Power Electronics Help. 2 Similar Problems

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

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010

    #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...

    where is T defined in this problem? is it 377? so would it be something like this?

    if so how is that even computed? thanks
  2. Ghar

    Active Member

    Mar 8, 2010
    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
    ok so Vrms = Vpeak / Sqrt(2)
    170/Sqrt(2) = 120.208 = Vrms

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