# Finding the instantaneous power

1. Feb 2, 2014

### theBEAST

1. The problem statement, all variables and given/known data

2. Relevant equations
P = IV

3. The attempt at a solution
The solution is there and they did it by multiplying everything in the time domain.

However, I decided to try to multiply the phasor forms to get the power but that did not work...

V*I = sqrt(2)cos(377t) * sqrt(2)cos(377t+pi)
= (sqrt2<0)(sqrt2<180)
= 2<180
= 2cos(377t+pi)

What is wrong with this?

2. Feb 2, 2014

### vanceEE

We want to simplify I first..
$√(2)cos(337t + \pi) \equiv -√(2)cos(337t)$
Therefore, $I = -√(2)cos(337t)$
$P = IV = (-√(2)cos(337t))(√(2)cos(337t)) = -2cos^2(337t)$
By the double-angle formulas:
$cos 2u \equiv 2cos^2u - 1$
$-cos 2u - 1 \equiv -2cos^2u$

$$-2cos^2(337t) \equiv -cos(754t) - 1$$
$$-cos(754t) - 1 \equiv cos(754t + \pi) -1$$

The key to this problem is recognizing that $$√(2)cos(337t + \pi) \equiv -√(2)cos(337t)$$
Leaving your answer as $-2cos^2(337t)$would be fine since $-2cos^2(337t) \equiv -1+1cos(754t + \pi).$

Last edited: Feb 2, 2014