Hi Petrucciowns,
I don't think I can spend too much more time on this thread, so I'll go through a bunch of stuff, just to make sure we're on the same page. It could be that you already know all of this stuff, but I'll go through it anyway. In an AC circuit, quantities are time-varying in a sinusoidal way. For example, the source voltage might be described using the following functional form:
[tex]v(t) = V_0 \cos(\omega t + \phi)[/tex]
In this case, [itex]V_0[/itex] describes the amplitude of the oscillation, and [itex]\phi[/itex] describes the initial phase of the oscillation. For mathematical convenience (and this does afford a
great deal of convenience), we can describe the oscillating quantity using a complex number [itex]\tilde{V}[/itex] such that:
[tex]\tilde{V}(t) \equiv V_0 e^{i (\omega t + \phi)}[/tex]
where this complex quantity has been defined in such a way that its
real part is the measurable physical quantity of interest:
[tex]v(t) = Re[\tilde{V}(t)][/tex]
This last relation just comes from the definition of a complex exponential. We can rewrite the complex quantity as:
[tex]\tilde{V}(t) =[V_0 e^{i \phi}]e^{i \omega t}[/tex]
The quantity in square brackets is a complex number in polar form called a phasor. Like any complex number, a phasor is a vector in the complex plane. Also like any complex number, it can be thought of as two real numbers, the magnitude and the phase (which descibe its length and its angle from the positive real axis respectively). These two numbers are exactly the two numbers that we need to fully-describe the sinusoidal osciallation; the magnitude of the phasor is the amplitude of the oscillation, and the phase of the phasor is the phase of the oscilllation. So we're expressing the information in the most compact possible way. By the way, the time-varying exponential factor in the expression above simply has the effect of causing the phasor to rotate counterclockwise in the complex plane with angular speed [itex]\omega[/itex]. As a result, its real part (i.e. the projection of the vector onto the real axis) will vary in length sinusoidally exactly as expected. One final note about phasors: sometimes, instead of writing them the way I did in the square brackets above, they are written as follows:
[tex]V_0 \angle \phi[/tex]
Now that all of that is out of the way, we see that in order to calculate the power, we must calculate the power. As far as I have been able to determine, there are two ways to do this. An easy way and a hard way.
The Easy Way
We can use another complex quantity, the apparent power,
S, as a convenient formulation of what is going on.
S is defined as
P +
iQ, where
P is the true power (the power available to do work on a load), and
Q is the reactive power. Reactive power arises from the fact that the voltage and current are not in phase, and are therefore sometimes in the opposite direction, meaning that energy is being stored rather than used up. This energy is just being shuffled back and forth between reactive elements in the circuit, not doing anything useful. I have found out that in terms of the complex voltage and current, the definition of S is:
[tex]S = \frac{\tilde{V}\tilde{I}^*}{2}[/tex]
Here the * denotes the complex conjugate. Let's say the phase difference between the voltage and current is [itex]\Delta \phi[/itex]. Therefore:
[tex]S = \frac{V_0 e^{i\omega t} I_0 e^{-i(\omega t + \Delta \phi)}}{2} = \frac{V_0 I_0}{2}e^{-i \Delta \phi}[/tex]
We know that the power, P, is the real part of S. Hence,
[tex]P = \frac{V_0 I_0}{2} \cos(\Delta \phi)[/tex]
Here, [itex]\cos(\Delta \phi)[/itex] is the power factor. The problem is that when you try these with the given numbers, you don't get the right answer:
P = (100 V)(2.32 A)(0.661)/2 = 76.676 W
The only thing I can think of is that maybe the given numbers are NOT amplitudes. Sometimes the stated voltage and current values are RMS values. In other words, the phasor is being stated in the form:
[tex]V_{\textrm{RMS}} \angle \phi[/tex]
Although RMS values are useful, I think that using them with the phasor notation is confusing and doesn't make sense when you think about what the magnitude of the phasor is supposed to mean. In any case, since these are sinusoids, the RMS values are smaller than the amplitudes by a factor of root 2.
[tex]P = \frac{V_0 I_0}{2} \cos(\Delta \phi) = \frac{V_0}{\sqrt{2}} \frac{I_0}{\sqrt{2}} \cos(\Delta \phi) \equiv V_{\textrm{RMS}} I_{\textrm{RMS}} \cos(\Delta \phi)[/tex]
So we are left with the standard result the the real power is the product of the RMS voltage with the RMS current, multiplied by the power factor. So, redoing the calculation, assuming the values given are RMS values already, we get:
P = (100 V)(2.32 A)(0.661) = 153.352 W
This is the closest I can get to the answer you have written.