P is the average value of the instantaneous power R*i^2 on [0,T]. Compute the power for R=2,5, i=10*sin(377t), T=2pi/377

To compute the instantaneous power, we'll ahve to
determine the definite integral of the function Ri^2, from t = 0 to t =
T.

P = (1/T) Int
(Ri^2)dt

We'll substitute the values for R, T and
i.

P = (377/2pi)*Int
2.5*100*[sin(377t)]^2dt

P = 377*250/2pi)*Int
[sin(377t)]^2dt

We'll have to use the
formula:

(sin x)^2 = [1 -
cos(x/2)]/2

We'll integrate both
sides:

Int [sin(377t)]^2dt = Int [1 -
cos(377t/2)]dt/2

We'll use the additive property of
integral:

Int [1 - cos(377t/2)]dt2 = Int dt/2 - Int
cos(377t/2)dt/2

Int dt/2 = (1/2)/Int
dt

Int dx/2 = (t)/2 + C
(1)

Int cos(377t/2)dt/2 = (1/2)*Int
cos(377t/2)dt

(1/2)*Int cos(377t/2)dt = (1/2)*
sin(377t/2)/(377/2) + C

(1/2)*Int cos(377t/2)dt =
sin(377t/2)/377 + C (2)

Int [sin(377t)]^2dt =
(t)/2 + sin(377t/2)/377 + C