# 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 $$\bbox[10px,border:1px solid grey]{Power \, Factor = F_P = \cos \theta} \tag{1}$$
Fig. 1: Purely resistive load with $F_p = 1$
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
$$({V_mI_m \over 2}) \cos \theta = ((100 V)(5 A)/2)(1) = 250 W$$
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 $$Fp = \cos \theta = \cos 90^\circ = 0$$ 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.
Fig. 2: Purely inductive load with $F_p = 0$
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, $$\bbox[10px,border:1px solid grey]{Power \, Factor = F_P = \cos \theta= {P \over V_{eff} I_{eff}}}\tag{2}$$ 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,
capacitive networks have leading power factors, and inductive networks have lagging power factors.
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.
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
Fig. 3: Example 1 (a)
Fig. 4: Example 1 (b)
Fig. 5: Example 1 (c)
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. $F_p = \cos \theta = {P \over V_{eff}I_{eff}} = {100 \over (20V)(5A)} = 1$