# 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

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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**

Let v(t) = Ri^2 = R*(10sin377t)^2. .

To find the average value P of v(t) on (0, T) where T = 2pi/377.

We know that average value of f(x) on (a,b) is F(t) = Int x*f(x)/ (b-a) dx from x = a to x = b.

Therefore F(t) = Int t *R ((10 sinkt)^2 /k)dt , on (0, 2pi/k) , k = 377.

F(t) = (100R/k) Int {t(1-sin2kt)/2}dt

F(t) = (100R/2k) { t^2/2- Int tsin2kt dt}.........(1)

Int x sin2kt dt = t (-cos2kt/2k) - Int { (t)' (-cos2kt/2k)}dt

Int tsin2kt = - (t*cos2kt)/2k +(1/4k^2)sin 2kt

Substituting the above result in (1) we get:

F(t) = (100R/2k){ (1/2)t^2 +(tcos2kt)/2k -(sin2kt)/4k^2}

F(2pi/k) - F(0) = (100R/2k) { (1/2)(2pi/k)^2 +(2pi/k)(cos2pi)/2k -sin(2pi)/4k^2}

F(2pi/k) - F(0) = (100R/2k) { 2pi^2/k^2 +pi/k^2}, where k = 377.

Therefore the average length = (50R/){2pi^2+pi}/(377^3)

Average length = 50(2pi^2+pi)R/377^3.