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580

    Cool =-22.5[j]

     <br />
\int_{t=0}^3(-90e^{-4t})<br />
\Rightarrow<br />
\left(\frac{45}{2}\left(\frac{1}{e^{12}}-1\right)\right)<br />
\Rightarrow<br />
-22.49986<br />
\Rightarrow -22.5[J]<br />

    i should know but why was it necessary to take the \int_{t=0}^3
    Last edited by bigwave; December 30th 2010 at 11:56 AM. Reason: latex
    Follow Math Help Forum on Facebook and Google+

  10. #10
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Your computations look correct to me.

    Power is equal to the time derivative of the work. Work has the same units as energy, and since there isn't any other energy floating around that I'm aware of, I postulated that the work must be equal to the integral of the power, by the fundamental theorem of the calculus. So that's why I thought the integral was the right thing to do. In addition, there was a clue: the "energy absorbed in 3 seconds". Whenever anyone uses a phrase like that, my mind automatically goes towards an integral, since the rate at which the energy is absorbing might vary over the 3 seconds. In that case, you must do an integral.

    Make sense in a sort of hand-waiving manner? Maybe someone else here on the MHF could produce a more rigorous proof.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580
    your answer is rigorous enough for me...
    sure apprecaite the help .... looks like I am getting the picture on this slowy..
    Follow Math Help Forum on Facebook and Google+

  12. #12
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Slow in, slow out, I always say. Keep them coming!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find power and energy delivered
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: December 28th 2010, 12:14 PM
  2. how much charge and energy
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: December 26th 2010, 11:33 PM
  3. Power and Energy
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: October 30th 2009, 01:38 PM
  4. work , energy and power
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: October 2nd 2007, 12:10 PM
  5. Power and energy
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: September 4th 2007, 01:06 PM

Search Tags


/mathhelpforum @mathhelpforum