Question

make me understand this ​

Answer 1

We know that

For DC supply

power = Current *voltage

voltage = current *resistance

So

power = i^2R

Now For AC supply

suppose AC current ocilates Sine function with phase angle Omega and time period T

So

I = Im Sin(omega *t)

Now we know that

p = I^2R

So

power = I^2 RSin^2(omega *t) ----(1

Now let us understand

we know that

cos(2x ) = 1 - 2sin^2x

Sin^2x = (cos2x - 1)/2

On putting in equation -(1)

power = i^2R(1 - cos(2omega*t )/2

We also write

power = i^2R/2 - i^2Rcos(2omega *t)/2

Now let us understand

Since cos(2omega*t) is form half positive(+1) and half negative (-1) so it will 0

Now

power = i^2R/2------(2)

Now on equating equation (1) and (2) we get

I^2R/2 = i^2Rsin^2(omega *t)

sin^2(omega *t) = 1/2

Hope helps you

Answer 2

Explanation:

Refer to attachment....