Power Electronics Help. 2 Similar Problems

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

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    #1.)[​IMG]

    #2.)The voltage across a 10-ohm resistor is v(t) = 170 sin(377t) V. Determine the average power absorbed by the resistor.


    here is what I have so far.
    #1.) R=10ohms, V(t)=170sin(377t)v... therefore use ohms law to find I(t)=17sin(377t)
    Pr=i(t)*v(t) = (17sin(377t))*(170sin(377t)) = 2890sin^2(377t)
    so that would be answer F

    #2.) not quite as easy. my book has this equation for average power...

    [​IMG]
    where is T defined in this problem? is it 377? so would it be something like this?

    [​IMG]
    if so how is that even computed? thanks
     
  2. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    I'm giving you your options, worst choice first:

    1) You could use a table of integrals to get the integral of sin^2.

    2) You can use sin(x)^2 = 1/2 - 1/2 cos(x) and then do the integral.

    3) You can convert to rms voltage. For a v(t) = Vpeak sin(wt), you have:
    Vrms = Vpeak / Sqrt(2)

    RMS calculations are like DC; power is simply Vrms^2 / R

    4) You can accept the fact that average power for a pure sinusoid v(t) = Vpeak sin(wt) across a resistor is;

    P = Vpeak^2 / (2R)

    Options 3) and 4) are essentially the same thing.

    You only need to delve into the integral equation if the waveform is something more complicated than a sinusoid(s).
     
  3. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    ok so Vrms = Vpeak / Sqrt(2)
    170/Sqrt(2) = 120.208 = Vrms

    so now power = Vrms^2 / R
    P = 1445w

    thanks
     
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