AC Power Factor
In the equation $P = (V_mI_m/2)\cos \theta$, the factor that has significant control over the delivered power level is the $\cos \theta$. No matter how large the
voltage or current, if $\cos \theta = 0$, the power is zero; if $\cos \theta = 1$, the
power delivered is a maximum. Since it has such control, the expression
was given the name power factor and is defined by
For a purely resistive load such as the one shown in [Fig. 1], the
phase angle between v and i is $0^\circ$ and $Fp = \cos \theta= cos 0^\circ = 1$. The power
delivered is a maximum of
For a purely reactive load (inductive or capacitive) such as the one
shown in [Fig. 2], the phase angle between v and i is $90^\circ$ and
The power delivered is then the minimum value
of zero watts, even though the current has the same peak value as
that encountered in [Fig. 1].
For situations where the load is a combination of resistive and
reactive elements, the power factor will vary between 0 and 1. The
more resistive the total impedance, the closer the power factor is to
1; the more reactive the total impedance, the closer the power factor
is to 0.
In terms of the average power and the terminal voltage and current,
The terms leading and lagging are often written in conjunction with
the power factor. They are defined by the current through the load. If
the current leads the voltage across a load, the load has a leading
power factor. If the current lags the voltage across the load, the load
has a lagging power factor. In other words,
The importance of the power factor to power distribution systems is
examined in the Ac power chapter. In fact, one section is devoted to power-factor
correction.
$$ \bbox[10px,border:1px solid grey]{Power \, Factor = F_P = \cos \theta} \tag{1}$$
Fig. 1: Purely resistive load with $F_p = 1$
$$({V_mI_m \over 2}) \cos \theta = ((100 V)(5 A)/2)(1) = 250 W$$
$$Fp = \cos \theta = \cos 90^\circ = 0$$
Fig. 2: Purely inductive load with $F_p = 0$
$$ \bbox[10px,border:1px solid grey]{Power \, Factor = F_P = \cos \theta= {P \over V_{eff} I_{eff}}}\tag{2}$$
Example 1: Determine the power factors of the following loads,
and indicate whether they are leading or lagging:
a. Fig. 3
b. Fig. 4
c. Fig. 5
Solution:
a. $F_p = \cos \theta = \cos |40^\circ - (-20^\circ)|$ $= \cos 60^\circ = 0.5\,\text{leading}$
b. $F_p = \cos \theta = \cos |80^\circ - 30^\circ)| $ $= \cos 50^\circ = 0.6428\,\text{lagging}$
b. $F_p = \cos \theta = {P \over V_{eff}I_{eff}} = {100 \over (20V)(5A)} = 1$
The load is resistive, and Fp is neither leading nor lagging.
a. Fig. 3
b. Fig. 4
c. Fig. 5
Fig. 3: Example 1 (a)
Fig. 4: Example 1 (b)
Fig. 5: Example 1 (c)
a. $F_p = \cos \theta = \cos |40^\circ - (-20^\circ)|$ $= \cos 60^\circ = 0.5\,\text{leading}$
b. $F_p = \cos \theta = \cos |80^\circ - 30^\circ)| $ $= \cos 50^\circ = 0.6428\,\text{lagging}$
b. $F_p = \cos \theta = {P \over V_{eff}I_{eff}} = {100 \over (20V)(5A)} = 1$
The load is resistive, and Fp is neither leading nor lagging.
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