#1.)

#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

 

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).

 

ok so Vrms = Vpeak / Sqrt(2)
170/Sqrt(2) = 120.208 = Vrms

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

thanks