Power (AC)
The discussion of power in Chapter 12 (Average Power) included only the average power
delivered to an ac network. We will now examine the total power equation in a slightly different form and will introduce two additional types
of power: apparent and reactive.
For any system such as in Fig. 1, the power delivered to a load at
any instant is defined by the product of the applied voltage and the
resulting current; that is,
In this case, since v and i are sinusoidal quantities, let us establish a
general case where
The chosen v and i include all possibilities because, if the load is purely
resistive, $\theta = 0$. If the load is purely inductive or capacitive, $\theta = 90$
or $\theta = -90$, respectively. For a network that is primarily inductive, v
is positive (v leads i), and for a network that is primarily capacitive, v is
negative (i leads v).
Substituting the above equations for v and i into the power equation
will result in
If we now apply a number of trigonometric identities, e.g,
replacing $a + b$ by $wt+\theta$, we get
now multiplying eq(1) by $\sin wt$,
Where
and
Hence eq(3) becomes,
Now the power equation can be written as,
where $V$ and $I$ are the rms values.
It would appear initially that nothing has been gained by putting the
equation in this form. However, the usefulness of the form of Eq. (4)
will be demonstrated in the following sections.
Fig. 1: Defining the power delivered to a load.
$$ p = vi$$
$$v = V_m \sin (wt + \theta)$$
$$i = I_m \sin (wt)$$
$$ p = V_m I_m \sin (wt + \theta) \sin (wt)$$
$$ \sin(a+b) = \sin a \cos b + \sin b \cos a \tag{1}$$
$$ \sin(wt+\theta) = \sin wt \cos \theta + \sin \theta \cos wt \tag{2}$$
$$ \sin wt \sin(wt+\theta) = \sin^2 wt \cos \theta + \sin \theta \sin wt \cos wt \tag{3}$$
$$ \sin^2 wt = {1 - \cos 2wt \over 2}$$
$$ \sin wt \cos wt = {1 \over 2} \sin 2wt$$
$$ \sin wt \sin(wt+\theta) = {1 - \cos 2wt \over 2} \cos \theta + \sin \theta {1 \over 2} \sin 2wt \tag{4}$$
$$ \begin{split}
p &= V_m I_m \sin (wt) \sin (wt + \theta)\\
&= V_m I_m [{1 - \cos 2wt \over 2} \cos \theta + \sin \theta {1 \over 2} \sin 2wt]\\
&= {V_m I_m \over 2} [(1 - \cos 2wt) \cos \theta + \sin \theta \sin 2wt]\\
&= VI(1 - \cos 2wt) \cos \theta + VI \sin \theta (\sin 2wt)\\
\end{split}
$$
$$ \bbox[10px,border:1px solid grey]{p = VI(1 - \cos 2wt) \cos \theta + VI \sin \theta (\sin 2wt)} \tag{4}$$
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