Finding Power (Simple Question)

Discussion in 'Homework Help' started by Student01, Mar 4, 2010.

  1. Student01

    Thread Starter Active Member

    Apr 15, 2009
    35
    0
    Hi all,

    I'm stuck on the following two questions:


    1)An eletrical element draws a current of i(t) = 10cos4t A and has a voltage of v(t) = 120cos4t V across it. Find energy absorbed by the element in 2s.

    2)The voltage across a device and the current through it are
    v(t) = 5cos2t V i(t) = 10 (1 - e^0.5t) A

    Calculate the power consumed by this device at t = 1 s.


    I know P = intregral( i(t) x v(t) dt) but I don't know how to integrate the product of two trigonometric functions, or the product of a trigonometric and exponential function. Am I going about this the wrong way?
     
  2. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    I don't know if integration is the correct method since I haven't started AC yet. It does sound logical though. I can tell you how to integrate the first part. If you get it in the form of cos^2(θ), you can use the half angle formula to create a relatively simple integral. You will end up using two substitutions, don't forget to update your limits of integration accordingly. Try it and you will see.
     
  3. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    The way the second part is worded, it sounds like they are looking for an instantaneous value. However, if they want from 0 to 1, it can be integrated, but it gets hairy. You end up using integration by parts twice and some creative algebra (as well as just slightly less than a tree's worth of paper).
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    In the first case you would need to integrate the instantaneous power function p(t)=v(t)*i(t) over the 2 second interval.

    I agree with Heavydoody that in the second case you only need to compute the instantaneous power at time t=1sec. No integration is required for that case.
     
  5. Eng_Bandar

    Active Member

    Feb 27, 2010
    50
    1
    Q1

    I think the problem in integration
    I will give the key and easy to complete it
    you can see Attachments

     
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    Last edited: Apr 15, 2010
  6. Student01

    Thread Starter Active Member

    Apr 15, 2009
    35
    0
    Ok, thanks for the help guys. So it turns out I just needed the right trigonometric identity to simplify the first problem, and was mis-interpreting the second question.
     
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