alternating current (ac) vi circuits containing inductance

466

SECTION

VIAlternating Current (AC)Circuits ContainingInductance

Cou

rtes

y of

Get

ty I

ma

ges,

Inc.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 466

OUTLINE

17–1 Inductance17–2 Inductive Reactance17–3 Schematic Symbols17–4 Inductors Connected

in Series17–5 Inductors Connected

in Parallel17–6 Voltage and Current

Relationships in anInductive Circuit

17–7 Power in an InductiveCircuit

17–8 Reactive Power17–9 Q of an Inductor

KEY TERMS

Current lags voltageInduced voltageInductance (L)Inductive reactance, (XL)Quality (Q)ReactanceReactive power (VARs)

Why You Need to Know

Inductance is one the three major types of loads found in alternating current circuits. Electricians needto understand the impact on a pure inductive circuit and how current lags voltage when the effect is

applied in an AC circuit. This unit■ explains how properties other than resistance can limit the flow of current.■ introduces another measurement called impedance. Impedance is the total current-limiting effect in an

AC circuit and can be comprised of more than one element, such as resistance and inductance. With-out an understanding of inductance, you will never be able to understand many of the concepts to fol-low in later units.

UNIT 17Inductance

in AC Circuits

Courtesy of Niagara Mohawk

Power Corporation.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 467

468 SECTION VI ■ Alternating Current (AC) Circuits Containing Inductance

OBJECTIVES

After studying this unit, you should be able to

■ discuss the properties of inductance in an AC circuit.

■ discuss inductive reactance.

■ compute values of inductive reactance and inductance.

■ discuss the relationship of voltage and current in a pure inductive circuit.

■ be able to compute values for inductors connected in series or parallel.

■ discuss reactive power (VARs).

■ determine the Q of a coil.

PREVIEW

This unit discusses the effects of inductance on AC circuits. The unitexplains how current is limited in an inductive circuit as well as the effect

inductance has on the relationship of voltage and current. ■

17–1

InductanceInductance (L) is one of the primary types of loads in AC circuits. Someamount of inductance is present in all AC circuits because of the continuallychanging magnetic field (Figure 17–1). The amount of inductance of a single

Magnetic field

Collapse of magnetic field

Expansion of magnetic field

FIGURE 17–1 A continually changing magnetic field induces a voltage into any conductor.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 468

conductor is extremely small, and, in most instances, it is not considered incircuit calculations. Circuits are generally considered to contain inductancewhen any type of load that contains a coil is used. For circuits that contain acoil, inductance is considered in circuit calculations. Loads such as motors,transformers, lighting ballast, and chokes all contain coils of wire.

In Unit 14, it was discussed that whenever current flows through a coil ofwire, a magnetic field is created around the wire (Figure 17–2). If the amountof current decreases, the magnetic field collapses (Figure 17–3). Recall fromUnit 14 several facts concerning inductance:

1. When magnetic lines of flux cut through a coil, a voltage is induced inthe coil.

2. An induced voltage is always opposite in polarity to the applied voltage.This is often referred to as counter-electromotive force (CEMF).

3. The amount of induced voltage is proportional to the rate of change ofcurrent.

4. An inductor opposes a change of current.

The inductors in Figure 17–2 and Figure 17–3 are connected to an alter-nating voltage. Therefore the magnetic field continually increases, decreases,and reverses polarity. Since the magnetic field continually changes magnitudeand direction, a voltage is continually being induced in the coil. This inducedvoltage is 180� out of phase with the applied voltage and is always in oppo-sition to the applied voltage (Figure 17–4). Since the induced voltage isalways in opposition to the applied voltage, the effective applied voltage isreduced by the induced voltage. For example, assume an inductor is con-nected to a 120-volts AC line. Now assume that the inductor has an inducedvoltage of 116 volts. Since the induced voltage subtracts from the applied volt-age, there are only 4 volts to push current through the wire resistance of thecoil (120 V � 116 V � 4).

UNIT 17 ■ Inductance in AC Circuits 469

FIGURE 17–2 As current flows through a coil,a magnetic field is created around the coil.

FIGURE 17–3 As current flow decreases, themagnetic field collapses.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM 69

Computing the Induced VoltageThe amount of induced voltage in an inductor can be computed if the resis-tance of the wire in the coil and the amount of circuit current are known. Forexample, assume that an ohmmeter is used to measure the actual amount ofresistance in a coil and the coil is found to contain 6 ohms of wire resistance(Figure 17–5). Now assume that the coil is connected to a 120-volts AC circuitand an ammeter measures a current flow of 0.8 ampere (Figure 17–6). Ohm’slaw can now be used to determine the amount of voltage necessary to push


Applied voltage

Induced voltageApplied voltage

Induced voltage

FIGURE 17–4 The applied voltage and induced voltage are 180 degrees out of phase with each other.

6

Ohmmeter

FIGURE 17–5 Measuring the resistance of a coil.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 470

0.8 ampere of current through 6 ohms of resistance:

E � I � R

E � 0.8 A � 6

E � 4.8 V

Since only 4.8 volts are needed to push the current through the wire resis-tance of the inductor, the remainder of the 120 volts is used to overcome thecoil’s induced voltage of 119.904 volts (��(120� V�)2� �� (�4.8� V)� 2�� 119.904 volts).

17–2

Inductive ReactanceNotice that the induced voltage is able to limit the flow of current through the cir-cuit in a manner similar to resistance. This induced voltage is not resistance, but itcan limit the flow of current just as resistance does. This current-limiting propertyof the inductor is called reactance and is symbolized by the letter X. This reac-tance is caused by inductance, so it is called inductive reactance and is symbol-ized by XL, pronounced “X sub L.” Inductive reactance is measured in ohms justas resistance is and can be computed when the values of inductance and fre-quency are known. The following formula can be used to find inductive reactance:

XL � 2 � fL

where

XL � inductive reactance

2 � a constant

� � 3.1416

f � frequency in hertz (Hz)

L � inductance in henrys (H)


120 VAC0.8 A

Ammeter

FIGURE 17–6 Measuring circuit current with an ammeter.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 471

The inductive reactance, in ohms, is caused by the induced voltage and is,therefore, proportional to the three factors that determine induced voltage:

1. the number of turns of wire;

2. the strength of the magnetic field;

3. the speed of the cutting action (relative motion between the inductor andthe magnetic lines of flux).

The number of turns of wire and the strength of the magnetic field are deter-mined by the physical construction of the inductor. Factors such as the size ofwire used, the number of turns, how close the turns are to each other, and thetype of core material determine the amount of inductance (in henrys, H) of thecoil (Figure 17–7). The speed of the cutting action is proportional to the fre-quency (hertz). An increase of frequency causes the magnetic lines of flux tocut the conductors at a faster rate and, thus, produces a higher induced voltageor more inductive reactance.


Coil with more inductance

Coil with less inductance

FIGURE 17–7 Coils with turns closer together produce more inductance than coils with turns far apart.

■ EXAMPLE 17–1

The inductor shown in Figure 17–8 has an inductance of 0.8 H and is con-nected to a 120-V, 60-Hz line. How much current will flow in this circuit ifthe wire resistance of the inductor is negligible?

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 472


0.8 H120 VAC60 Hz

FIGURE 17–8 Circuit current is limited by inductive reactance.

SolutionThe first step is to determine the amount of inductive reactance of the inductor:

XL � 2 � fL

XL � 2 � 3.1416 � 60 Hz � 0.8

XL � 301.594 �

Since inductive reactance is the current-limiting property of this circuit, it canbe substituted for the value of R in an Ohm’s law formula:

If the amount of inductive reactance is known, the inductance of the coilcan be determined using the formula

■

LX

fL�

�2

IEX

IV

I A

L�

��

�

120301 594

0 398

.

.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 473


■ EXAMPLE 17–2

Assume an inductor with a negligible resistance is connected to a 36-V, 400-Hz line. If the circuit has a current flow of 0.2 A, what is the inductanceof the inductor?

SolutionThe first step is to determine the inductive reactance of the circuit:

Now that the inductive reactance of the inductor is known, the inductancecan be determined:

■

LX

f

LHz

L H

L��

��

� �

�

2180

2 3 1416 400

0 0716

.

.

XEI

XVA

X

L

L

L

�

�

� �

360 2

180

.

■ EXAMPLE 17–3

An inductor with negligible resistance is connected to a 480-V, 60-Hz line.An ammeter indicates a current flow of 24 A. How much current will flowin this circuit if the frequency is increased to 400 Hz?

SolutionThe first step in solving this problem is to determine the amount of induc-tance of the coil. Since the resistance of the wire used to make the induc-tor is negligible, the current is limited by inductive reactance. The inductivereactance can be found by substituting XL for R in an Ohm’s law formula:

XEI

XVA

X

L

L

L

�

�

� �

48024

20

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 474


Now that the inductive reactance is known, the inductance of the coil can befound using the formula

NOTE: When using a frequency of 60 Hertz, 2 � � � 60 � 376.992. To sim-plify calculations, this value is generally rounded to 377. Since 60 hertz isthe major frequency used throughout the United States, 377 should be mem-orized because it is used in many calculations:

Since the inductance of the coil is determined by its physical construction,it does not change when connected to a different frequency. Now that theinductance of the coil is known, the inductive reactance at 400 Hertz can becomputed:

XL � 2�fL

XL � 2 � 3.1416 � 400 Hz � 0.053

XL � 133.204 �

The amount of current flow can now be found by substituting the value ofinductive reactance for resistance in an Ohm’s law formula:

■

IEX

IV

I A

L�

��

�

480133 204

3 603

.

.

LHz

L H

��

�

20377

0 053.

LX

fL�

�2

17–3

Schematic SymbolsThe schematic symbol used to represent an inductor depicts a coil of wire. Sev-eral symbols for inductors are shown in Figure 17–9. The symbols shown withthe two parallel lines represent iron-core inductors, and the symbols withoutthe parallel lines represent air-core inductors.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 475

17–4

Inductors Connected in SeriesWhen inductors are connected in series (Figure 17–10), the total inductance ofthe circuit (LT) equals the sum of the inductances of all the inductors:

LT � L1 � L2 � L3

The total inductive reactance (XLT) of inductors connected in series equalsthe sum of the inductive reactances for all the inductors:

XLT � XL1 � XL2 � XL3


Air-core inductors

Iron-core inductors

FIGURE 17–9 Schematic symbols for inductors.

FIGURE 17–10 Inductors connected in series.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 476

17–5

Inductors Connected in ParallelWhen inductors are connected in parallel (Figure 17–11), the total inductancecan be found in a manner similar to finding the total resistance of a parallelcircuit. The reciprocal of the total inductance is equal to the sum of the reci-procals of all the inductors:

or

L

L L L

T �

� �

11 1 1

1 2 3

1 1 1 1

1 2 3L L L LT� � �


■ EXAMPLE 17–4

Three inductors are connected in series. Inductor 1 has an inductance of 0.6 H, Inductor 2 has an inductance of 0.4 H, and Inductor 3 has an induc-tance of 0.5 H. What is the total inductance of the circuit?

Solution

LT � 0.6 H � 0.4 H � 0.5 H

LT � 1.5 H■

■ EXAMPLE 17–5

Three inductors are connected in series. Inductor 1 has an inductive reactanceof 180 �, Inductor 2 has an inductive reactance of 240 �, and Inductor 3 hasan inductive reactance of 320 �. What is the total inductive reactance of thecircuit?

Solution

XLT � 180 � � 240 � � 320 �

XLT � 740 �■

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 477

Another formula that can be used to find the total inductance of parallel induc-tors is the product-over-sum formula:

If the values of all the inductors are the same, total inductance can be found bydividing the inductance of one inductor by the total number of inductors:

Similar formulas can be used to find the total inductive reactance of induc-tors connected in parallel:

or

or

or

XXN

LTL�

XX XX X

LTL L

L L�

�

�

1 2

1 2

X

X X X

LT

L L L

�

� �

11 1 1

1 2 3

1 1 1 1

1 2 3X X X XLT L L L� � �

LLN

T �

LL LL L

T ��

�

1 2

1 2


FIGURE 17–11 Inductors connected in parallel.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 478

17–6

Voltage and Current Relationships in an Inductive CircuitIn Unit 16, it was discussed that when current flows through a pure resistive cir-cuit, the current and voltage are in phase with each other. In a pure inductivecircuit, the current lags the voltage by 90�. At first this may seem to be an impossible condition until the relationship of applied voltage and inducedvoltage is considered. How the current and applied voltage can become 90� outof phase with each other can best be explained by comparing the relationshipof the current and induced voltage (Figure 17–12). Recall that the induced volt-age is proportional to the rate of change of the current (speed of cutting action).At the beginning of the waveform, the current is shown at its maximum value inthe negative direction. At this time, the current is not changing, so induced volt-age is zero. As the current begins to decrease in value, the magnetic field pro-duced by the flow of current decreases or collapses and begins to induce a volt-age into the coil as it cuts through the conductors (Figure 17–3).


■ EXAMPLE 17–6

Three inductors are connected in parallel. Inductor 1 has an inductance of2.5 H, Inductor 2 has an inductance of 1.8 H, and Inductor 3 has an induc-tance of 1.2 H. What is the total inductance of this circuit?

Solution

■

L

H H H

L

L H

T

T

T

�

� �

�

�

11

2 51

1 81

1 2

11 788

0 559

. . .

( . ).

Circuit current

Induced voltage

FIGURE 17–12 Induced voltage is proportional to the rate of change of current.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 479

The greatest rate of current change occurs when the current passesfrom negative, through zero and begins to increase in the positive direction(Figure 17–13). Since the current is changing at the greatest rate, the inducedvoltage is maximum. As current approaches its peak value in the positive direc-tion, the rate of change decreases, causing a decrease in the induced voltage.The induced voltage will again be zero when the current reaches its peak valueand the magnetic field stops expanding.

It can be seen that the current flowing through the inductor is leading theinduced voltage by 90�. Since the induced voltage is 180� out of phase with theapplied voltage, the current lags the applied voltage by 90� (Figure 17–14).


Point of no current change

Induced voltage

Circuit current

Greatest rate of current change

FIGURE 17–13 No voltage is induced when the current does not change.

Applied voltage

Current flow

FIGURE 17–14 The current lags the applied voltage by 90�.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 480

17–7

Power in an Inductive CircuitIn a pure resistive circuit, the true power, or watts, is equal to the product of thevoltage and current. In a pure inductive circuit, however, no true power, or watts,is produced. Recall that voltage and current must both be either positive or neg-ative before true power can be produced. Since the voltage and current are 90�out of phase with each other in a pure inductive circuit, the current and voltagewill be at different polarities 50% of the time and at the same polarity 50% of thetime. During the period of time that the current and voltage have the same po-larity, power is being given to the circuit in the form of creating a magnetic field.When the current and voltage are opposite in polarity, power is being given backto the circuit as the magnetic field collapses and induces a voltage back into thecircuit. Since power is stored in the form of a magnetic field and then given back,no power is used by the inductor. Any power used in an inductor is caused bylosses such as the resistance of the wire used to construct the inductor, generallyreferred to as I2R losses, eddy current losses, and hysteresis losses.

The current and voltage waveform in Figure 17–15 has been divided intofour sections: A, B, C, and D. During the first time period, indicated by A, thecurrent is negative and the voltage is positive. During this period, energy isbeing given to the circuit as the magnetic field collapses. During the second timeperiod, B, both the voltage and current are positive. Power is being used to pro-duce the magnetic field. In the third time period, C, the current is positive andthe voltage is negative. Power is again being given back to the circuit as the field


A B C D

FIGURE 17–15 Voltage and current relationships during different parts of a cycle.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 481

collapses. During the fourth time period, D, both the voltage and current arenegative. Power is again being used to produce the magnetic field. If theamount of power used to produce the magnetic field is subtracted from thepower given back, the result will be zero.

17–8

Reactive PowerAlthough essentially no true power is being used, except by previously men-tioned losses, an electrical measurement called volt-amperes-reactive (VARs) isused to measure the reactive power in a pure inductive circuit. VARs can becomputed in the same way as watts except that inductive values are substitutedfor resistive values in the formulas. VARs is equal to the amount of current flow-ing through an inductive circuit times the voltage applied to the inductive partof the circuit. Several formulas for computing VARs are

where

EL � voltage applied to an inductor

IL � current flow through an inductor

XL � inductive reactance

17–9

Q of an InductorSo far in this unit, it has been generally assumed that an inductor has no resis-tance and that inductive reactance is the only current-limiting factor. In reality,that is not true. Since inductors are actually coils of wire, they all contain someamount of internal resistance. Inductors actually appear to be a coil connected inseries with some amount of resistance (Figure 17–16). The amount of resistancecompared with the inductive reactance determines the quality (Q) of the coil.Inductors that have a higher ratio of inductive reactance to resistance are consid-ered to be inductors of higher quality. An inductor constructed with a large wirewill have a low wire resistance and, therefore, a higher Q (Figure 17–17). Induc-tors constructed with many turns of small wire have a much higher resistance

VARs E I

VARsE

X

VARs I x

L L

L

L

L L

� �

�

� �

2

2


65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 482

and, therefore, a lower Q. To determine the Q of an inductor, divide the induc-tive reactance by the resistance:

QXR

L�


Pure inductance

Internal resistance

FIGURE 17–16 Inductors contain internal resistance.

Large wire produces an inductorwith a high Q.

Small wire produces an inductorwith a low Q.

FIGURE 17–17 The Q of an inductor is a ratio of inductive reactance as compared to resistance.The letter Q stands for quality.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 483


Total coil impedance(Z)

Inductive reactance(XL)

Resistance(R)

FIGURE 17–18 Coil impedance is a combination of wire resistance and inductive reactance.

Although inductors have some amount of resistance, inductors that have a Qof 10 or greater are generally considered to be pure inductors. Once the ratio ofinductive reactance becomes 10 times as great as resistance, the amount of resis-tance is considered negligible. For example, assume an inductor has an inductivereactance of 100 ohms and a wire resistance of 10 ohms. The inductive reactivecomponent in the circuit is 90� out of phase with the resistive component. Thisrelationship produces a right triangle (Figure 17–18). The total current-limitingeffect of the inductor is a combination of the inductive reactance and resistance.This total current-limiting effect is called impedance and is symbolized by the let-ter Z. The impedance of the circuit is represented by the hypotenuse of the righttriangle formed by the inductive reactance and the resistance. To compute thevalue of impedance for the coil, the inductive reactance and the resistance mustbe added. Since these two components form the legs of a right triangle and theimpedance forms the hypotenuse, the Pythagorean theorem discussed in Unit 15can be used to compute the value of impedance:

Z R X

Z

Z

Z

L� �

� �

�

� �

2

2 2

2

10 100

10 100

100 499

,

.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 484

Notice that the value of total impedance for the inductor is only 0.5 ohmgreater than the value of inductive reactance.

If it should become necessary to determine the true inductance of an induc-tor, the resistance of the wire must be taken into consideration. Assume that aninductor is connected to a 480-volt 60-Hz power source and that an ammeterindicates a current flow of 0.6 ampere. Now assume that an ohmmetermeasures 150 � of wire resistance in the inductor. What is the inductance ofthe inductor?

To determine the inductance, it will be necessary to first determine theamount of inductive reactance as compared to the wire resistance. The totalcurrent-limiting value (impedance) can be found with Ohm’s law:

The total current-limiting effect of the inductor is 800 �. This value is acombination of both the inductive reactance of the inductor and the wire re-sistance. Since the resistive part and the inductive reactance part of theinductor are 90� out of phase with each other, they form the legs of aright triangle with the impedance forming the hypotenuse of the triangle(Figure 17-19). To determine the amount of inductive reactance, use thefollowing formula:

Now that the amount of inductive reactance has been determined, the induc-tance can be computed using the formula

LX

f

L

L henrys

L�

�

��

�

2785 812

2 3 1416 60

2 084

.

.

.

X Z R

X

X

L

L

L

� �

� �

�

2 2

2 2800 150

785 812. �

ZE

I

Z

Z.

�

�

�

480

0 6800 �


65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 485

Summary■ Induced voltage is proportional to the rate of change of current.

■ Induced voltage is always opposite in polarity to the applied voltage.

■ Inductive reactance is a countervoltage that limits the flow of current, asdoes resistance.


Z (Impedance)

800 W

R (Resistance) 150 W

XL

(Inductive

reactance)

FIGURE 17–19 The inductive reactance forms one leg of a right triangle.

65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 486

■ Inductive reactance is measured in ohms.

■ Inductive reactance is proportional to the inductance of the coil and the fre-quency of the line.

■ Inductive reactance is symbolized by XL.

■ Inductance is measured in henrys (H) and is symbolized by the letter L.

■ When inductors are connected in series, the total inductance is equal to thesum of all the inductors.

■ When inductors are connected in parallel, the reciprocal of the total induc-tance is equal to the sum of the reciprocals of all the inductors.

■ The current lags the applied voltage by 90� in a pure inductive circuit.

■ All inductors contain some amount of resistance.

■ The Q of an inductor is the ratio of the inductive reactance to the resistance.

■ Inductors with a Q of 10 are generally considered to be “pure” inductors.

■ Pure inductive circuits contain no true power or watts.

■ Reactive power is measured in VARs.

■ VARs is an abbreviation for volt-amperes-reactive.

Review Questions1. How many degrees are the current and voltage out of phase with each

other in a pure resistive circuit?

2. How many degrees are the current and voltage out of phase with eachother in a pure inductive circuit?

3. To what is inductive reactance proportional?

4. Four inductors, each having an inductance of 0.6 H, are connected inseries. What is the total inductance of the circuit?

5. Three inductors are connected in parallel. Inductor 1 has an inductanceof 0.06 H; Inductor 2 has an inductance of 0.05 H; and Inductor 3 hasan inductance of 0.1 H. What is the total inductance of this circuit?

6. If the three inductors in Question 5 were connected in series, whatwould be the inductive reactance of the circuit? Assume the inductorsare connected to a 60-Hz line.

7. An inductor is connected to a 240-V, 1000-Hz line. The circuit current is0.6 A. What is the inductance of the inductor?


65803_17_ch17_p466-489.qxd 2/15/08 9:53 AM Page 487

8. An inductor with an inductance of 3.6 H is connected to a 480-V, 60-Hz line. How much current will flow in this circuit?

9. If the frequency in Question 8 is reduced to 50 Hz, how much currentwill flow in the circuit?

10. An inductor has an inductive reactance of 250 � when connected to a60-Hz line. What will be the inductive reactance if the inductor is con-nected to a 400-Hz line?


Practical Applications

You have the task of ordering a replacement inductor for one that has becomedefective. The information on the nameplate has been painted over and can-

not be read. The machine that contains the inductor operates on 480 V at a fre-quency of 60 Hz. Another machine has an identical inductor in it, but its nameplatehas been painted over also. A clamp-on ammeter indicates a current of 18 A, and avoltmeter indicates a voltage drop across the inductor of 324 V in the machine thatis still in operation. After turning off the power and locking out the panel, you dis-connect the inductor in the operating machine and measure a wire resistance of1.2 � with an ohmmeter. Using the identical inductor in the operating machine asan example, what inductance value should you order and what would be theminimum VAR rating of the inductor? Should you be concerned with the amountof wire resistance in the inductor when ordering? Explain your answers. ■

Practical Applications

You are working as an electrician installing fluorescent lights. You notice thatthe lights were made in Europe and that the ballasts are rated for operation on

a 50-Hz system. Will these ballasts be harmed by overcurrent if they are connectedto 60 Hz? If there is a problem with these lights, what will be the most likely causeof the trouble? ■

Practice Problems

Inductive Circuits

1. Fill in all the missing values. Refer to the following formulas:

X L

LX

f

fX

L

L

L

L

� �

��

��

2

2

2

f

65803_17_ch17_p466-489.qxd 2/15/08 9:54 AM Page 488


2. What frequency must be applied to a 33-mH inductor to produce aninductive reactance of 99.526 �?

3. An inductor is connected to a 120-volt, 60-Hz line and has a current flowof 4 amperes. An ohmmeter indicates that the inductor has a wireresistance of 12 �. What is the inductance of the inductor?

4. A 0.75-henry inductor has a wire resistance of 90 �. When connected toa 60-Hz power line, what is the total current-limiting effect of theinductor?

5. An inductor has a current flow of 3 amperes when connected to a 240-volt, 60-Hz power line. The inductor has a wire resistance of 15 �.What is the Q of the inductor?

Inductive Inductance (H) Frequency (Hz) Reactance (�)

1.2 60

0.085 213.628

1000 4712.389

0.65 600

3.6 678.584

25 411.459

0.5 60

0.85 6408.849

20 201.062

0.45 400

4.8 2412.743

1000 40.841

65803_17_ch17_p466-489.qxd 2/15/08 9:54 AM Page 489

alternating current (ac) vi circuits containing inductance

Documents