Gibilisco S. Teach Yourself Electricity and Electronics

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the circuit conductors. In either case, the situation can be represented by a resistance, R, in series

with a capacitive reactance, X

(Fig. 14-10).

If you know the values of X

and R, you can find the angle of lead, also called the RC phase angle,

by plotting the point R − jX

on the RC plane, drawing the vector from the origin 0 − j0 out to that

point, and then measuring the angle of the vector clockwise from the R axis. You can use a protrac-

tor to measure this angle, as you did in the previous chapter for RL phase angles. Or you can use

trigonometry to calculate the angle.

As with RL circuits, you need only know the ratio of X

to R to determine the phase angle. For

example, if X

=−4 Ω and R = 7 Ω, you’ll get the same angle as with X

=−400 Ω and R = 700 Ω,

or with X

=−16 Ω and R = 28 Ω. The phase angle will be the same whenever the ratio of X

to R

is equal to −4:7.

Current Leads Voltage 221

14-8 In a pure capacitive

reactance, the current

leads the voltage

by 90°.

14-9 In a circuit with

capacitive reactance

and resistance, the

current leads the

voltage by less

than 90°.

Pure Resistance

As the resistance in an RC circuit gets large compared with the absolute value of the capacitive reac-

tance, the angle of lead becomes smaller. The same thing happens if the absolute value of X

gets

small compared with the value of R.

When R is many times larger than the absolute value of X

, whatever their actual values, the

vector in the RC plane points almost along the R axis. Then the RC phase angle is close to 0°. The

voltage comes nearly into phase with the current. The plates of the capacitor do not come anywhere

near getting fully charged with each cycle. The capacitor is said to “pass the ac” with very little loss,

as if it were shorted out. But it will still have an extremely high X

for any ac signals at much lower

frequencies that might exist across it at the same time. (This property of capacitors can be put to use

in electronic circuits. An example is when an engineer wants to let radio-frequency signals get

through while blocking signals at audio frequencies.)

Ultimately, if the absolute value of the capacitive reactance gets small enough, the circuit acts as

a pure resistance, and the current is in phase with the voltage.

How Much Lead?

If you know the ratio of capacitive reactance to resistance, or X

/R, in an RC circuit, then you can

find the phase angle. Of course, you can find this angle if you know the precise values, too.

Pictorial Method

You can use a protractor and a ruler to find phase angles for RC circuits, just as you did with RL cir-

cuits in the previous chapter, as long as the angles aren’t too close to 0° or 90°. First, draw a line

somewhat longer than 10 cm, going from left to right on the paper. Then, use the protractor to con-

struct a line going somewhat more than 10 cm vertically downward, starting at the left end of the

horizontal line. The horizontal line is the R axis of an RC plane. The line going down is the X

axis.

If you know the actual values of X

and R, divide or multiply them by a constant, chosen to

make both values fall between −100 and 100. For example, if X

=−3800 Ω and R = 7400 Ω, di-

vide them both by 100, getting −38 and 74. Plot these points on the lines. The X

point goes 38

mm down from the intersection point between your two axes. The R point goes 74 mm to the right

of the intersection point. Next, draw a line connecting the two points, as shown in Fig. 14-11. This

line will be at a slant and will form a triangle along with the two axes. This is a right triangle, with

the right angle at the origin of the RC plane. Measure the angle between the slanted line and the R

axis. Use the protractor for this. Extend the lines, if necessary, using the ruler, to get a good reading

222 Capacitive Reactance

14-10 Schematic

representation of a

circuit containing

resistance and

capacitive reactance.

on the protractor. This angle will be between 0 and 90°. Multiply this reading by −1 to get the RC

phase angle. That is, if the protractor shows 27°, the RC phase angle is −27°.

The actual vector is found by constructing a rectangle using the origin and your two points,

making new perpendicular lines to complete the figure. The vector is the diagonal of this rectangle,

running out from the origin (Fig. 14-12). The phase angle is the angle between the R axis and this

vector, multiplied by −1. It will have the same measure as the angle of the slanted line you con-

structed in Fig. 14-11.

How Much Lead? 223

14-11 Pictorial method

of finding phase

angle in a circuit

containing resistance

and capacitive

reactance.

14-12 Another pictorial

method of finding

phase angle in a

circuit containing

resistance and

capacitive reactance.

This method shows

the actual impedance

vector.

Trigonometric Method

Using trigonometry, you can determine the RC phase angle more precisely than the pictorial

method allows. Given the values of X

and R, the RC phase angle is the arctangent of their ratio.

Phase angle in RC circuits is symbolized by the lowercase Greek letter φ, just as it is in RL circuits.

Here are the formulas:

φ=tan

−1

/R)orφ=arctan (X

/R)

When doing problems of this kind, remember to use the capacitive reactance values for X

, and not

the capacitance values. This means that, if you are given the capacitance, you must use the formula

for X

in terms of capacitance and frequency and then calculate the phase angle. You should get an-

gles that come out negative or zero. This indicates that they’re RC phase angles rather than RL phase

angles (which are always positive or zero).

Problem 14-4

Suppose the capacitive reactance in an RC circuit is −3800 Ω and the resistance is 7400 Ω. What is

the phase angle?

Find the ratio X

/R =−3800/7400. The calculator display should show you something like

−0.513513513. Find the arctangent, or tan

−1

, getting a phase angle of −27.18111109° on the cal-

culator display. Round this off to −27.18°.

Problem 14-5

Suppose an RC circuit works at a frequency of 3.50 MHz. It has a resistance of 130 Ω and a capac-

itance of 150 pF. What is the phase angle?

First, find the capacitive reactance for a capacitor of 150 pF at 3.50 MHz. Convert the capaci-

tance to microfarads, getting C = 0.000150 µF. Remember that microfarads go with megahertz (mil-

lionths go with millions to cancel each other out). Then:

=−1/(6.28 × 3.50 × 0.000150)

=−1/0.003297 =−303 Ω

Now you can find the ratio X

/R =−303/130 =−2.33. The phase angle is equal to the arctangent

of −2.33, or −66.8°.

Problem 14-6

What is the phase angle in the preceding circuit if the frequency is raised to 7.10 MHz?

You need to find the new value for X

, because it will change as a result of the frequency change.

Calculating:

=−1/(6.28 × 7.10 × 0.000150)

=−1/0.006688 =−150 Ω

The ratio X

/R in this case is equal to −150/130, or −1.15. The phase angle is the arctangent of

−1.15, which turns out to be −49.0°.

224 Capacitive Reactance

Quiz

Refer to the text in this chapter if necessary. A good score is at least 18 correct. Answers are in the

back of the book.

1. As the size of the plates in a capacitor increases, all other things being equal,

(a) the value of X

increases negatively.

(b) the value of X

decreases negatively.

does not change.

(d) we cannot say what happens to X

without more data.

2. If the dielectric material between the plates of a capacitor is changed, all other things being

equal,

(a) the value of X

increases negatively.

(b) the value of X

decreases negatively.

does not change.

(d) we cannot say what happens to X

without more data.

3. As the frequency of a wave gets lower, all other things being equal, the value of X

for a

capacitor

(a) increases negatively.

(b) decreases negatively.

(d) depends on the current.

4. What is the reactance of a 330-pF capacitor at 800 kHz?

(a) −1.66 Ω

(b) −0.00166 Ω

(d) −603 kΩ

5. Suppose a capacitor has a reactance of −4.50 Ω at 377 Hz. What is its capacitance?

(a) 9.39 µF

(b) 93.9 µF

(d) 74.2 µF

6. Suppose a 47-µF capacitor has a reactance of −47 Ω. What is the frequency?

(a) 72 Hz

(b) 7.2 MHz

(d) 7.2 Hz

Quiz 225

7. Suppose a capacitor has X

=−8800 Ω at f = 830 kHz. What is C?

(a) 2.18 µF

(b) 21.8 pF

(d) 2.18 pF

8. Suppose a capacitor has C = 166 pF at f = 400 kHz. What is X

(a) −2.4 kΩ

(b) −2.4 Ω

−6

Ω

(d) −2.4 MΩ

9. Suppose a capacitor has C = 4700 µF and X

=−33 Ω. What is f ?

(a) 1.0 Hz

(b) 10 Hz

(d) 10 kHz

10. Each point in the RC plane

(a) corresponds to a unique inductance.

(b) corresponds to a unique capacitance.

(d) corresponds to a unique combination of resistance and reactance.

11. If R increases in an RC circuit, but X

is always zero, the vector in the RC plane will

(a) rotate clockwise.

(b) rotate counterclockwise.

(d) always point straight down.

12. If the resistance R increases in an RC circuit, but the capacitance and the frequency are

nonzero and constant, then the vector in the RC plane will

(a) get longer and rotate clockwise.

(b) get longer and rotate counterclockwise.

(d) get shorter and rotate counterclockwise.

13. Each complex impedance value R − jX

(a) represents a unique combination of resistance and capacitance.

(b) represents a unique combination of resistance and reactance.

(d) All of the above are true.

226 Capacitive Reactance

14. In an RC circuit, as the ratio X

/R approaches zero, the phase angle

(a) approaches −90°.

(b) approaches 0°.

(d) cannot be found.

15. In a purely resistive circuit, the phase angle is

(a) increasing.

(b) decreasing.

(d) −90°.

16. If X

/R =−1, then what is the phase angle?

(a) 0°

(b) −45°

(d) Impossible to find because there’s not enough data given

17. In Fig. 14-13, the impedance shown is

(a) 8.02 + j323.

(b) 323 + j8.02.

(d) 323 − j8.02.

Quiz 227

14-13 Illustration for Quiz

Questions 17 and 18.

18. In Fig. 14-13, note that the R and X

scale divisions are not the same size. What is the actual

phase angle?

(a) −1.42°

(b) About −60°, from the looks of it

(d) −88.6°

19. Suppose an RC circuit consists of a 150-pF capacitor and a 330-Ω resistor in series. What is

the phase angle at a frequency of 1.34 MHz?

(a) −67.4°

(b) −22.6°

(d) −65.6°

20. Suppose an RC circuit has a capacitance of 0.015 µF. The resistance is 52 Ω. What is the

phase angle at 90 kHz?

(a) −24°

(b) −0.017°

(d) None of the above

228 Capacitive Reactance

IN THIS CHAPTER, A COMPLETE, WORKING DEFINITION OF COMPLEX IMPEDANCE IS DEVELOPED. YOU’LL

also get acquainted with admittance, the extent to which an ac circuit allows (or admits) current

flow, rather than impeding it. As we develop these concepts, let’s review, and then expand on, some

of the material presented in the previous couple of chapters.

Imaginary Numbers

Have you been wondering what j actually means in expressions of impedance? Well, j is nothing but

a number: the positive square root of −1. There’s a negative square root of −1, too, and it is equal

to −j. When either j or −j is multiplied by itself, the result is −1. (Pure mathematicians often denote

these same numbers as i or −i.)

The positive square root of −1 is known as the unit imaginary number. The set of imaginary

numbers is composed of real-number multiples of j or −j. Some examples are j4, j35.79, −j25.76,

and −j25,000.

The square of an imaginary number is always negative. Some people have trouble grasping this,

but when you think long and hard about it, all numbers are abstractions. Imaginary numbers are no

more imaginary (and no less real) than so-called real numbers such as 4, 35.79, −25.76, or −25,000.

The unit imaginary number j can be multiplied by any real number on a conventional

real number line. If you do this for all the real numbers on the real number line, you get an imag-

inary number line (Fig. 15-1). The imaginary number line should be oriented at a right angle to

the real number line when you want to graphically portray real and imaginary numbers at the

same time.

In electronics, real numbers represent resistances. Imaginary numbers represent reactances.

Complex Numbers

When you add a real number and an imaginary number, you get a complex number. In this context,

the term complex does not mean “complicated.” A better word would be composite. Examples are

229

CHAPTER

Impedance and Admittance

4 + j5, 8 − j 7, −7 + j13, and −6 − j 87. The set of complex numbers needs two dimensions—a

plane—to be graphically defined.

Adding and Subtracting Complex Numbers

Adding complex numbers is just a matter of adding the real parts and the complex parts separately.

For example, the sum of 4 + j 7 and 45 − j 83 works out like this:

(4 + 45) + j(7 − 83)

= 49 + j(−76)

= 49 − j 76

Subtracting complex numbers is a little more involved; it’s best to convert a difference to a sum. For

example, the difference (4 + j 7) − (45 − j 83) can be found by multiplying the second complex

number by −1 and then adding the result:

(4 + j 7) − (45 − j83)

= (4 + j 7) + [−1(45 − j83)]

= (4 + j 7) + (−45 + j83)

=−41 + j90

230 Impedance and Admittance

15-1 The imaginary

number line.