In an ac circuit or system containing nonzero resistance and nonzero reactance, the relationships among true power PT, apparent (VA) power PVA, and imaginary (reactive) power PX are as follows:

**PVA
2 = PT
2 + PX
2
PT < PVA
PX < PVA**

If there is no reactance in the circuit or system, then PVA = PT, and PX = 0. Engineers strive to minimize, and if possible eliminate, the reactance in power-transmission systems.

#### Power Factor

In an ac circuit, the ratio of the true power to the VA power, PT/PVA, is called the power factor. If there is no reactance, the ideal case, then PT = PVA, and the power factor (PF ) is equal to 1. If the circuit contains all reactance and no resistance of any significance (that is, zero or infinite resistance), then PT = 0, and therefore PF = 0. When a load, or a circuit in which you want power to be dissipated, contains resistance and reactance, then PF is between 0 and 1. That is, 0 < PF < 1. The power factor can also be expressed as a percentage between 0 and 100, written PF%. Mathematically, we have these formulas for the power factor:
**PF = PT/PVA
PF% = 100PT/PVA **

When a load has some resistance and some reactance, then some of the power is dissipated as true power, and some is rejected by the load as imaginary power. In a sense, this imaginary power is sent back to the power source. There are two ways to determine the power factor in an ac circuit that contains reactance and resistance. One method is to find the cosine of the phase angle. The other method involves the ratio of the resistance to the absolute-value impedance.

#### Cosine of Phase Angle

Recall that in a circuit having reactance and resistance, the current and the voltage are not in phase. The phase angle (φ) is the extent,expressed in degrees, to which the current and the voltage differ in phase. If there is no reactance, then φ = 0°. If there is a pure reactance, then either φ = +90° (if the reactance is inductive) or else φ = −90° (if the reactance is capacitive). The power factor is equal to the cosine of the phase angle:

**PF = cos φ**

#### Problem 1

**Vector diagram showing the phase angle for a purely resistive impedance of 600 + j 0. The R and jX scales are relative.**

**Suppose a circuit contains no reactance, but a pure resistance of 600 Ω. What is the power factor?**

Without doing any calculations, it is evident that PF = 1, because PVA = PT in a pure resistance. That means PT/PVA = 1. But you can also look at this by noting that the phase angle is 0°, because the current is in phase with the voltage. Using your calculator, you can see that cos 0° = 1. Therefore, PF = 1 = 100%. The vector for this case is shown in above figure.

#### Problem 2

**Vector diagram showing the phase angle for a purely capacitive impedance of 0 − j40. The R and jX scales are relative.**

**Suppose a circuit contains a pure capacitive reactance of −40 Ω, but no resistance. What is the power factor?**

Here, the phase angle is −90° (above figure). A calculator will tell you that cos −90° = 0. Therefore, PF = 0, and PT/PVA = 0 = 0%. None of the power is true; all of it is reactive.

#### Problem 3

**Vector diagram showing the phase angle for a complex impedance of 50 + j50. The R and jX scales are relative.**

**Suppose a circuit contains a resistance of 50 Ω and an inductive reactance of 50 Ω in series. What is the power factor?**

The phase angle in this case is 45° (above figure). The resistance and reactance vectors have equal lengths and form two sides of a right triangle, with the complex impedance vector forming the hypotenuse. To determine the power factor, you can use a calculator to find cos 45° = 0.707. This means that PT/PVA = 0.707 = 70.7%.

#### The Ratio R/Z

The second way to calculate the power factor is to find the ratio of the resistance R to the absolutevalue impedance Z. In Fig. above figure, this is visually apparent. A right triangle is formed by the resistance vector R (the base), the reactance vector jX (the height), and the absolute-value impedance Z (the hypotenuse). The cosine of the phase angle is equal to the ratio of the base length to the hypotenuse length; this represents R/Z.

#### Problem 4

**Suppose a circuit has an absolute-value impedance Z of 100 Ω, with a resistance R = 80 Ω. What is the power factor?**

Simply find the ratio PF = R/Z = 80/100 = 0.8 = 80%. Note that it doesn’t matter whether the

reactance in this circuit is capacitive or inductive.

#### Problem 5

**Suppose a circuit has an absolute-value impedance of 50 Ω, purely resistive. What is the power factor?**

Here, R = Z = 50 Ω. Therefore, PF = R/Z = 50/50 = 1 = 100%.

#### Problem 6

**Suppose a circuit has a resistance of 50 Ω and a capacitive reactance of −30 Ω in series. What is the power factor?**

Use the cosine method. First, find the phase angle. Remember the formula: φ = arctan (X/R), where X is the reactance and R is the resistance. Therefore, φ = arctan (−30/50) = arctan (−0.60) = −31°. The power factor is the cosine of this angle; PF = cos (−31°) = 0.86 = 86%.

#### Problem 7

**Illustration for Problem 7. (The vertical and horizontal scale increments differ; this is a common practice in graphs, often done for illustration convenience.)**

**Suppose a circuit has a resistance of 30 Ω and an inductive reactance of 40 Ω. What is the power factor? Use the R/Z method.**

Find the absolute-value impedance: Z2 = R2 + X 2 = 302 + 402 = 900 + 1600 = 2500. Therefore, Z = 25001/2 = 50 Ω, so PF = R/Z = 30/50 = 0.60 = 60%. This problem can be represented vectorially by a 30:40:50 right triangle, as shown in above figure.

#### How Much of the Power Is True?

The preceding formulas allow you to figure out, given the resistance, reactance, and VA power, how many watts are true or real power, and how many watts are imaginary or reactive power. This is important in RF equipment, because some RF wattmeters display VA power rather than true power. When there is reactance in a circuit or system, the wattage reading is therefore exaggerated.

#### Problem 8

**Suppose a circuit has 50 Ω of resistance and 30 Ω of inductive reactance in series. A wattmeter shows 100 W, representing the VA power. What is the true power?**

First, calculate the power factor. Suppose you use the phase-angle method. Then:

**φ = arctan (X/R )
= arctan (30/50) = 31° **

The power factor is the cosine of the phase angle. Thus:

**PF = cos 31° = 0.86 = 86%**

Remember that PF = PT/PVA. This formula can be rearranged to solve for true power:

**PT = PF × PVA**

= 0.86 × 100

= 86 W

= 0.86 × 100

= 86 W

#### Problem 9

**Suppose a circuit has a resistance of 1000 Ω in parallel with a capacitance of 1000 pF. The frequency is 100 kHz. If a wattmeter designed to read VA power shows a reading of 88.0 W, what is the true power?**

This problem is rather complicated because the components are in parallel. To begin, be sure the units are all in agreement so the formulas will work right. Convert the frequency to megahertz: f = 100 kHz = 0.100 MHz. Convert capacitance to microfarads: C = 1000 pF = 0.001000 μF. From the previous topics, recall the formula for capacitive susceptance, and calculate it for this situation:

The conductance of the resistor, G, is the reciprocal of the resistance, R, as follows:

**G = 1/R
= 1/1000
= 0.001000 S **

Now, use the formulas for calculating resistance and reactance in terms of conductance and susceptance in parallel circuits. First, find the resistance:

**R = G/(G2 + B 2)**

= 0.001000/(0.0010002 + 0.0006282)

= 0.001000/0.000001394

= 717 Ω

= 0.001000/(0.0010002 + 0.0006282)

= 0.001000/0.000001394

= 717 Ω

Then, find the reactance:

**X = −B/(G2 + B2)**

= −0.000628/0.000001394

= −451 Ω

= −0.000628/0.000001394

= −451 Ω

Next, calculate the phase angle:

**φ = arctan (X/R)**

= arctan (−451/717)

= arctan (−0.629)

= −32.2°

= arctan (−451/717)

= arctan (−0.629)

= −32.2°

The power factor is found from the phase angle as follows:

**PF = cos φ**

= cos (−32.2°)

= 0.846 = 84.6%

= cos (−32.2°)

= 0.846 = 84.6%

The VA power, PVA, is given as 88.0 W. Therefore:

**PT = PF × PVA**

= 0.846 × 88.0

= 74.4 W

= 0.846 × 88.0

= 74.4 W