Average power in a DC source

Discussion in 'Homework Help' started by notoriusjt2, Sep 17, 2010.

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    [​IMG]

    the formula in the book states
    Pdc = Vdc *Iavg

    since DC voltage is already stated i just need to find Iavg

    so how the heck do i do that? the book only gives me formulas for average power, and if i base my equation of their equation it will like like this

    [​IMG]
    now that just doesnt look right. what am i doing wrong? and yea i accidentally put that negative sign in front of the 5dt
     
  2. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    You really need to brush up on calculus.

    \int_{0}^{T} A dt = AT

    Your equation is wrong, you need to divide by the period.

    I_{avg} = \frac{1}{T}\int_{t_0}^{t_0 + T} i(t) dt
     
  3. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    so that equation with a 1/.02 out in front of it would be correct?
     
  4. Ghar

    Active Member

    Mar 8, 2010
    655
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    Almost, you need 1/20 since you kept everything in milliseconds :p

    Also, I'm assuming this is a periodic waveform, where it repeats every 20 ms.
    If not the average will be different.
    If it doesn't repeat at all the average is 0, since T goes to infinity.
     
  5. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    good call on the milliseconds. ;)

    and your right i need to brush up on this calculus stuff. so with this integral now. just give me a basic idea how to calculate it. Its been so long since I have attempted one
     
  6. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    Ok, well the most important point is that integrals are linear...

    \int_{0}^{T}x(t)dt = \int_{0}^{T_a}x(t)dt + \int_{T_a}^{T}x(t)dt

    This is how you can break up the equation for average into what you wrote in the first post.

    \frac{1}{T}\int_{0}^{T}i(t)dt = \frac{1}{20}[\int_{0}^{6}i(t)dt + \int_{6}^{10}i(t)dt + \int_{10}^{20}i(t)dt]

    Just plug stuff in... I gave the integral identity you need to use in my first reply.
    You should really figure out how to do integrals though, that's the easiest one you can do.
     
  7. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    [​IMG]

    ok i get how to set up the equation. i would substitute the current values into i(t) dt. but im having trouble actually doing the integrating. i can probably plug this into some website and have it computed but i want to learn how to do it by hand.

    so how do i actually do the integration
     
  8. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    The integral is most easily done by the power rule:

    \int x^n dx = \frac{1}{n + 1} x^{n+1}

    And using the concept of linearity again:

    \int Af(x) dx = A \int f(x)dx

    And of course;

    \int_{a}^{b}f(x)dx = F(b) - F(a)

    Where F(x) is the integral of f(x)

    So, combine that stuff together:

    \int_{0}^{6}7dt = 7 \int_{0}^{6}dt

    In this case, n = 0, since t^0 = 1

    \int dt = \frac{1}{0 + 1} t^{0 + 1} = t

    Use that fact:

    7 \int_{0}^{6}dt = 7(6 - 0) = 42 A\cdot ms

    Keep doing that... sum it... and then divide by 20ms.

    You'll get:
    \frac{1}{20}(42 - 20 + 40) = 3.1 A

    The alternative is you forget about the integral and realize what it is:
    area under the curve.

    It's 3 rectangles.

    7A for 6ms, -5A for 4ms, and 4A for 10ms.

    Sum your rectangles... and the average will be that sum divided by the total time, 20 ms.
     
  9. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    Thank you this really helped out a lot
     
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