23.Reactance and Impedance

Study Unit

Reactance andImpedance

By

Joseph A. Risse, P.E.

and

J. A. Sam Wilson, M.S., CET

Resistors, capacitors, and inductors are the basic building blocks of all electronic systems. And,in spite of the fact that there has been a radical change in amplifying devices—and a completerevision of the appearance of systems—these basic building blocks that make up all circuits arethe same.

In this study unit, you’ll learn how resistors, capacitors, and inductors behave when connected toform basic circuits.

When you complete this study unit, you’ll be able to

Explain how resistors, capacitors, and inductors work in DC (direct current) circuits

Calculate time relationships in circuits

Determine the reactance of a capacitor or inductor in an AC (alternating current) circuit

Calculate the impedance of series RLC (resistive-inductive-capacitive) circuits

Find the phase angle between the voltage and current in parallel RC (resistive-capacitive),RL (resistive-inductive), and series RLC circuits

Work with j operators

Preview

iii

BACKGROUND INFORMATION . . . . . . . . . . . . . . . . . . . . . . 1Some Important DefinitionsPhysical Laws and EffectsUsing the Scientific CalculatorReview of Radian MeasurementSpeed, Velocity, Vector, and Phasor

DETERMINING INDUCTIVE AND CAPACITIVE REACTANCE . . . . . . . . 15Omega (Angular Velocity)Working with Sine WavesInstantaneous Sine Wave Voltage and CurrentHow a Capacitor Opposes a Change of VoltageCapacitors with a Vacuum DielectricResistance and Capacitance in DC CircuitsResistance and Inductance in DC CircuitsUniversal Time-Constant CurvesDelay-Before-Start CircuitsDelay-Before-Stop CircuitsCapacitive ReactanceHow Does an Inductor Oppose a Change in Current?Faraday’s LawContact ProtectionCounter VoltageAn Inductor Static Curve for High InductanceInductive Reactance

DETERMINING IMPEDANCE AND PHASE ANGLES FOR RC,RL, AND RLC CIRCUITS. . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Review of Right Triangle CalculationsReactances and Impedances as PhasorsWorking with Math SymbolsThe j OperatorCoordinatesPower in AC CircuitsA Special Series RLC Circuit

POWER CHECK ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . 67

EXAMINATION. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

Contents

v

BACKGROUND INFORMATION

Some Important DefinitionsBefore beginning our work with reactance and capacitance, let’s re-view some important terms. Some of these definitions are based uponmodels that technicians find useful for analyzing AC circuits.

Components that do not generate a voltage or current are called passivecomponents. A resistor is an example of a passive component. Compo-nents that do generate a voltage or current are called active components.A transistor in an amplifier circuit is called an active component, be-cause it can be considered to be the source of the signal voltage in theequivalent circuit of the amplifier.

In this study unit, you’ll learn about three very important passivecomponents: resistors, capacitors, and inductors. You’ll also learnhow resistors, capacitors, and inductors are used together in variousDC and AC circuit combinations. Figure 1 shows a circuit that containsthese passive components. In order to understand the type of circuitshown in Figure 1, you must first understand how each of the compo-nents performs in both DC and AC circuits. You must also learn howsimple two-component combinations perform in circuits.

C2

R1

R2

L

C1

ACSIGNALOUTPUT

ACSIGNALINPUT

DC VOLTAGE

DC VOLTAGE

FIGURE 1—A basic“coupling circuit” is usedto deliver a signal from oneamplifier to another. This typeof circuit is made withresistors, capacitors, and aninductor.

Reactance and Impedance

1

Let’s take a brief look at resistors, capacitors, and inductors.

A resistor is a component that opposes the flow of current in bothDC and AC circuits. Resistors radiate energy in the form of heat,but they don’t store energy. For the purposes of this study unit, theresistors that will be considered are the carbon-composition type.The value of resistance in these resistors is the same in both ACand DC circuits. This is an important point, because some resistors,such as wire-wound types, can have inductance. Resistors thathave inductance don’t act like carbon-composition resistors insome AC circuits.

A capacitor is a component that opposes any change in voltageacross its terminals in both AC and DC circuits. Capacitors storeenergy in their dielectric region.

An inductor is a component that opposes any change of currentthrough it in both AC and DC circuits. Inductors store energy ina surrounding magnetic field.

Physical Laws and EffectsThere are some basic physical laws and effects that affect the behaviorof resistors, capacitors, and inductors in circuits.

Every time there’s a current flow, there’s always a magnetic fieldaround that current. The strength of the magnetic field dependsdirectly upon the strength of the current. So, if you have an ACcurrent in which the strength of the current is always changing,there’s a continuous change of magnetic field strength aroundthe current.

The direction of the magnetic field around a current is directlydependent upon the direction of electron current. This importantlaw is illustrated in Figure 2.

Every time a conductor moves through a magnetic field, there’salways a voltage induced across the conductor.

Every time an alternating current flows through a resistor, thereare two effects that always occur

1. There’s always heat generated.

2. There’s always an AC voltage drop across the resistor.

2 Reactance and Impedance

A resistor does not cause the voltage and current in an AC circuitto get out of phase. Figure 3 shows the waveforms for voltage andcurrent in a resistor circuit. Observe that the waveforms for voltageand current both go through their maximum values and throughzero at the same instant of time. When this occurs, the AC voltageand current are “in phase.” Although we’re considering AC currenthere, keep in mind that heat and voltage drop also occur in resistiveDC circuits.

Every time there’s a flow of AC current through a capacitor, there arealways two effects that must be considered:

There’s always a shift in phase between the voltage and current.The voltage and current get out of step, and the current leads thevoltage (Figure 4). If there’s no other passive component in thecapacitive circuit, the current leads the voltage by 90° as shownin Figure 4.

Reactance and Impedance 3

FLUX LINES

ELECTRON FLOW

FIGURE 2—When electroncurrent flows through awire, there’s always asurrounding magnetic field.When the thumb points inthe direction of electroncurrent, the fingers circlein the direction of themagnetic field. Observethat the field lines are lessconcentrated and there-fore the field is weaker,as the distance from thecurrent is increased.

0

E

E

V

I

F t

THE CIRCUIT

T

V R

FIGURE 3—The sine wavevoltage and current in aresistive circuit are inphase. They go throughmaximum points, E, andzero values point, F, atthe same time, and theytake the same amount oftime for one cycle. T is thetime for one cycle. Timealways gets later as youmove along the time line,t. So, F is later than E.

There’s always an AC voltage drop across the capacitor when anAC current is flowing through it.

Every time there’s a flow of AC current through an inductor, there arealways two effects that must be considered:

There’s always a shift in phase between the voltage and current.In this case, the current lags behind the voltage (Figure 5). Asshown in this illustration, if there’s no other passive component inthe inductive circuit, the current lags behind the voltage by 90°.

There’s always an AC voltage drop across the inductor whenthere’s an AC current flowing through it.


0m n

y

V V C

I

t

x THE CIRCUIT

T

FIGURE 4—The sine wavevoltage in a capacitivecircuit is 90° out of phase.Note that the current(point x) goes throughmaximum one-fourthcycle (90°) before thevoltage maximum at pointy. T is the time for onecycle, which is also calledthe period of the wave.

0

m

V V L

I

n

t

THE CIRCUIT

T

FIGURE 5—The sine wavevoltage and current inan inductive circuit are 90°out of phase. Notice thatthe current peak at ncomes later than thevoltage peak at m.

Using the Scientific CalculatorIn this lesson, you’ll find that certain problems can be solved morequickly and accurately using a scientific calculator. These calculatorsare widely available at a modest price.

Throughout this study unit, you’ll find instructions for solving certainproblems using a scientific calculator. The symbols in brackets ([ ])represent actual keys on the calculator. The symbols on your calculatormight be slightly different.

Let’s try the following example problem.

What is the square root of 42 + 32?

Keystrokes Display Description

[4] [X2] [=] 16 Begin by takingthe square of 4.

[+] [3] [X2] [=] 25 Then, add thesquare of 3.

[ ] [=] 5 Finally, take the

square root of the sum(25). Answer: 5.

Review of Radian MeasurementSometime in elementary school, students are usually introduced to thefact that a circle can be divided into 360 even segments called degrees.By that measurement, a 90° angle (also called a right angle) makes upone-fourth of a circle. There are four 90-degree quarters in a circle. Theconcept of degree measurement is reviewed in Figure 6A.

The measurement of angles in degrees is a popular idea with peoplewho don’t work in technology. For people working in technology, thefavored angle measurement is the radian. Figure 6B shows the basicconcept of measurement using radians. If you take the radius of a circleand lay it along the circumference as shown, then draw lines from thecenter of the circle to each end of that radius located on the circumfer-ence, the angle between the two lines is one radian. One definition fora radian (sometimes called a rad) is the angle included within an arcequal to the radius of the circle.


If you continue to divide the half circle into radian arcs, you’ll findthat you can draw a little more than three radian arcs in a half circle,which is 180 degrees. In fact, you can draw 3.14 radian ( radians) arcsin a half circle (Figure 6C).

In equation form, radians = 180°.


FIGURE 6—(A) A circle can be divided into 360 equal parts called degrees. In scientific work, the degreesare counted counterclockwise. You can see that 0° and 360° are the same point. Lines are also drawn at30°, 45°, 60°, 90°, 180°, and 270°. The Roman numerals I, II, III, and IV mark the four quadrants. (B) A radiancan be an angle or part of a circle. It takes a little over radians (3.14 radians) to make a half circle (180°).(C) This illustration demonstrates that there are radians in 180°.

270°

(A)

90°

60°45°

30°

180°

II I

III IV

0°(360°)

RADIUS

RADIUS

ONE RADIAN

ONE RADIAN

rr

r

(B)

1 RADIAN

0

2 RADIANS

3 RADIANS

π/2 RADIANS

3π/2 RADIANS

π RADIANS

(C)

There are two basic equations you can use to convert from knownradians to unknown degrees and to convert from known degrees tounknown radians:

Equation 1: To get the number of degrees when you know the numberof radians, multiply the number of radians 180 degrees/( radians):

number of radians180 degrees

radians= number of degrees

where 180 degrees/ radians is the multiplier.

Equation 2: To get the number of radians when you know the numberof degrees, multiply the number of degrees by ( radians)/(180 degrees):

number of degreesradians

180 degrees= number of radians

where ( radians/180 degrees) is also called the multiplier.

How do you remember which equation to use? Remember that

radians = 180°, and 180° = radians. Therefore, both of the multipliersin the above equations have a value of 1.0 as shown here:

180

radians=1 Also,

radian

180=1

A scientific calculator can be used to easily find the number of degreesin a known number of radians. It can also be used to find the numberof radians when the number of degrees are known.

To find the number of degrees there are in 5 radians using the scien-tific calculator, perform the following operations.


[5] [] [180] [=] 900 Multiply 5 180.

[] [2nd] [] [=] 286.4788976 Divide by .Answer: 286.5.

To find the number of radians there are in 200 degrees using thescientific calculator, perform the following operations.



[200] [] [2nd][] [=] 628.318307 Multiply 200 .

[] [180] [=] 3.49068504 Divide by 180 degrees.Answer: 3.49.

Remember that in both of the equations above, you’re only using amultiplier that has a value of 1, and multiplying by 1 doesn’t changethe values in the equation. What you are changing is the units of anglemeasurement.

To summarize, when you use the equations to convert units, alwaysuse the multiplier that has the unit you don’t want in the denominatorand the unit you do want in the numerator.

Here are a few sample problems for converting radians to degrees anddegrees to radians. Use the keystrokes on your scientific calculator asdescribed above. The answers shown at the end of this study unitshow how to use strikeouts to obtain the correct answers.

Questions 1–10: Convert the values in the following problems.

1. How many degrees are there in one radian? In this problem, youhave the number of radians (which you don’t want). What youwant to find is the number of degrees.

1 radian180 degrees

radians= 57.3 degrees

2. How many radians are there in 90°? In this problem you have thenumber of degrees (which you don’t want). What you want to findis the number of radians.

90 degreesradians

180 degrees=

2radians

3. 270° = ______ radians

270 degreesradians

180 degrees= 4.71 radians

4. 1° = ______ radians

1 degreeradians

180 degrees= 0.0175

5. 30° = ______ radians

30 degreesradians

180 degrees=

6radians


6. 45° = ______ radians

45 degreesradians

180 degrees=

45

180=

1

4=

4radians

7. /4 radians = ______ degrees

4radians

180 degrees

radians=

180

4degrees = 45 degrees

8. 3 radians = ______ degrees

3 radians180 degrees

radians= 540 degrees

9.

radians degrees

3radians

180 degrees

radians=

180

3degrees = 60 degrees

10.3

4radians = _____ degrees

3

4radians

180 degrees

radians= 42.97 degrees

Speed, Velocity, Vector, and PhasorAs a technician, you need to be able to distinguish between the termsspeed, velocity, vector, and phasor.

Speed is the rate of change of distance. Speed doesn’t have a direction.For example, you might say, “The speed of the motorcycle is 40 milesper hour.”

Velocity is a vector value. Velocity has magnitude and direction. So, youmight say, “The velocity of the motorcycle is 40 miles per hour north.”Technically, it would not be correct to say, “The velocity of the motor-cycle is 40 miles per hour.”

A vector is a quantity that illustrates both magnitude and direction. Aphasor is an arrow that represents magnitude and an angle from agiven starting position. The drawings in Figure 7 illustrate two waysto draw a phasor.


Our interest at this time is in the use of phasors to describe what’shappening in AC circuits. A good example is the rotating phasor inFigure 8. If you project the tip of the arrow in Figure 8 along a linethat represents degrees of rotation, and if the phasor is turning at aconstant rate, the projection will be a sine wave.


6 UNITS

6 UNITS

6 UNITSAT 45° USUALLYWRITTEN AS6 45°

45°

π/4 RADIANS

6 UNITSAT π/4 RADIANSSOMETIMESWRITTEN AS6 π/4

FIGURE 7—Two ways torepresent a phasorrotated to 45 degrees.Both phasors are inidentical positions.

FIGURE 8—A sine waveresults when the position of aphasor that’s rotating at aconstant rate is projected.

The fact that a sine wave is generated by a rotating phasor is a veryimportant point. A sine wave voltage or current can be represented by arotating phasor. The voltage or current at any instant can be representedby the position of the phasor at that instant. The amplitude of the sinewave is represented by the length of the phasor. The frequency of thesine wave is determined by the angular velocity of the phasor.

These facts allow us to describe voltages and current by phasors, andthat will greatly simplify the descriptions of circuit behavior.

Earlier in this lesson, we showed that a current in an inductive circuitlags behind the applied voltage by 90°. This was illustrated in Figure 5and is repeated in Figure 9 along with the phasors that represent thesame situation. When you look at the phasor representation, imaginethe two phasors are like the hands of a clock that are welded together.

When the two phasors turn counterclockwise, they turn together. Thestandard direction of phasor rotation in the United States is counterclockwise.In some other countries, phasor rotation is clockwise.

In Figure 9, the phasors have been stopped in a position where the in-stantaneous voltage is zero volts, and the current at that instant is at itsmost negative value.


FIGURE 9—As the twophasors rotate counter-clockwise, two sinewaves are generated.The current sine waveis 90 degrees behindthe voltage sine wave.

As you work with rotating phasors, you’ll become more familiarwith the advantages of this method of representing sine waves. As anexample, using sine waveforms, try to draw two sine waves that startwhen one is at +23 degrees and the other is at –37 degrees. You can seehow easy this is to do with phasors by referring to Figure 10.


+23

37

FIGURE 10—The method ofrepresenting sine wavesthat’s depicted in thisfigure is easier than tryingto draw sine waveforms atspecific phase angles.


Power Check 1

At the end of each section in Reactance and Impedance, you’ll be asked to pause and checkyour understanding of what you’ve just read by completing a “Power Check.” Writing theanswers to these questions will help you review what you’ve studied so far. Please completePower Check 1 now.

1. True or False? The statement, “The current leads the voltage” is the same as, “The voltagelags the current.”

2. Where is the energy stored in an inductor?

____________________________________________________________________________

3. List three passive components.

____________________________________________________________________________

4. Which component opposes any change in current through it?

____________________________________________________________________________

5. Components that generate a voltage or can be considered to be the source of a signalvoltage are called _______ components.

6. Which type of resistor may sometimes act like an inductor?

____________________________________________________________________________

7. Where is the energy stored in a capacitor?

____________________________________________________________________________

8. Name two effects of an AC current flowing through a resistor.

____________________________________________________________________________

9. Which component opposes any change in voltage across its terminals?

____________________________________________________________________________

10. True or False? The AC sine wave voltage across an inductor lags behind the current.

11. Speed at a given direction is called _______.

12. The favored angle measurement for people working in technology is _______.

(Continued)


Power Check 1

Questions 13–18: Make the following conversions.

13. 90° = _______ radians

14. 270° = _______ radians

15. 120° = _______ radians

16.

6radians = _______ degrees

17. 2 radians = _______ degrees

18. One radian = _______ degrees

Check your answers with those on page 67.

DETERMINING INDUCTIVE AND CAPACITIVEREACTANCE

Omega (Angular Velocity)You’ve seen how a sine wave can be considered to be generated by arotating phasor. The faster the phasor rotates, the higher the frequencybeing generated. In other words, the frequency of the sine wave is thedirect result of the angular velocity of the rotating phasor. Therefore,we need to review some of the basics of angular velocity and the directrelationship of angular velocity to frequency.

The angular velocity, represented by omega, , of a phasor is equal toone revolution divided by the time, t, that it takes the phasor to makeone revolution:

=one revolution

time (t) for one revolution

For example, if it takes 0.5 seconds to make one complete revolution(360°), the angular velocity is

=degrees

time=

360

0.5 sec= 720 per second

As mentioned before, one revolution can be written a number of dif-ferent ways. The one we’re interested in at this time is the use of 2

radians to represent one complete revolution.

=one revolution

time (t)=

2 (radians)

t (seconds)

Two of the most important equations you’ll use in electronics are

t(frequency)

( )time for one cycle 1

f

also, frequency)time for one cycle)

ft

((

1

where t is the time of one cycle of waveform in seconds, and thefrequency, f, is the frequency of the waveform in hertz. This equationis sometimes written as t = 1/f. Therefore, the latter equation for canbe written by substituting 1/f for t:


2 2

12

1

t f f

Remember that whenever you divide by a fraction, you must invertthe fraction and multiply. So, the above equation becomes

21

12

f

ff2

Working with Sine WavesIn your work with AC circuits, you’ll be working with some veryimportant equations dealing with omega (). Remember that omegarepresents the angular velocity of a rotating phasor. Remember alsothat the angular velocity of the rotating phasor determines the frequencyof the projected sine wave.

You may wonder why the sine wave is such an important part of elec-tronics technology. One reason is that every periodic wave, no matterhow complicated its waveform, can be constructed with a combinationof sine waves. Therefore, you can think of a sine wave as being a perfectwaveform. Figure 11 gives two examples.


SQUARE WAVE

SQUARE WAVE

A

A

B B

C

C

C

C

TRIANGULAR WAVE(B)

SQUARE WAVE(A)

C

C

D

D

E

E

E

E

F

F

FIGURE 11—Sine waves Zcan be used to constructnon-sinusoidal waveforms.Figure 11A depicts a squarewave; Figure 11B a triangularwave. Both are constructedby combining several sinewaves.

Another reason that the sine wave is important in electronics technologyis that any periodic wave that’s not a sine wave contains harmonicfrequencies. Harmonic frequencies are multiples of the basic waveform.

Look again at Figure 11. Observe that the non-sinusoidal waveformsare actually constructed of a basic sine wave and multiple frequenciesof that sine wave.

This information isn’t trivial—it has many applications in theory andin troubleshooting electronic equipment. We’ll momentarily step asidefrom our discussion of omega to show how this important informationcan be used in troubleshooting. (We’ll return to study these applicationsin greater detail later.)

Figure 12A shows a test being performed on an amplifier. A pure sinewave is introduced into an amplifier, and the output waveform istested with an instrument called a distortion analyzer. Assuming theamplifier isn’t distorting the waveform, the distortion analyzer willshow that the output waveform is a pure sine wave. More impor-tantly, the distortion analyzer will show how much distortion theamplifier is causing.

Figure 12B shows a second way that sine waves are used in electronicstechnology. The example shows a square wave introduced to the inputof an amplifier. The output should be a square wave that contains allof the sine waves that make up the square wave. Observe that theamplifier cannot pass all sine wave frequencies. (An oscilloscope canbe used to display waveforms of voltages.)


SINEWAVEINPUT

(A)

AMPLIFIER

DISTORTIONANALYZER

INPUTWAVEFORM

OUTPUTWAVEFORM

(B)

FIGURE 12—In Figure 12A,a test is being performedon an amplifier. Figure 12Bshows a second way thatsine waves are helpful introubleshooting anamplifier. Here thetechnician can observewhich, if any, frequenciesare not passed throughthe amplifier.

If you refer back to Figure 11, you’ll see that a square wave is madeup of a basic sine wave and multiple frequencies (called harmonicfrequencies) of that basic wave. If the amplifier can pass all of thoseharmonic frequencies, the output will be a square wave. If the outputisn’t a square wave, it means the amplifier isn’t passing the necessaryharmonic frequencies. As a technician, you’ll know how to interpretthe shapes of the output waveforms.

Instantaneous Sine Wave Voltage and CurrentThere are applications where you need to know the voltage value ofa sine wave voltage waveform at any instant of time, or the currentvalue of a sine wave current waveform at any instant of time. A basicequation for instantaneous voltage is

v = VM sin

where v is the voltage at the instant the rotating phasor (that createdthe sine wave voltage) reaches a chosen angle

A similar equation for instantaneous current is

i = IM sin

where i is the current at the instant the rotating phasor (that createdthe sine wave current) reaches the chosen angle

Let’s take a look at a problem using the two formulas just given.

Assume that a sine wave voltage having a peak voltage of 100 voltsand a frequency of 60 hertz has been stopped at 45 degrees. What isthe value of voltage at that instant?

In Figure 13A, the sine wave is shown on a degree axis. In Figure 13B,the sine wave is shown on a radian axis. To get a better understandingof the point we’re interested in, refer to Figure 14. The illustrationshows that a sine wave can be drawn on a degree or radian axis. Bothsine waves are produced by a rotating phasor having a length equalto the maximum voltage or current. When the phasor has rotated 45degrees, we want to stop it and determine the value of voltage at thatpoint. In our example, we’re going to determine that “instantaneousvoltage” three different ways.

First, looking at the problem graphically, you can see that the phasorthat produces the sine wave is drawn to scale at an angle of 45 degrees.The vertical distance from the end of the phasor is measured on thesame scale—that is, the voltage at the instant the phasor is stopped atan angle of 45 degrees.


Secondly, to solve the problem using the equation above, you see that:

v = VM sin = 100 sin 45

where VM = 100 volts, and is 45 degrees.

Use the degree axis to represent the sine wave with a rotating phasor.In Figure 15, the phasor has been drawn for the 45 degree point.


360°270°180°

(A)

POINT OF INTEREST = 2π/8 radians

v = Vm Sin 45° = 100 Sin 45° = (100)(0.707)

= 70.7V

v = Vm Sin θ

Sin 2π/8 rad = 0.707+

v = Vm Sin ωt + θ

90°

45°

0°

(B)

π/2 π 3π/2 2π0°

FIGURE 13—In Figure 13A,a sine wave is shown on adegree axis. In Figure 13B,a sine wave is shown on aradian axis.

RADIANS0

100V

π/2

v

π 3π/2 2π

DEGREES0

100V

90

v

180 270 360

FIGURE 14—This illustrationshows that sine waves canbe drawn on either a degreeor a radian axis. In bothcases, the sine waves aregenerated by a rotatingphasor.

To solve the above equation using a scientific calculator, perform thefollowing operations.


[100] [ [45] [SIN] [=] 70.71067812 Multiply 100 volts

the sine of 45 degrees.Answer: 70.7 volts.

Before we go on to the third method of solving the sample problem,we’ll introduce two equations for finding the instantaneous value ofsine wave voltage or current at any instant of time.

The first equation is

v = VM sin (t )

where v is the instantaneous value of voltage at any time t, VM is themaximum, or peak, voltage of the sine wave, w is the angular velocitymultiplied by the instant of time of interest (multiplying by t givesyou an angle in radians), and is the phase angle in radians (if it exists).

The second equation is

i = IM sin (t )

where i is the instantaneous value of current at any time t, IM is themaximum (or, peak) current of the sine wave, w is the angular velocitymultiplied by the instant of time of interest (again, this multiplicationproduces an angle in radians), and is the phase angle in radians(if it exists).


MAXIMUM VOLTAGE10 INCHES

45°

O DEGREE AXIS

7+ IN

CH

ES

70. 7VVOLTAGE AT THE

INSTANT THE ROTATINGPHASER HAS REACHED

45 DEGREES

FIGURE 15—This illustrationshows that instantaneousvoltage can bedetermined graphicallywhen drawn to scale.

There’s a tradition of writing the phase angle in degrees and the angu-lar velocity in radians. When that’s done, you must change the phaseangle (when it exists) into radians before it can be combined with t.

Note: When you’re using a scientific calculator to find the sine (or anyother trig function such as cosine or tangent) in radians, you mustswitch the calculator to the radians mode of operation. To do so, pushthe degree key or follow the instructions that came with your scientificcalculator.

Assuming you’ve changed to radian mode, let’s find the sine of 0.5radians on the scientific calculator.


[.5] [sin] [=] 0.4794255386 Find the sine of0.5 radians.

Answer: 0.479.

Note: A very common mistake made by students is to forget to switchthe calculator back to the degree mode when working problems in-volving angles in degrees. In some scientific calculators, some othercalculations won’t work properly unless the calculator is in the degreemode.

We’ll now continue with the third solution for the problem usingradian measurement.

Given: VM = 100 volts (maximum)

frequency = 60 hertz

Since the solution that involves radian measurement requires the timet, our first step is to find the time t it takes to get to the point where thephasor has turned 45 degrees. We know that 45 degrees is one-eighthof a complete revolution:

one-eighth revolution 45

360

1

8

degrees

degrees

The time for one complete cycle t can be obtained from the frequency(60 hertz).

tf

1 1

600 016666667.


For 18 cycle, the time is

time cycle secfor.

.1

8

1

8

0 016666667

80 002083333 t

You know that angular velocity = 2f. So,

angular velocity = = 2f = 2 3.14 60 = 377

Substituting the determined time (0.002083333) for t in the equationfor instantaneous voltage, 377 for , and 100 volts for VM gives

v = 100 [sin (377 sin 0.785416541 = 70.7 volts

Remember when using the scientific calculator that the sine of anangle expressed in radians must be determined in the RAD mode.The numbers are carried out in the solution to an unusual number ofplaces to avoid an error when rounding off values.

How a Capacitor Opposes a Change of VoltageRemember that a capacitor opposes any change in voltage across itsterminals. That concept is easier to understand by connecting a DCvoltage across a capacitor as shown in Figure 16A.

The capacitor, in this case, has a plastic dielectric. The switch is usedto deliver the DC voltage across the capacitor. The capacitor can’timmediately assume the same voltage as the battery. You can see thatfrom the charging curve in Figure 16B.

Scientists use the following model to explain what’s happening. Theytell us that dielectric materials have many tiny electric dipoles in them,and each dipole has a positive and a negative end (Figure 16C). Noticethat the switch is open.

Before a voltage is applied across the plates of the capacitor, the dipolesare arranged in a random direction as shown in Figure 16C. When thevoltage is applied, current begins to flow into one plate of the capacitorand out of the other (Figure 16D). That flow produces an electric fieldbetween the plates. The greater the difference in the number of electronsbetween one plate and another, the stronger the electric field.

The dipoles begin to align themselves with the electric field. Note thatthe negative sides of the dipole turn toward the positive charge on oneplate. At the same time, the positive sides of the dipoles turn towardthe negative capacitor plate—that is, the plate with the greatest numberof electrons. This action is based upon the very basic principle thatopposite charges attract.


The dipoles don’t turn freely. In fact, there’s opposition within thedielectric that opposes the realignment of the dipoles. However, theelectric field eventually succeeds in turning all of the dipoles.

If there was no opposition to the dipoles turning, they would turnvery rapidly, and the field would turn all of the dipoles at nearly thesame time.

Now you can understand the reason for the curve of the chargingcapacitor. The full voltage across the capacitor can only occur whenall of the dipoles are aligned. The current must flow into the platesto create the field between the plates, and the alignment of dipolesfollows the establishment of the field. The time it takes the capacitor


FIGURE 16—The behavior of a capacitor can be observed by connecting a DC voltage as shown in 16A. InFigure 16B, you can see if takes time for the charging capacitor to reach the same voltage level as the DCsource. In Figure 16C, you can see that the dipoles are in random directions. In Figure 16D, you can seethat the dipoles are now aligned in the same direction opposite to the current flow.

V

SWITCH

(A)

SWITCHCLOSED

(B)

TIME INCREASE

VOLTAGEACROSS

CHARGINGCAPACITOR

VO

LTA

GE

SWITCH (OPEN)

DIPOLES AREIN RANDOMDIRECTIONS

DIPOLE

(C) (D)

V

–

+

SWITCH (CLOSED)

DIPOLES ARE

ALIGNED

ELECTRONS

V

to reach the full voltage is very short, and it’s difficult to observe withthe best electronic equipment. However, the voltage across the capaci-tor doesn’t reach the maximum value instantly. It’s correct to say thatthe capacitor opposes the change in voltage across its terminals.

If you remove the applied voltage across the capacitor—by openingthe switch in Figure 16A—the capacitor doesn’t immediately lose itscharge. The reason for that is that some of the dipoles remain alignedfor a short period of time. That explains why the discharge curvedoesn’t immediately drop to zero volts.

Again, it’s proper to say that the capacitor opposes the change involtage across its terminals.

Capacitors with a Vacuum DielectricThe discussion on capacitors assumes there’s a mylar (or other non-conductor) between the plates. However, there are capacitors madewith a vacuum dielectric. That means there are no dipoles in thedielectric. So, how is the capacitor charged? Refer again to Figure 16D.

There are actually two mechanisms that must be explained regardinga charging capacitor. You’ve just studied the charging of a capacitorwhen there’s a dielectric present.

As shown in the illustration, the voltage across the capacitor doesn’tinstantly reach the applied voltage. As electrons move into the negativeplate, a negative charge accumulates in the plate. As that charge be-comes more and more negative, it opposes the flow of electrons intothe plate. That explains why the voltage across a capacitor with avacuum dielectric can’t immediately reach the applied voltage.

In a capacitor when a mylar (or other type of dielectric) is charged, thesame thing happens as with the vacuum dielectric. However, the effectof the negative accumulation of negative charges in a capacitor with avacuum dielectric is small compared to the effect of dipole action in acapacitor with a dielectric material.

Therefore, when explaining the action of a capacitor, the usual proce-dure is to disregard the effect of accumulating negative charges on thenegative plate and the depletion of electrons on the positive plate. Theamount of energy stored in a capacitor with a vacuum dielectric is smallcompared to capacitors with a dielectric material. However, the voltageat which the capacitor breakdowns is higher in a vacuum dielectric.


Resistance and Capacitance in DC CircuitsYou’ll recall that capacitors are components that oppose any change involtage across their terminals, and that inductors are components thatoppose any change in current through them. This opposing feature ofboth inductors and capacitors is the heart of many circuits. The circuitsin Figure 17 define the time-constant relationship for RC (resistive-capacitive) and RL (resistive-inductive) circuits having a DC-appliedvoltage. Let’s take a look at these circuit actions and time-constantequations.


R1

R1

R2

R2

RC CIRCUIT

(A)

RL CIRCUIT

(B)

E

E

L

2

2

3

3

1

1

S

S

CEL

ECTR

ON

CH

ARG

ING

CUR

REN

T

FIGURE 17—In Figure 17A,the capacitor can becharged and dischargedby this circuit. Whenswitch S is in position 2,the capacitor chargesthrough R1. The solid arrowshows the charge path.When the switch is inposition 3, the capacitordischarges through R2.The broken-line arrowshows the discharge path.Figure 17B represents anRL circuit where currentflow through the inductor,L, increases when theswitch is in position 2 anddecreases when S ismoved to position 3.

The time constant t for the charging capacitor is given by the formula

= RC

in which stands for time in seconds required for the capacitor tocharge to 63% of applied voltage, R stands for resistance, in ohms,through which the capacitor charges, C stands for capacitance, infarads.

When the capacitor is charging, the voltage across it changes, asshown by curve A in Figure 18.

When the switch in Figure 17A is moved to position 3, the capacitordischarges through R2. (The broken-line arrow shows the path of thedischarge current.) The time constant, which is the time required for thevoltage across the capacitor to decay to 37% of its original chargedvalue, is calculated using the same formula.

When the capacitor is discharging (switch position 3), the voltageacross it decreases along the curve marked B in Figure 18.

Resistance and Inductance in DC CircuitsThe subject of inductance will be taken up in greater detail later in thislesson. It’s introduced here for comparison with capacitors in DC circuits.When switch S is moved to position 2 in the circuit of Figure 17B, thecurrent through the inductor L starts to increase. However, since theinductor opposes this increase in current, it doesn’t reach its maximum


100

90

80

70

60

50

40

30

20

10

00 1 2

TIME CONSTANTS

CURVE A

CURVE B

PERC

ENT

OF

MA

XIM

UM V

OLT

AG

E O

R C

URRE

NT

3 4 5

FIGURE 18—These areuniversal time-constantpaths that show the chargingof the capacitor along thetime-constant curve markedA. The capacitor dischargecurve is marked B. Thesecurves also represent thebehavior of current in an RLcircuit.

value instantly. As a matter of fact, it doesn’t reach 63% of its maximumvalue until one time constant has elapsed. The time-constant formula is

L

R

in which stands for time constant, in seconds, L stands for inductance,in henries, R stands for resistance, in ohms.

The increasing current follows curve A of Figure 18. If switch S inFigure 17B is moved to position 3 after the current reaches its maxi-mum value, the collapsing magnetic field around the inductor willinduce a voltage that tries to keep the current flowing. This inducedcurrent will flow through R2 and the short circuit through switch S.The current through the inductor and R2 will decay along curve B inFigure 18.

Universal Time-Constant CurvesThe increase and decrease of voltage across a capacitor in an RC circuitfollow the graphs of Figure 18. Likewise, the increase and decrease ofcurrent in an RL circuit follow the same graphs.

An important feature of these curves is the fact that they don’t go beyondthe fifth time-constant points. The reason is that a capacitor is consid-ered to be fully charged (or fully discharged) after five time constants.Similarly, the current through an inductor reaches its maximum (orminimum) value after five time constants.

The following example problem shows you how to use the curves todetermine voltage at a specific time constant.

Example: Assume that a series RC circuit is connected across a 50-voltDC circuit. How much voltage will appear across the capacitor at atime equal to two time constants?

Solution: The universal time-constant curve (Figure 18) shows that, attwo time constants, the voltage across the capacitor is about 86% of theapplied voltage:

86% of 50 V = 0.86 V

Answer: 43 V


Delay-Before-Start CircuitsMany circuits used in industrial electronic systems operate on thetime-constant principle. The RC or RL circuit is actually part of a morecomplex electronic circuit, but the basic principles just studied applyto all such systems. Figure 19 gives two examples with the electronicportion of the system shown in block-diagram form.

The circuit in Figure 19A is designed to begin its operation at sometime after the switch S is opened. In other words, you have to openthe switch, and then wait for a certain period of time before the circuitbegins to operate.

Note that the capacitor is short-circuited by the switch in the closedposition. At such time, the input to the circuit is grounded; soVin = 0 V. The circuit can’t operate because it requires an inputvoltage of 10 V.


TIME

15V

10V

5V

+ 15V

CIRCUIT ISENERGIZED

WHEN Vin = 10V

R

C S

Vin

Vin

Vout

CIRCUITENERGIZED

CIRCUITDE ENERGIZED

SOPENED

TIME

15V

10V

5V

+ 15V

CIRCUIT ISDE ENERGIZEDWHEN Vin = 5V

S

R C

Vin

Vin

Vout

(B)(A)

SOPENED

FIGURE 19—This figure shows block diagrams and graphs showing how time-constant circuits are used todelay before start (Figure 19A) and before stop (Figure 19B).

The instant the switch is opened, capacitor C begins to charge throughresistor R. When the voltage across the capacitor reaches 10 V, thecircuit begins to operate and to produce an output signal at Vout.

The graph in Figure 19A shows that the charging capacitor causes theinput voltage (Vin) to increase along a curved path. Since Vin must be10

15 (or 6623%) of the applied voltage to operate the circuit, it follows

that the delay between opening the switch and obtaining an output isonly slightly longer than one time constant. (You’ll recall that one timeconstant occurs at 63% of the applied voltage.)

Delay-Before-Stop CircuitsIn the circuit of Figure 19B, the input to the circuit must drop to 5 V orless before the circuit is de-energized and the output is cut off. Whenthe switch is opened, C discharges through R, and the voltage across Rdecreases as the capacitor discharges.

This concept is easy to understand. When the switch is closed, thevoltage across both R and C, which is Vin, is equal to 15 V. When theswitch opens, C discharges through R, and Vin decreases as shown inthe graph in Figure 19B. If C discharges completely, Vin will go to 0 V.

The circuit in Figure 19B is actually used in a power supply protectioncircuit. The switch opens when the supply voltage drops. If the dropis only momentary, the supply isn’t shut off by Vout. However, if thevoltage Vin reaches 5 V before the supply voltage gets back to normaland closes the switch, the Vout will shut the supply off.

Note that Vin must drop to 515 of the supply voltage before the voltage

at Vout is cut off. (This value 515 is 331

3% of the supply, which is onlyslightly less than the voltage at one time constant, or 37% of V.)

For practical purposes, both circuits in Figure 19 can be considered asoperating one time constant after the switch is opened.

Capacitive ReactanceIf you apply an AC voltage across the plates of a capacitor, the capacitorwill continually oppose the changing voltage, and will also continuallyoppose the current. This opposition to the changing current is calledthe capacitive reactance. By definition, capacitive reactance is the oppo-sition of a capacitor to an AC current. It’s represented by the symbol XC.Capacitive reactance is measured in ohms.


How much reactance (opposition) does a capacitor offer to the flowof AC current? In a lab experiment, it can be shown that higher capaci-tance values offer lower opposition to AC current flow. It can also beshown that currents at higher frequencies encounter less opposition toAC current flow.

To summarize, the capacitive reactance decreases when the capacitanceincreases. Capacitive reactance also decreases when the frequencyincreases. Capacitive reactance and capacitance are said to be inverselyrelated. The equation for capacitive reactance is given here:

XfC C

c

1

2

1

where XC is the capacitive reactance in ohms, f is the applied ACfrequency in hertz, is the angular velocity, and equal to 2f is theangular velocity, and C is the capacitance in farads.

That equation is sometimes given as:

XfC

c 0 159.

because 12 is equal to 0.159.

Remember that the values of the frequency must be stated in hertz,and the capacitance must be stated in farads. That’s accomplished bywriting higher or lower values with scientific notation (such as milliand mega) or by a power of ten. For example, capacitance values areusually given in microfarads—abbreviated F or 10 –6 farads.

A scientific calculator doesn’t have words like micro, pico, kilo, andmega, so those notations must be converted to powers of 10 in order tobe used in calculations. A scientific calculator makes the conversioneasy.

The following powers of 10 represent the engineering notations:

106 = mega

103 = kilo

10–3 = milli

10–6 = micro

10–9 = nano

10–12 = pico


Let’s take a look at the following conversion examples:

2,100 nanofarads = 2100 10–9 farads

33 picofarads = 33 10–12 farads

123 milliamperes = 123 10–3 amperes

270 kilohms = 270 103 ohms

Writing prefixes in powers of 10 (as in the above examples) is calledengineering notation.

When a number is written as a number between 1 and 10 multipliedby a power of 10, it’s called scientific notation. Let’s take a look at someexamples of scientific notation:

0.000746 = 7.46 10–4

37,100,000 = 3.71 107

Some scientific calculators can convert numbers directly to engineeringor scientific notation. Follow these examples on your scientificcalculator for conversion to engineering notation.

Converting a number to engineering notation. Observe that the lettersENG may be located above a key. That means it can be accessed by the2nd key. Enter the number, then push the [2nd] key, and then [ENG].Next, push the [=] key. Let’s try doing the following conversions toengineering notation using the scientific calculator:

0.000746 farads = 746 10–6 farads = 746 microfarads


[0.000746] 0.000746 Enter the number to beconverted.

[2nd] [ENG] [=] 746–06 Convert to engineeringnotation.Answer: 746 10–6 farads= 746 microfarads.

Let’s try another example.


5,280 meters = 5.28 103 meters = 5.28 kilometers


[5280] 5280 Enter the number to beconverted.

[2nd] [ENG] [=] 5.2803 Convert to engineeringnotation.Answer: 5.28 103 meters

= 5.28 kilometers.

You should observe that each answer in the examples above has apower of ten that’s given for a prefix. For example: 300 106 hertzmeans 300 megahertz.

Converting a number to scientific notation using a scientific calculator.You should locate the letters SCI on your calculator. If the letters SCIare above a key, a number can be converted to scientific notation bypushing [2nd] and then [SCI].

Let’s try doing the following conversion to scientific notation using ascientific calculator:

300,000,000 = 3 108


[300000000] 300000000 Enter the number to beconverted.

[2nd] [SCI] = 308 Convert to scientificnotation. Answer: 3 108.

Remember, every number written with scientific notation is a numberbetween 1 and 10 times a power of 10; units like micro, pico, kilo, andmega are called engineering notation.

For example, in engineering notation an equation is expressed as:

0.000027 farads = 27 10–6 farads = 27 microfarads

In scientific notation, the same equation is expressed as:

0.000027 farads = 2.7 10–5 farads


The scientific calculator will process either the numerical value or thepowers of ten in calculations. If there’s room, the calculations mayrevert to numerical values. If you prefer to have the answer expressedin powers of 10, push the [EE] key.

Use your scientific calculator to perform the following typical operation:

Find the value of 27 megohms 10 picofarads.


[27] [EE] [06] 2706 Express as a power often.

[] 27000000 Pressing the [] keyshows you’re preparingto multiply.

[10] [EE] [] [12] 10–12 Express 10 picofarads asa power of ten by press-ing [10] [EE], then [] (toinput the minus sign).

[=] 0.00027 Complete the operation.Answer: R C = 0.00027.

When a number is expressed in scientific notation, it’s some numberbetween 1 and 10 multiplied by a power of 10.

To convert 0.00027 to scientific notation, enter that number and push[2nd] [SCI] = 2.7 10–4.

Remember that when using your scientific calculator to solve problemsusing the EE key, always use the key for negative exponents. Whenthe exponent is positive, the + is understood, so don’t try to obtain itwith the key.

Let’s solve the following problem.

Example: A 1.5-microfarad capacitor is connected across a 0.5-kilohertz sine wave generator (Figure 20A). What is the value ofcapacitive reactance in ohms?


Solution: Write the equation.

XfC

C

1

2

Convert microfarads to farads and kilohertz to hertz:

f = 0.5 kilohertz = 0.5 103 hertz

C = 1.5 microfarads = 1.5 10–6 farads

Then, using standard calculation:

XfC

c

1

2

1

2 0 5 10 1 5 10212 2

3 6( ) ( ( . ) ( . ).

–ohms

Or, solve for the denominator (2fC) on the scientific calculator:

Keystroke Display Description

[2] [] [] [] 6.283185307 Multiply 2 Enter []to prepare for the nextstep.

[.5] [EE] [03] [] 3141.592654 Continue the operationby multiplying by 0.503.Enter [ to prepare forthe next step.

[1.5] [EE] [ 06] [=] 0.004712389 Find the reciprocal.

[1/x] [=] 212.2065908 Answer: The capacitivereactance is 212.2 ohms.


FIGURE 20—Use Figure20A to determine thevalue of capacitivereactance. Use Figure20B to find the currentflow in the circuit.

C

(A)

2πfC1

f = 0.5 kHz

C = 1.5 µf

XC =

By way of review, (0.004712389) is the denominator of the answer.The [1/x] key puts it in the numerator, so XC = 212.2 ohms.

The illustration in Figure 20A has been redrawn in Figure 20B with thegenerator voltage added. The capacitive reactance of the capacitor hasalso been added. We’ll now determine the current in the circuit.

You’ll note that the RMS value of voltage—also called the “effectivevalue of voltage”—is given in the illustration. The RMS value isindicated by the use of V for RMS voltage.

Note: The maximum voltage (also called the peak voltage) is usuallyindicated with VM, and an instantaneous value of voltage is shownwith a lowercase letter v.

The equation for current is given here:

I V

X C

10

212 20 047 47

.. amps milliamps

How Does an Inductor Oppose a Change in Current?Earlier in this study unit, it was explained that motion between aconductor and a magnetic field causes a voltage to be generatedacross the conductor. There are many applications of this law.

An interesting example is the attempt to generate electricity by usingthe space shuttle to sweep a long cable through the earth’s magneticfield. In accordance with the rule about a voltage generated by a con-ductor and magnetic field having motion between them, that cablemoving through the earth’s magnetic field will produce a voltageacross it. An early attempt failed because the cable broke, but thetheory is still correct.

Remember, any time there’s relative motion between a conductor anda magnetic field, there’s always a voltage induced across the conductor.The amount of voltage depends upon how fast the magnetic field andconductor move in relation to each other, and the number of conductorsbeing moved.

The term relative motion needs further explanation. It doesn’t matter ifthe magnetic field is stationary and the conductor is moving throughthe field, or the conductor is stationary and the magnetic field is movingthrough the conductor.

In fact, the conductor and the magnetic field can both be moving aslong as they’re moving with relation to each other.


Faraday’s LawThe law relating a voltage generated by relative motion between aconductor and a magnetic field is called Faraday’s law. Faraday’s lawis written as an equation this way:

v Nd

dt

where v is the induced voltage at any instant of time, N is the numberof conductors involved, and is the symbol used for magnetic flux.

The expression d/dt means “the rate of change of flux (relative to theconductor) with respect to time.” In other words, it means the relativespeed between the conductor and the magnetic flux ().

The negative sign in the equation is very important. The negative signis due to Lenz’ law, which states that whenever there’s an induced volt-age, it will produce a current that opposes any change in currentthrough the conductor.

Lenz’ law can be demonstrated using the circuit in Figure 21A. Acurrent is presumed to be flowing through the coil, and the switch isclosed.

In Figure 21B, the switch is opened in an attempt to stop the currentflow. That produces an induced voltage. According to Lenz’ law, thatinduced voltage will oppose the change and try to keep the currentflowing.


FIGURE 21—Lenz’ law can be demonstrated using the circuits in Figure 21A. When the switch is open, a highcounter voltage appears at the switch terminals in Figure 21B.

L

(A)

L

SPARK

(B)

If there are enough turns of wire in the inductor, the induced voltagewill be so high that it will force an arc across the switch contacts.

In a short time, the arc of current across the switch contacts will stopbecause the induced voltage will decrease as the coil current eventu-ally stops flowing. (It eventually stops flowing because the switch isopen.)

The arc across the switch contacts is very important because it willeventually destroy the switch contacts with repeated operation of the switch!

Contact ProtectionThere are a number of ways to stop the destruction of the switchcontacts.

Figure 22 shows one way to protect the switch contacts. A capacitor isconnected across the switch. When the counter voltage from the coilappears across the capacitor, it stores the high energy and then slowlyleaks it back into the circuit.

You’ll remember that a capacitor is a component that opposes anychange in voltage across its terminals. So, you can say the capacitoropposes the change in voltage across its terminals and prevents arcingacross the switch contacts.


LV

C

FIGURE 22—The capacitorsuppresses the arcdischarge across theswitch terminals.

Counter VoltageThe induced voltage in a coil was, in earlier times, called a counterEMF. (EMF is a term that was used earlier in this study unit to meanvoltage.) In the latest use of terms according to the Institute of Electricaland Electronic Engineers (IEEE), counter EMF is now called countervoltage. You’ll sometimes see it referred to as kickback voltage.

You’ll still see EMF and counter EMF used in some texts because they’regood models for circuit descriptions. However, as an up-to-date tech-nician, you should remember that the term EMF isn’t considered to bea correct representation of voltage.

Consider the coil in Figure 23. Assume the current through the coil isincreasing. The magnetic field around each turn of the coil increasesand cuts through nearby turns. That induces a counter voltage in thoseadjacent turns. So, the counter voltages induced in all turns combine toproduce a total counter voltage in the coil.

The inductance of a coil is directly related to the number of turns of acoil. In other words, if you increase the number of turns, you increasethe coil inductance; when you decrease the number of turns, youdecrease the induction.

Also, the rate of change of magnetic flux around a current-carryingwire is directly related to the rate of change of current. From thisinformation, you can understand the second mathematical versionof Faraday’s and Lenz’ laws, which is

v Ndi

dt


FIGURE 23—A magneticfield shown with brokenarrows cuts acrossadjacent turns.

where v is the counter voltage that is induced at any instant of time t,N is the number of turns of wire in the inductor, and di/dt is the rateof change of current through the inductor.

When the current through a coil is decreasing, the magnetic field aroundeach turn is collapsing. That induces voltage in adjacent windings withan opposite polarity compared with the induced voltage with anincreasing current.

Think of it this way: an inductor opposes any change in currentthrough the coil.

The overall result is that the inductor causes an opposition to alternatingcurrent flowing in a coil. That opposition is called inductive reactance.

An Inductor Static Curve for High InductanceEarlier in this study unit, a basic circuit is shown with a battery, switchand inductor (Figure 24). The circuit illustrates what happens when aninductor is first charged.


TIMECONSTANT

CURVE

V

UNPREDICTABLE

TIME

I

FIGURE 24—The timeconstant for a largeinductor is different fromthe time constant for asmall inductor.

The charging curve in Figure 24 is different from the one shown previ-ously because a high inductance is presumed. The static curve for highinductance isn’t as easily drawn as for a capacitor. When the switch inFigure 24 is first closed, there’s a rapid change in current through theinductor.

Remember that the counter voltage is directly dependent upon therate of change of current and the number of turns in the coil. Since thecurrent changes rapidly when the switch is first closed, there’s a highcounter voltage generated. That high counter voltage can be verydestructive to the switch.

Look closely at the curve in Figure 24 and observe that the initial sectionof the curve is blank. That gap in the curve eliminates the first part ofthe curve because the amount of counter voltage—and its effect on thecurrent—can’t be predicted unless more is known about the inductanceof the coil.

If we move past that uncertainty, the curve in Figure 24 is a time-constant-charging curve. That means the current can’t rise to its finalvalue in an instant.

As with the circuit having only a capacitor, the time it takes for thecurrent to reach its maximum value is very short, with the staticcondition illustrated by the circuit having a low value of inductance.

Inductive ReactanceWhen an AC current flows through an inductor, it opposes the increasein current on one half cycle, and opposes the decrease in current onthe next half cycle. Again, the conductor opposes any change in current.

The overall result is that the inductor opposes the flow of AC current.That opposition is called inductive reactance, and it’s represented by thevariable XL.

The amount of opposition to a sine wave current is given by theequation

XL = 2fL

where f is the frequency of the current, L is the inductance of theinductor, and XL is the opposition the coil offers to the flow of ACcurrent (measured in ohms).

This equation is sometimes written as:

XL = L


Let’s solve the following problem.

Example: The inductance of the coil in Figure 25 is 1.5 millihenries,and the frequency of the current is 2.5 kilohertz. The applied AC voltageis 10 V. What is the inductive reactance of the coil? How much currentis flowing through the circuit? (Disregard any transient condition inthe startup of the circuit.)

Solution A: XL = 2fL

= 2 3.14 103 1.5 10–3

= 23.6 ohms

Solution B: I =V

XL

10

23 6.= 423 miliamperes


10V2.5 kHz 1.5 mH

FIGURE 25—Theinductance of thecoil is 1.5 millihenries.


Power Check 2

1. What is the symbol for angular velocity?

____________________________________________________________________________

2. What is the equation for angular velocity when the time, t, for one cycle is known?

____________________________________________________________________________

3. What is the equation for angular velocity when the frequency of the projected sine wave isknown?

____________________________________________________________________________

4. How does the length of a rotating phasor affect the projected sine wave?

____________________________________________________________________________

5. How many radians are there in one complete revolution of a phasor?

____________________________________________________________________________

6. Arcing at switch contacts in an inductive circuit can be greatly reduced by connecting a_______ across the switch contacts.

7. The dielectric of a capacitor is a material considered to be made up of many tiny _______.

8. True or False? In the region between the plates of a charged capacitor, there are electricfield lines.

9. Can a capacitor with a vacuum dielectric be charged?

____________________________________________________________________________

10. The opposition that a capacitor offers to the flow of alternating current is calledcapacitive _______.

11. True or False? XC = 12fc is the correct equation for the opposition a capacitor offers to

the flow of AC current.

12. Convert 165 kilohertz to hertz.

____________________________________________________________________________

13. Convert 1.50 microfarads to farads.

____________________________________________________________________________

(Continued)


Power Check 2

14. The opposition that an inductor offers to the flow of alternating current is calledinductive _______.

15. Anytime there’s relative motion between a conductor and a magnetic field, there’s a voltagegenerated across the conductor. This is a statement of _______ law.

16. The opposition offered by an inductor to the flow of alternating current is called _______.

17. How much AC current flows in a 2-henry inductive circuit if the applied voltage is 120 voltsat 60 hertz?

____________________________________________________________________________

18. What is the capacitive reactance of a 27 microfarad capacitor in a 400 hertz circuit?

____________________________________________________________________________


DETERMINING IMPEDANCE AND PHASE ANGLES FORRC, RL, AND RLC CIRCUITS

Review of Right Triangle CalculationsIn the next section, you’ll find a number of applications that requireyour understanding of right triangle calculations. It’s a good idea toreview these calculations at this time.

A right angle is defined as a 90-degree angle. Any triangle that has aninside angle of 90° is called a right triangle. Remember, in any triangle,all of the inside angles of the triangle will add to 180°. In the right tri-angle of Figure 26, angles A and B will add to equal 90° so that—whenadded to right angle C—the total number of degrees in the triangleis 180°.

It’s important to understand the Pythagorean theorem for right trian-gles. It states that the sum of the squares of sides equals the square ofthe hypotenuse. In Figure 26, the hypotenuse is marked with the letter c.Notice that the sides of the triangle are marked with lowercase letters,and the angles are marked with capital letters.

In the form of an equation, the Pythagorean theorem is

a 2 + b 2 = c 2

where a, b, and c are the lengths of the sides of a right triangle.

It’s very likely that you’ll be asked to find the length of the hypotenusewhen you know the lengths of the sides. Here’s the equation that youwould use:

c a b 2 2


A C

B

ac

b

SYMBOL FOR 90

FIGURE 26—A + B = 90degrees in any righttriangle.

There are three relationships based upon trigonometry that youshould know. They’re sine, cosine, and tangent. They’re ratios basedupon the lengths of the sides. When reviewing the following relation-ships in equation form, refer to the standard right triangle shown inFigure 26.

sin Aa

c

cos Ab

c

tan Aa

b

To practice solving problems using the equations above on a scientificcalculator, follow these steps:

Find the sine of 30 degrees.

Keystrokes Display Calculator

[30] [SIN] [=] 0.5 Answer: The sine of 30degrees is 0.5.

Find the cosine of 30.


[30] [COS] [=] 0.866025404 Answer: The cosine of 30degrees is 0.866.

Find the tangent of 30 degrees.


[30] [TAN] [=] 0.577350269 Answer: The tangent of30 degrees is 0.577.

Note: The [=] is optional in these examples, but it’s a good idea forsome calculations.

Now, let’s solve a problem using these functions.

Example: What is the sine of angle A of the triangle shown inFigure 27A?


Solution: Sine angle A = opposite side hypotenuse 49.4 = 0.526.

Remember that 0.526 is the sine of angle A. If you want the value ofangle A in degrees, you must take the inverse sine of 0.526, which iswritten sin–1 0.526. That does not mean sine raised to the –1 exponent.It’s simply a mathematical way of saying inverse sine. On a scientificcalculator, it’s usually found by taking the second function [2nd] ofthe sine.

Example: [0.526] [2nd] [SIN] = 31.7 degrees.

Figure 27B shows the triangle of Figure 27A with the values of thesides and the value of angle A. Use the Pythagorean theorem toshow that the correct value is shown for the hypotenuse.

c 42 26 49 3962 2 .

Reactances and Impedances as PhasorsFigure 28 is an illustration of how we use four quadrants to plotphasors. Observe in Figure 28A that the quadrants are numberedcounterclockwise. This arrangement is used in the United States, andit has several names: x-y plot, x-y coordinates, and Cartesian coordinates.


FIGURE 27—From theinformation in Figure 27A,you can find all othervalues of the triangle inFigure 27B.

You can use the information in Figure 28B to find the impedancephasor, Z

Z R XL2 2 2 2100 100 141 ohms

The phasors you’ll be dealing with will always be in the first or fourthquadrants. That includes the lines that are used for boundaries ofthose quadrants. The zero-degree line is sometimes called the referenceline, because all phasors are positioned with reference to that line.

In the series circuit of Figure 29A, the phasors are marked for thevoltages across the resistance, inductive reactance, and capacitivereactance. For instance, the voltage across the capacitor is I times XC,the capacitive reactance. The voltage across the resistance is in thezero position. In the phasor diagram, the current and voltage areseen to be in phase in that position.


+ y

y

(A)

(C)(B)

II I

III IV

+ x x

V XL = 100ΩZ

XL

R

R = 100Ω

Z = 141Ω

FIGURE 28—The RL phasorsrepresenting the series RLcircuit can be resolved asin (C).

The voltage across the inductive reactance (IXL) is leading the currentphasor because the current lags the voltage across an inductor.Remember that the positive direction of rotation of the phasors iscounterclockwise.

The voltage across the capacitive reactance (IXC) is lagging 90 degreesbehind the current. Another way of saying that the capacitive reactanceis lagging 90 degrees behind the current is that the current is leadingthe voltage across the capacitor.

In Figure 29C, each of the phasors in Figure 29B has been divided by I.The result shows that resistance, capacitive reactance, and inductivereactance can be represented by phasors.

As an example, the circuit of Figure 30A has only inductive reactanceand resistance. The phasor representation is shown in Figure 30B.


(A) (B)

XL

XC

R

RI V(VOLTS)

IR

ORIGIN

PHASOR DIAGRAM

IXL

IXL

IXC IXC

(C)

R

XL

XC

FIGURE 29—A series RLCcircuit with voltage formulasacross each component isshown in Figure 29A. InFigure 29B, a phasordiagram shows the currentresistance, and voltagephase relationship. In Figure29C, you’ll see that when thephasor diagram is dividedby I, you get a phasordiagram for components.

The phasors in Figure 30B can be combined into a single phasor asshown in Figure 30C. This is called vector addition, or resolving thephasors.

In order to visualize vector addition, refer to Figure 31A. Two dogsare pulling a sled. Their forces on the sled are vectors because theyrepresentforces applied at a distance. The vectors are marked on thedrawing. Note: The actual direction of the sled is the vector sum ofthe two vector forces produced by the dogs.

Figure 31B shows the vectors are added by the parallelogram method.Lines are drawn parallel to the vectors to form a parallelogram. Anarrow is drawn from the origin—which is the location of the sled—tothe points where the added lines meet. That line is called the resultantof the two vectors, or the vector sum.

This same technique was shown in Figure 30C for resolving phasors.In that case shown in Figure 30C, the parallelogram was actually arectangle.

The resultant of the phasors in Figure 30C represents the total opposi-tion to current flow in the circuit of Figure 30A.

That opposition is called the impedance. The impedance of an ACcircuit is the combined opposition due to all of the resistances andreactances. That combined opposition is obtained by resolving allof the phasors that represent opposition to current flow, and it’srepresented by an uppercase Z.


V

(A)

(B) (C)

XL = 100Ω

X L = 1

00

Ω

XL

R = 100Ω

R = 100Ω R

141+ Ω

FIGURE 30—In Figure 30A,a series RL circuit hasa phasor diagram ofFigure 30B, which can beresolved to show animpedance of 141 ohmsas in Figure 30C.

The length of the impedance phasor is measured in the same units asthe resistor and inductive reactance phasors. In other words, if thosephasors are drawn so that one inch represents 100 ohms, then thelength of the impedance phasor can be measured, and the impedance—in ohms—can be determined.

The impedance in any AC circuit can be obtained by drawing thephasors for R, L, and C on graph paper and by measuring the lengthof the resultant to determine the impedance value in ohms. Let’s workon a problem using the graphical method.

The answer to the graphical solution can’t be determined as accuratelyas a mathematical solution. However, a graphical solution can be usedas a method of checking a math solution. Also, it’s easier in most casesthan a math solution.

Example: What is the impedance of the circuit in Figure 32A, and howmuch current will flow in the circuit?

Solution: Figure 32B shows the R and XC phasors drawn to a scale of1 inch represents 100 ohms. The illustration has been reduced forpresentation here, but the inch markings are clearly shown. As youcan see, the phasor representing the resistor is three inches long torepresent 300 ohms. The phasor that represents a capacitive reactanceof 400 ohms is 4 inches long.


FIGURE 31—Two dogs pulling a sled in different directions introduce a vector problem in Figure 31A. InFigure 31B, the vectors can be resolved to show the resultant direction of the sled.

(A)

DOG #1

SLED

DOG #2

(B)

DOG #1

ACTUAL DIRECTIONOF SLED

SLED

DOG #2

The impedance, which is the resultant, is 5 inches long. Therefore, theimpedance, Z, of the circuit is 500 ohms. The current in the circuit isobtained by an Ohm’s law type of equation:

I =V

Z= =

500 volts

500 ohms1 ampere

Working with Math SymbolsAn important part of this study unit is the introduction of math symbolsthat you may encounter in technical literature. Your understandingof electronics is directly related to your understanding of what thesymbols mean, and your ability to work with the symbols.

In this section, we’ll introduce some additional math symbols that willbe used for working with right triangles that are related to impedancetriangle problems.

In any right triangle, the relationships of the sides to the angles is thesame regardless of the size of the triangle. In other words, if the anglesare the same, the relationships are the same whether the sides aremeasured in inches, centimeters, or miles.

The triangle in Figure 33 has sides in the ratio of 3 to 4 to 5. This typeof triangle is used so often it’s sometimes called a standard 3-4-5 righttriangle (Figure 33).


500V

R

(A) (B)

400ΩXC

300Ω

R = 300Ω

Z = 500Ω

XC

= 4

00

Ω

FIGURE 32—The series RCcircuit in Figure 32A can berepresented by R and XC

phasors. The resultingimpedance phasor can beobtained by resolving the Rand XC phasors (Figure 32B).

You’ve already learned the three basic right-angle relationships: sine,cosine, and tangent. There are three additional relationships for thestandard triangle. Here are those relationships and their names (thevariables used refer back to Figure 26):

Cosecant ( )csc Ac

a

Secant ( )sec Ac

b

Cotangent ( )cot Ab

a

Although you won’t see these relationships as often as sine, cosine,and tangent, you’ll encounter cosecant, secant, and cotangent intechnical literature, and you should know them.

In some cases, you’ll be given the value of a trigonometric relation-ship, such as the value of the sine, and you’ll need to be able to findthe angle. This angle is called the inverse sine, and it’s represented bythe symbol sin–1.

For example, suppose you know that sin A = 0.26. Now that youknow the sine of the angle, you can determine the angle. The equa-tion would be

sin–1 0.26 = number of degrees = 15.07°.

The inverse sine is sometimes identified that way on scientificcalculators.

Let’s solve the above problem on the scientific calculator.


[.26] [2nd] [SIN] [=] 15.07006214 Find the inverse sine of0.26. Answer: 15.07°.

The symbolism is the same for inverse cosine and inverse tangent.


4

53

FIGURE 33—This is a 3-4-5triangle.

Let’s practice finding the inverse cosine of the following problem usingthe scientific calculator. Again, we’re assuming that your calculator willperform the inverse cosine function by hitting the [2nd], then [COS]keys. Be sure this is the case.

Let’s find x when we’re given cos x = 0.555.


[.555] [] [2nd] [COS–1] [=] 56.28928522 Find the inverse cosineof 0.555. Answer: x = 56.3°.

The j OperatorIn the world of mathematics, an operator is a symbol that tells you whatto do with what follows. Some examples of operators are +, –, , and .In the days before World War II, a vacuum tube amplifier that wasdesigned to solve those arithmetic operations was called an operationalamplifier. Today, an operational amplifier is designed to perform manymore arithmetic operations.

The j operator is a special symbol that tells you which direction to turnon a graph. The expression a +j tells you to turn left 90 degrees. Theexpression a – j tells you to turn to the right 90 degrees.

Figure 34 shows examples of using +j and –j for locating points on agraph. Let’s try plotting two different points using a j operator.

Example 1: Plot the point 4+j3 on a graph.

Solution: Figure 35 shows how the plotting is done. The 4 is consideredto be positive. (The first number is always considered to be positive.) Inthis case, the positive first number tells you to move to the right. It tellsyou to start at the origin and move to the right 4 units. The +j3 tells youto turn left 90 degrees and move upward three units. The point 4+j3 ismarked on the graph.


O+j

j

FIGURE 34—This illustrationshows the graphicaldirection of +j and –j.

Example 2: Plot the point 3–j4 on a graph.

Solution: Figure 36 shows how the plotting is done. The positive 3tells you to move 3 units to the right. The –j4 tells you to turn to theright and go 4 units. The point 3–j4 is marked on the graph.

Both of the problem-solving examples show that j is an operator thattells you what to do with the next number. The 4+j3 and 3–j4 valuesare called complex numbers. The 4+j3 and 3–j4 values are the two legs ofa right triangle as shown in Figure 37.

You should think of the complex impedances as being complete de-scriptions of an impedance that form the legs of an impedance trian-gle. The symbol for an impedance described that way is , and you’llsometimes hear it referred to as z-dot. Here are some examples of theway z-dot is written:

= 5+j10

and

= 230–j124


10

2 3 4

1

2

3

4 + j3

4 + j3

FIGURE 35—Location of4+j3

10 2

2

3

4

1

3

3 j4

FIGURE 36—Location of3–j4

Remember this important thing about z-dot. Z-dot isn’t a value ofimpedance. Z-dot is a method of describing impedance. In the worldof mathematics, a z-dot is called “a complex number in rectangularform.” Later, we’ll discuss complex numbers in other forms.

An important thing to know is that complex impedances can betreated as numbers in computation. As an example, consider thefollowing two phasors: 7+j5 and 5–j3. If it becomes necessary to addthose phasors—because they represent complex impedances in se-ries—they can be added as numbers as shown in Figure 38. As youcan see, the j-numbers—sometimes called imaginary numbers—areadded separately to get the phasor sum.

Rectangular coordinates are important in electronics because theyestablish the position of Z, which is the magnitude of the impedance.Indirectly, Z also represents the angular position of the impedance byresolving the real and imaginary parts of the phasors. As shown inFigure 38, the real part of the rectangular impedance represents theresistive part, and the imaginary part represents the reactance. The termimaginary is unfortunate: it comes from the fact that it’s impossible totake the square root of –1. However, the term imaginary is useful fordescribing phasor impedances.


R

R

Z

Z

+jXL

jXC

FIGURE 37—The j operatorsrepresent legs onimpedance triangles.

CoordinatesPolar coordinates and Euler (pronounced oiler) notation are two addi-tional ways to describe phasor impedance. Polar coordinates andEuler notation are similar, and they can be studied together.

As shown in Figure 39, the magnitude and phase angle of an impedancecan be used to completely describe an impedance on a graph. Alwaysremember that the magnitude of an impedance in a series circuit isobtained from the equation:

Z R X XL c 2 2( )


7 + j5

7 + j5

5 j3

5 j3

7 + j5

5 j3

12 + j2 EQUALS PHASOR SUM

Z

V

Z

FIGURE 38—Addition ofRectangular Coordinates

=

FIGURE 39—TheIllustration ofImpedanceRepresented as aPolar Coordinate

Figure 40 shows two impedance (Z1 and Z2) connected in parallel. Todetermine the total impedance (ZT), use the standard formula given inFigure 40.

The solution is as follows:

ZT Z Z

Z + Z1 2

1 2

ZT ( )( )10 25 3 30

Z + Z1 2

ZT 30 55

Z + Z1 2

Calculate Z1 and Z2 as follows:

Z1 = 10 25°Z1 = 10 (cos 25° + j sin 25°)Z1 = 10 (0.9 + j0.4)Z1 = 9 + j4

Z2 = 3 30°Z2 = 3 (cos 30° + j sin 30°)Z2 = 3 (0.8660 + j0.5)Z2 = 2.5 + j1.5

Then, substitute the rectangular coordinate back into the formula:

Zj j

T

30 55

9 4 2 5 1 5

( ) ( . . )

ZT 30 55

11 5 5 5

. .


=

FIGURE 40—This illustrationshows how polarcoordinates and polar torectangular conversions canbe used to solve a parallelimpedance problem.

Convert the rectangular form 11.5 + j5.5 (a + jb) to polar form (r )as follows:

r a b2 2

r 11 5 5 52 2. .

r 12 75.

Phase angle: (pronounced theta) = tan-1 5 5

11 5

.

.

= 25.5°

Thus, in the polar form of r

Z1 + Z2 = 12.75 2.5

ZT =30 55

12 75 25 5

. .

ZT = 2.35 29.5°

As you can see from the preceding example, finding the combinedimpedance of two parallel impedances, it becomes necessary toconvert from rectangular form to polar form, and from polar form torectangular form.

Here are some important things to remember about the rectangularand polar forms:

Impedances can be easily added and subtracted when they’re inrectangular form. They cannot be added when they’re in the polaror Euler forms.

If impedances are to be multiplied or divided, it’s easiest to do ifthey’re in polar form.

Your scientific calculator should have a keypunch series thatpermits direct rectangular-to-polar and polar-to-rectangularconversions.

It’s helpful to know that capital letters from the beginning of thealphabet are used to represent numbers in rectangular form; capitalletters near the end of the alphabet are used to represent numbers inpolar and Euler forms.


Power in AC CircuitsIn a DC circuit, there are several ways to calculate power:

P = V I

P =V

R

2

P = I 2R

In an AC circuit in which there’s no phase angle between the voltageand current, the same equations are used. However, if there’s nophase angle between the voltage and current, the symbols V and I inthe above equations represent RMS values.

Remember that the RMS values of voltage or current and the averagevalues of voltage and current can be obtained from the maximumvalues:

VRMS = 0.707 VMAX

IRMS = 0.707 I MAX

If more accurate values are required, replace 0.707 with 1 divided bythe square root of 2:

1

2 0 707106781.

Note: Average values of voltage and current (which aren’t used in thisstudy unit) are determined from the maximum values:

VAVE = 0.636 V MAX

IAVE = 0.636 I MAX

If more accurate values are needed, replace 0.636 with 2 divided by

20.636619772

If you multiply the RMS voltage by the RMS current of a circuit thathas a phase angle, you won’t get the true power for that circuit. Thereason is that the RMS voltage and the RMS current maximum valuesdon’t occur at the same time, and the products don’t take the phaseangles into consideration. So, V I gives the apparent power, and it’smeasured in volt-amperes.


Refer to Figure 41A to see the phasor that represents apparent power.The phase angle () is the same as the phase angle between the voltageand current in the AC circuit.

If you complete the triangle that involves the apparent power and thephase angle, the result is the power triangle shown in Figure 41B. Youshould memorize the power triangle, because it defines all of the typesof power in an AC circuit that have a phase angle.

From this triangle and the definition of cosine given earlier, you canwrite the following equation:

cos true power

apparent power

From the equation above, you can write the following very importantequation:

true power = apparent power cos

You can think of cos as being the quantity needed to change theapparent power into true power. This quantity is called the power factor.

So, true power = apparent power power factor.

A power factor indicates that there’s a phase angle between thevoltage and the current in the circuit. A power generator will becomeoverheated when there’s a phase angle in the circuit the generatorsupplies with power.


(B)

APPAREN

T POWER

=V X I

θ

(A)

APPAREN

T POWER

=V X I

θ

TRUE POWER =V X I COS θ

VARS

(REA

CTI

VE V

OLT

-AM

PERE

S)

FIGURE 41—Figure 41Ashows that in an AC circuitthe apparent poweris V I.

The problem is so serious that power companies have the right to tax acompany that’s causing the voltage and current to get out of phase. Inorder to avoid the problem, power companies have automatic equip-ment that regulates the power and that introduces a phase angle ofzero degrees on the power line.

The cosine of zero degrees is equal to 1.0. So, the ideal value of thepower factor is 1.0.

A Z-angle meter is used to measure the phase angle between voltageand current. A version of the Z-angle meter automatically measuresthe power factor on the power line.

The actual power, also called the true power or resistive power, is equal tothe product of the RMS voltage and RMS current multiplied by thecosine of the phase angle:

true power = V I cos

In the power triangle, the term VARS stands for reactive volt-amperes.VARS is an imaginary power that would be dissipated by reactivecomponents (inductors and capacitors) in a circuit.

A Special Series RLC CircuitDo you remember the impedance equation we used earlier?

Z R X XL c 2 2 ( )

When a series AC circuit has resistance, capacitance, and inductance,you have to resolve the XC and XL phasors before solving for theimpedance value. As shown in Figure 42, these phasors are 180° outof phase. Their resultant is obtained by simple subtraction. The usualprocedure is to subtract the smaller from the larger and take the reac-tance of the larger.

Example: In a series AC circuit, R = 100 ohms, XL = 50 ohms, andXC = 50 ohms. What is the impedance of the circuit?

Solution: First, resolve the reactance phasors. X = XL – XC = 0.

Since the reactances cancel, there’s only a resistance of 100 ohmsserving as the impedance.

Note that it doesn’t matter whether XL or XC is larger—after thephasor values are subtracted, the result is squared, and a squaredvalue will always be positive.


The impedance formula we just used is called a general-case formula,because it can be used for any AC series circuit. If any one of the threecomponents (R, L, or C) is missing in the circuit, that component isgiven a value of zero.

Now, take time to reinforce your understanding of phasors for R, XC

and XL by completing Power Check 3.



X Z

Y

zx

y

Power Check 3

1. A right angle is an angle that measures _______ degrees.

2. A right triangle always has one angle that measures _______ degrees.

3. The sum of all of the inside angles in a triangle must equal _______ degrees.

4. Write the Pythagorean theorem for a right triangle having sides marked s, t, and u(where u is the hypotenuse).

____________________________________________________________________________

5. Write the equation for finding the length of the hypotenuse for the triangle described inQuestion 4 above. (Side u is the hypotenuse.)

____________________________________________________________________________

Refer to the triangle in Figure 42 to answer questions 6–12. Calculate the numericalvalues requested.

6. sin x = __________

7. sin y = __________

8. cos x = __________

9. cos y = __________

10. tan x = __________

11. tan y = __________

12. Calculate the value of the following: sin230° + cos230° = ________.

(Continued)


Power Check 3

Calculate the numerical values of the following:

13. If sin H =5

6 4., then H = _______ degrees.

14. If cos H =4

6 4., then H = _______ degrees.

15. If tan H =5

4, then H = _______ degrees.

16. Given: sin x = 0.445, find angle x.

____________________________________________________________________________

17. Given: cos x = 0.018, find x.

____________________________________________________________________________

18. Given: tan x = 1.5, find x.

19. Given: sin x = 1.0, find x.

____________________________________________________________________________

20. How are quadrants on an x-y graph numbered?

____________________________________________________________________________

21. True or False? The direction of rotation for phasors in the United States is counter-clockwise.

22. An arrow that represents magnitude of speed and direction on earth is called a _______.

23. The combined opposition to current flow (due to R, XC, and XL) in an AC circuit is called_______.

24. A series RCL circuit has an impedance (Z) of 240 ohms. An applied AC voltage of 8 volts isconnected across the circuit. How much current is flowing in the circuit?

____________________________________________________________________________

(Continued)


Power Check 3

25. If the sine of an angle is 0.345, what is the angle in degrees?

____________________________________________________________________________

26. In what parts of an x-y graph will you find impedance phasors?

____________________________________________________________________________

27. If a sine wave takes 0.001 second to complete one cycle, what is the frequency?

____________________________________________________________________________

28. If the angular velocity of a sine wave is 377, what is the frequency?

____________________________________________________________________________

29. What is the name given for the cosine of the phase angle between V and I in an ACcircuit?

____________________________________________________________________________


NOTES


1

1. True

2. The energy is stored in the surroundingmagnetic field. (The energy is returnedwhen the current through the inductoris interrupted.)

3. Resistors; capacitors; inductors

4. An inductor

5. active

6. Wire-wound resistors

7. In a capacitor, the energy is stored in thedielectric material between the plates.

8. Two effects of AC current flowingthrough a resistor are heat generationand an AC voltage drop across theresistor.

9. A capacitor

10. False

11. velocity

12. radians

13. /2 radians

14. 1.5 radians

15. 0.667 radians

16. 30°

17. 360°

18. 57.3°

2

1. (omega)

2. =2

t

3. = 2f

4. The length of the phasor determines theamplitude of the projected sine wave.

5. 2 radians

6. capacitor

7. electric dipoles

8. True

9. Yes

10. reactance

11. True

12. 1.65 103 hertz

13. 1.50 10–6 farads

14. reactance

15. Faraday’s

16. inductive reactance

17. First, find the inductive reactance:

XL = 2fL = 2 60 Hz ohms

Then: I = VX L

=120

754

V

ohms= 0.159 A =

159 mA

Power Check Answers

67

18. Start by writing the equation forcapacitive reactance:

XfC

c

1

2

Next, calculate the value of the denomi-nator.

2 2 400 0 000027 0 0679 fC . .

Then, take the reciprocal (1/x) of thedenominator to get XC:

X c 1

0 067914 7

.. ohms

3

1. 90

2. 90

3. 180

4. u2 = s2 + t2

5. u s t 2 2

6. sin x = xz

7. sin y = yz

8. cos x = yz

9. cos y = xz

10. tan x = xy

11. tan y = yx

12. In symbols, sin2 30° means to find thesin of 30 degrees and then square it.Likewise, cos2 30° means to find thecos of 30 degrees and then square it.

sin2 30° = 0.25 degrees, and cos2 30° =0.75 degrees. So, 0.25° + 0.75° = 1.

13. H = sin–1 56 4. = 51.38 degrees

14. H = cos–1 46 4. = 51.32 degrees

15. H = tan–1 54 = 51.34 degrees

16. Angle x = sin–1 0.445 = 26.42 degrees

17. Angle x = cos–1 0.018 = 88.97 degrees

18. Angle x = tan–1 1.5 = 56.31 degrees

19. Angle x = sin–1 1.0 = 90 degrees

20. I, II, III, and IV

21. True

22. vector

23. impedance

24. I = VZ 8

240= 0.033 amps = 33 mA

25. Angle x = sin–1 0.345 = 20.18 degrees

26. I and IV

27. 1,000 cycles per second

28. 60 hertz

29. Power factor

68 Power Check Answers

Reactance and Impedance

When you feel confident that you have mastered the material in this study unit, complete the followingexamination. Then submit only your answers to the school for grading, using one of the examination answeroptions described in your “Test Materials”envelope. Send your answers for this examination as soon as youcomplete it. Do not wait until another examination is ready.

Questions 1–20: Select the one best answer to each question.

1. Which of the following is not a passive component?

A. Transistor C. CapacitorB. Resistor D. Inductor

2. Which of the following is a component that opposes any change in current through it?

A. Air-dielectric capacitor C. CoilB. Mylar-dielectric capacitor D. Resistor

3. According to Faraday’s Law, every time there’s motion between a conductor and magnetic field, there’s

A. a current generated. C. friction.B. heat generated. D. a voltage generated.

Examination 69

EXAMINATION NUMBER:

08603700Whichever method you use in submitting your exam

answers to the school, you must use the number above.

For the quickest test results, go tohttp://www.takeexamsonline.com

http://www.takeexamsonline.com

4. When an AC sine wave current flows through an inductor, the current

A. leads the voltage. C. is a square wave.B. is in phase with the voltage. D. lags the voltage.

5. How many radians are equal to 57.3 degrees?

A. C. 1.0B. 2 D.

6. Which of the following is a component that opposes any change in voltage across its terminals?

A. Resistor C. InductorB. Capacitor D. Transistor

7. Components that do not generate a voltage are called _______ components.

A. intrinsic C. bilateralB. active D. passive

8. How many radians are there in 90 degrees?

A. radians C. /4 radiansB. 2 radians D. /2 radians

9. Which of the following is a correct statement concerning omega ()?

A. is no longer being used. C. must be converted to radians.B. is equal to 2f. D. must be converted to degrees.

10. Capactive reactance is measured in

A. ohms. C. microfarads.B. farads. D. henries.

11. On a graph of a sine wave, time

A. gets later as you move to the right.B. stands still if the graph is plotted on a degree axis.C. gets later as you move to the left.D. stands still if the graph is plotted on the radian axis.

12. Which of the following is correct?

A. radians = 90° C. radians = 270°B. radians = 180° D. radians = 360°

13. How long does it take a capacitor to charge through a resistor to 63% of the applied DC voltage?

A. 50% of one time constant C. one time constantB. 0.63 of one time constant D. 0.63 seconds

14. Which of the following is correct for scientific notation?

A. 600,000 Hertz C. 600 103 HertzB. 6000 102 Hertz D. 6 105 Hertz

70 Examination

15. Which of the following is the meaning of di/dt?

A. Magnitude of current C. Current at some instant of timeB. Change in current D. Rate of change of current

16. Which of the following is the same as 6 nanofarads?

A. 6 10–9 farads C. 6 10–12 faradsB. 6 10–6 farads D. 6 106 farads

17. Which of the following is correct for determining capacitive reactance?

A. Xc = 2f/t C. Xc = 0.159/fc

B. Xc = 1/C D. Xc = 0.159 R C

18. Any periodic wave, regardless of its wave shape, can be constructed with a combination of

A. square waves. C. triangular waves.B. sine waves. D. angles.

19. In the time-constant curve of a resistor and large inductor, the initial part of the curve that representsthe start-up of current is blank. That blank space is due to

A. inductive reactance. C. kickback voltage.B. slow switching speed. D. switching delay.

20. Which of the following is correct for calculating inductive reactances?

A. XL = Ldi

dt

C. XL = time constant L

B. XL = 2fL D. XL = 12fL

Examination 71

Documents

23.Reactance and Impedance