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FIG. 12.61 Using PSpice to determine the transient response for a circuit in which the inductive element has an initial condition. FIG. 12.62 A plot of the applied pulse and resulting current for the circuit of Fig. 12.61. COMPUTER ANALYSIS ⏐⏐⏐ 509
510 ⏐⏐⏐ INDUCTORS Electronics Workbench The transient response of an R-L network can also be obtained using Electronics Workbench. The circuit to be examined appears in Fig. 12.63 with a pulse voltage source to simulate the closing of a switch at t � 0 s. The source, referred to as PULSE – VOLTAGE – SOURCE in the Source listing, is the near the bottom left of the Sources parts bin. When selected, it will appear with a label, an initial voltage, a step voltage, and a frequency. All can be changed by simply double-clicking on the source symbol to obtain the Pulse Voltage dialog box. As shown in Fig. 12.63, the Pulsed Value will be set at 20 V, and the Delay Time to 0 s. The Rise Time and Fall Time will both remain at the default levels of 1 ns. For our analysis we want a Pulse Width that is at least twice the 5t transient period of the circuit. For the chosen values of R and L, t � L/R � 10 mH/100 � � 0.1 ms � 100 ms. The transient period of 5t is therefore 500 ms or 0.5 ms. Thus, a Pulse Width of 1 ms would seem appropriate with a Period of 2 ms. The result is a frequency of f � 1/T � 1/2 ms � 500 Hz. When all have been set and selected, the parameters of the pulse source will appear as shown in Fig. 12.63. Next the resistor, inductor, and ground are placed on the screen to complete the circuit. This time we will want to see the node names so that we can call for them when we set up the simulation process. This is accomplished through Options-Preferences-Show node names. In this case we have two—one at the positive terminal of the supply (1) and the other at the top end of the inductor (2) representing the voltage across the inductor. The simulation process is initiated by the following sequence: Simulate-Analyses-Transient Analysis. The result is the Transient Analysis dialog box in which Analysis Parameters is chosen first. Under FIG. 12.63 Using Electronics Workbench to obtain the transient response for an inductive circuit.
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1 Introduction 1.1 THE ELECTRICAL/E
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� S I mind that similar progressi
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� S I denser, which we refer to t
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� S I decades will probably conta
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� S I TABLE 1.1 Comparison of the
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� S I The second was originally d
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� S I A quick method of determini
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� S I revealing that the operatio
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� S I 1 1 2300 � � 3.333E�1
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� S I Since the power of ten will
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� S I EXAMPLE 1.14 a. Determine t
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� S I Initial Settings Format and
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� S I c. Since the division will
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� S I Three software packages wil
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� S I SECTION 1.8 Conversion with
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2 Current and Voltage 2.1 ATOMS AND
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I e V CURRENT ⏐⏐⏐ 33 the dist
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I e V The chemical activity of the
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I e V erence plane. If the weight i
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I e V 2.4 FIXED (dc) SUPPLIES The t
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I e V FIG. 2.14 Maintenance-free 12
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I e V batteries are relatively warm
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I e V EXAMPLE 2.5 a. Determine the
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I e V TABLE 2.1 Relative conductivi
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I e V be accomplished is to open th
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I e V (a) (c) Contact Sliding switc
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I e V Control switch Meter leads (a
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I e V tant to disconnect the charge
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I e V GLOSSARY ⏐⏐⏐ 57 28. Fin
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3 Resistance 3.1 INTRODUCTION The f
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R G Note that the area of the condu
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R G EXAMPLE 3.3 What is the resista
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R G RESISTANCE: METRIC UNITS ⏐⏐
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R G The conversion factor between r
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R G Absolute zero -273.15°C Inferr
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R G Since R 20 of Eq. (3.8) is the
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R G result is a tremendous saving i
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R G TYPES OF RESISTORS ⏐⏐⏐ 75
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R G TYPES OF RESISTORS ⏐⏐⏐ 77
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R G a 1% failure rate would reveal
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R G gaps. Dropping to the 10% level
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R G THERMISTORS ⏐⏐⏐ 83 Prelim
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R G APPLICATIONS ⏐⏐⏐ 85 3.14
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R G longer heating element in stand
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R G cally this law relates voltage,
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R G MATHCAD ⏐⏐⏐ 91 key at the
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R G *13. What is the new resistance
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R G SECTION 3.13 Varistors 58. a. R
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98 ⏐⏐⏐ OHM’S LAW, POWER, AN
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100 ⏐⏐⏐ OHM’S LAW, POWER, A
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5 Series Circuits 5.1 INTRODUCTION
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S the same through series elements
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S Solution: RT � R1 � R2 � R3
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S type of element. In other words,
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S c. V1 � IR1 � (2 A)(4 �)
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S In the above discussion the curre
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S Voltage Sources and Ground Except
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S EXAMPLE 5.14 Find the voltage Vab
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S EXAMPLE 5.19 Using the voltage di
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S ∆V L 120 V 100 V 0 V L voltage
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S EXAMPLE 5.24 Determine the voltag
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S 5.11 APPLICATIONS Holiday Lights
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S wiring, you will find that since
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S and right-click on the mouse. A l
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S FIG. 5.68 Applying Electronics Wo
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S also be used to remind the progra
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S 3. Find the applied voltage E nec
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S 9. Determine the current I and th
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S 40 V I + R 3 R 2 R 1 V 3 - 30 �
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S *28. For the network of Fig. 5.97
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6 Parallel Circuits 6.1 INTRODUCTIO
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P rent level. In other words, as th
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P EXAMPLE 6.4 a. Find the total res
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P R T R 6 � R′ T � ��
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P d. RT � 30 � � 30 � � 0
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P 0.25 S � 0.1 S � 0.05 S � 0
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P EXAMPLE 6.13 Determine the curren
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P I2 � I4 � I5 12 A � I4 �
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P For the particular case of two pa
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P and (R 1 � R 2)I 1 � R 2I R 1
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P A short circuit is a very low res
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P must therefore be zero volts, as
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P + E - A + - I s + - ensure a posi
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P 12-gage fuse link Filter capacito
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P car such as the lights, air condi
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P careful, you can work with one li
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P In particular, note that m (or M)
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P This time, rather than using mete
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P PROBLEMS ⏐⏐⏐ 205 *6. Determ
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P 14. Using the information provide
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P *21. Find the unknown quantities
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P SECTION 6.8 Open and Short Circui
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7 Series-Parallel Networks 7.1 SERI
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S S P P For parallel resistors R 1
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S S P P a I A E 16.8 V R 1 9 � R
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S S P P I s R T E 24 V R 6 � R1
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S S P P R 3 R 2 7 � 5 � I 3 R 4
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S S P P E 72 V 72 V I5 ��
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S S P P (6 �)I3 6 I6 ��
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S S P P To demonstrate the validity
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S S P P 7.5 POTENTIOMETER LOADING F
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S S P P The Ammeter The maximum cur
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S S P P ammeter or voltmeter becaus
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S S P P 120 V + - VR1 R1 + - 1 �
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S S P P would be created between th
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S S P P Note also that the current
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S S P P 7.9 COMPUTER ANALYSIS PSpic
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Heading Preprocessor directive Defi
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S S P P 3. For the network of Fig.
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S S P P 10. For the network of Fig.
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S S P P 17. For the configuration o
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S S P P 26. For the ladder network
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S S P P 34. Using a 50-mA, 1000-�
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8 Methods of Analysis and Selected
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N A current-source networks, it wil
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N A excellent approximation to drop
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N A EXAMPLE 8.7 Reduce the network
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N A of series elements. Figure 8.20
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N A appearing in Solution 1 in a ve
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N A R 1 E 1 - + + - 4 � 15 V Appl
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N A EXAMPLE 8.11 Consider the same
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N A EXAMPLE 8.13 Find the branch cu
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N A Node a is then used to relate t
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N A 1. Assign a loop current to eac
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N A EXAMPLE 8.18 Find the current t
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N A The nodal analysis method is ap
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N A or I � � � and V2� �
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N A Step 3: Included in Fig. 8.49 f
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N A I 3 V 1 I 1 R 3 10 � 6 A R1 4
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N A EXAMPLE 8.23 Write the nodal eq
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N A EXAMPLE 8.25 Using nodal analys
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N A Solution: The nodal voltages ar
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N A with the bottom of the determin
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N A any unknown quantities if mesh
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N A To obtain the relationships nec
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N A Solution: RB RC (20 �)(10 �
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N A 8.13 APPLICATIONS The Applicati
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N A or outside forces such as light
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N A Schematic with Nodal Voltages W
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N A A photograph of the outside and
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N A PROBLEMS SECTION 8.2 Current So
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N A SECTION 8.4 Current Sources in
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N A *16. For the transistor configu
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N A SECTION 8.8 Mesh Analysis (Form
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N A *37. Using the supernode approa
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N A *49. Repeat Problem 48 for the
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9 Network Theorems 9.1 INTRODUCTION
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Th This final relationship between
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Th The total current through the 4-
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I Th R 1 6 mA R 3 Current divider r
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Th E 1 12 V 6 � 4 V E 2 4 � (a)
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Th R 1 3 � R 2 6 � a R Th b b (
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Th R 1 R 1 6 � 6 � R 4 a 3 �
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Th 6 � R 1 R 2 12 � b R3 a R4 3
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Th Direct Measurement of E Th and R
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Th Conclusion: 5. Draw the Norton e
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Th NORTON’S THEOREM ⏐⏐⏐ 341
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Th E Th I N R N FIG. 9.78 Defining
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Th Note, in particular, that P L is
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Th When R L � R Th, R L h% ��
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Th EXAMPLE 9.14 A dc generator, bat
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Th Note Fig. 9.91, where V1 � V3
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Th that of E 1 and E 3. The total c
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Th combination of elements that wil
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Th RT � R1 � R2 � (R3 � R4)
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Th are presented with a bundle of r
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Th FIG. 9.116 Using PSpice to deter
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Th FIG. 9.119 Using PSpice to deter
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Th FIG. 9.121 Plot resulting from t
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Th 2. Using superposition, find the
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Th *8.Find the Thévenin equivalent
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Th 21. For each network of Fig. 9.1
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Th SECTION 9.8 Reciprocity Theorem
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10 Capacitors 10.1 INTRODUCTION Thu
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The attraction and repulsion betwee
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w � Q, so the dielectric is also
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d � o d d C = 5 µF A (a) C = 0.1
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EXAMPLE 10.4 Find the maximum volta
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Dipped phenolic coating Ceramic die
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lower potential. This capacitor can
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Type: Miniature Axial Electrolytic
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The factor e �t/RC is an exponent
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after five time constants of the ch
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which employs the function e �x a
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Solutions: a. Charging phase: vC
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t2 � (R2 � R3)C � (1 k��3
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a. Find the mathematical expression
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and C t ��t loge�1 � �v
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10.11 THÉVENIN EQUIVALENT: t � R
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Solution: The network is redrawn in
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�v 4 v 3 v2 0 v C (V) 1 t2 t3 �
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and substituting for C T: 1/C 1 V 1
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Solution: As previously discussed,
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Flash Lamp The basic circuitry for
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light in parallel with the capacito
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Black (feed) Ground Black White Gro
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sufficiently large to be considered
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FIG. 10.77 Using PSpice to investig
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Settings-AverageIC dialog box. Anal
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16. Find the distance in millimeter
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29. For the situation of Problem 25
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*40. The capacitor of Fig. 10.100 i
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*47. For the network of Fig. 10.107
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11 Magnetic Circuits 11.1 INTRODUCT
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Magnetic flux lines I Conductor FIG
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EXAMPLE 11.1 For the core of Fig. 1
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Substituting, we have (11.4) The ma
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- H s - B max e Saturation B R d -
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1.4 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6
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above is evidenced by the fact that
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An approach frequently employed in
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and the magnetizing force is H (she
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The flux density of the air gap in
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� � 0 Hbelbe � Hbcdelbcde �
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EXAMPLE 11.9 Find the magnetic flux
Page 460 and 461: Speakers and Microphones Electromag
Page 462 and 463: 4,000,000,000,000 bits of informati
Page 464 and 465: ingly enough one that perhaps will
Page 466 and 467: Hall Effect Sensor The Hall effect
Page 468 and 469: ciently close to establish contact
Page 470 and 471: SECTION 11.8 Hysteresis 9. For the
Page 472 and 473: *18. For the series-parallel magnet
Page 474 and 475: 12 Inductors 12.1 INTRODUCTION We h
Page 476 and 477: Inductors are coils of various dime
Page 478 and 479: (a) (d) (b) FIG. 12.10 Various type
Page 480 and 481: ciated with the applied ac signal m
Page 482 and 483: the voltage across the coil is not
Page 484 and 485: y 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0
Page 486 and 487: 12.8 INITIAL VALUES This section wi
Page 488 and 489: Let us now test the validity of the
Page 490 and 491: The mathematical expression for the
Page 492 and 493: v R1 R1 Defined polarity + vL - v L
Page 494 and 495: iL � (1 � e �t/t ) t � �
Page 496 and 497: 12.12 INDUCTORS IN SERIES AND PARAL
Page 498 and 499: Applying the current divider rule,
Page 500 and 501: EXAMPLE 12.12 Find the energy store
Page 502 and 503: + Feed 120 V ac - Return ��
Page 504 and 505: would need for 15 W, not to mention
Page 506 and 507: will scatter to all sides of the mo
Page 508 and 509: ing trace appears in the bottom of
Page 512 and 513: Parameters, use 0 s as the Start ti
Page 514 and 515: *11. Find the waveform for the curr
Page 516 and 517: *19. For the network of Fig. 12.76:
Page 518 and 519: *29. The switch of Fig. 12.83 has b
Page 520: 37. Find the voltage V 1 and the cu
Page 523 and 524: 522 ⏐⏐⏐ SINUSOIDAL ALTERNATIN
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560 ⏐⏐⏐ SINUSOIDAL ALTERNATIN
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14 The Basic Elements and Phasors 1
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� RESPONSE OF BASIC R, L, AND C E
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� The opposition established by a
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� Therefore, RESPONSE OF BASIC R,
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� is directly related to the stra
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� applied. In general, therefore,
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� θv the voltage or current. The
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� EXAMPLE 14.11 Determine the ave
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� any number not on the real axis
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� Polar to Rectangular X � Z co
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� Reciprocal The reciprocal of a
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� - Multiplication To multiply tw
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� Solutions: a. By Eq. (14.34), b
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� often frustrating if one lost m
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� CALCULATOR AND COMPUTER METHODS
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� PHASORS ⏐⏐⏐ 611 is entere
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� PHASORS ⏐⏐⏐ 613 6 A v 1 =
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� PHASORS ⏐⏐⏐ 615 - 2 p 41.
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� COMPUTER ANALYSIS ⏐⏐⏐ 617
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� PROBLEMS ⏐⏐⏐ 625 *20. For
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� PROBLEMS ⏐⏐⏐ 627 *47. a.
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15 Series and Parallel ac Circuits
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a c v � 100 sin qt ⇒ phasor for
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a c v � 24 sin qt ⇒ phasor form
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a c v � 15 sin qt ⇒ phasor nota
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a c real and imaginary axes and fin
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a c In rectangular form, and VR �
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a c Time domain: In the time domain
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a c or E � V R � V L � V C wh
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a c (6�0)*(50�30)/((6�0)�(9
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a c Therefore, E � �(1�3�.3
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a c f � 1 kHz XC � � �15.92
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a c f � 0 Hz XC � � ⇒ very
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a c ohms and the short-circuit equi
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a c elements by 90°, and leads the
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a c EXAMPLE 15.12 For the network o
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a c EXAMPLE 15.14 Find the admittan
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a c I Admittance diagram: As shown
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a c E Admittance diagram: As shown
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a c Phasor notation: As shown in Fi
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a c Using Eq. (15.32), we obtain G
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a c on the total impedance at that
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a c 90° 60° 45° 30° 0° θ T In
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a c 0° -30° -45° -60° -90° θL
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a c The total impedance at the freq
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a c Solution: Rp � 8k� Xp (resu
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a c I = 12 A ∠ 0° FIG. 15.95 Ser
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a c + e - a + v R - R impedance. Th
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a c Switched outlets Parallel outle
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a c transfer of power (see Section
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a c Consequently, the sound generat
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a c V applied 170 80.24 V(volts) V
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a c Run PSpice key. The result will
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a c Electronics Workbench We will n
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a c PROBLEMS SECTION 15.2 Impedance
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a c 7. For the circuit of Fig. 15.1
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a c + E = 20 V ∠ 70° - 20 � (a
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a c SECTION 15.7 Admittance and Sus
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a c 31. Repeat Problem 30 for the c
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a c 41. For the network of Fig. 15.
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a c GLOSSARY Admittance A measure o
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710 ⏐⏐⏐ SERIES-PARALLEL ac NE
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744 ⏐⏐⏐ METHODS OF ANALYSIS A
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18 Network Theorems (ac) 18.1 INTRO
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Th Z 1 E 1- + Z 1 I s1 j 4 � Z1
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Th EXAMPLE 18.4 For the network of
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Th pendent sources. The solution ob
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Th is the replacement of the term r
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Th Z1 � R1 � j XL1 � 6 ��
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Th will behave like the actual tran
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Th obtained by applying a source of
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Th Method 3: See Fig. 18.49. Eg Ig
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Th or Eoc�1 � � � k1k2R2
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Th Independent Sources The procedur
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Th Solution: Steps 1 and 2 (Fig. 18
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Th The Norton equivalent circuit ap
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Th so Eg(1 � h) � Ig[R1 � (1
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Th + E = 9 V ∠ 0° - Z1Z 2 (10
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Th (8.54 V) 72.93 and Pmax � �
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Th ��
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Th For 100 W, Is � � Np �Ip
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Th Redrawing the network as shown i
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Th 0.5 -0.5 v s (volts) 0 T /2 T -T
Page 832 and 833:
Th culation of �32.74° of Exampl
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Th FIG. 18.101 Using PSpice to dete
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Th FIG. 18.104 The output file for
Page 838 and 839:
Th FIG. 18.107 Using PSpice to dete
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Th *5. Using superposition, find th
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Th PROBLEMS ⏐⏐⏐ 841 11. Find
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Th PROBLEMS ⏐⏐⏐ 843 20. Find
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Th PROBLEMS ⏐⏐⏐ 845 *40. Find
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Th GLOSSARY ⏐⏐⏐ 847 49. Using
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19 Power (ac) 19.1 INTRODUCTION The
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P q s Power delivered to element by
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P q s The reason for rating some el
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P q s If the average power is zero,
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P q s 2 V and QC � �� (VAR) (
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P q s j I 2 X L = Q L I 2 X C = Q C
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P q s Thus, PT 600 W Fp ��
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P q s Motor: h � Pi � � �45
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P q s + E = E ∠0° - I L Solving
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P q s The equivalent parallel load
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P q s As the name implies, power-fa
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P q s duced, the eddy current loss
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P q s The vast majority of generato
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P q s 349.2 kW � 240 kW � 109.2
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P q s Peak icon to the right of the
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P q s small region below the axis i
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P q s 6. For the circuit of Fig. 19
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P q s *14. For the circuit of Fig.
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P q s SECTION 19.10 Effective Resis
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888 ⏐⏐⏐ RESONANCE E s + - Sou
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890 ⏐⏐⏐ RESONANCE Q L = I 2 X
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892 ⏐⏐⏐ RESONANCE R R( f ) 0
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894 ⏐⏐⏐ RESONANCE f < f s: ne
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896 ⏐⏐⏐ RESONANCE Solving the
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898 ⏐⏐⏐ RESONANCE As Q s of t
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900 ⏐⏐⏐ RESONANCE b. Since Qs
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902 ⏐⏐⏐ RESONANCE I Z T Y T R
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904 ⏐⏐⏐ RESONANCE R l Z Tm Z
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906 ⏐⏐⏐ RESONANCE Setting the
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908 ⏐⏐⏐ RESONANCE Inductive R
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910 ⏐⏐⏐ RESONANCE For an idea
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912 ⏐⏐⏐ RESONANCE TABLE 20.1
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914 ⏐⏐⏐ RESONANCE Example 20.
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916 ⏐⏐⏐ RESONANCE I e. Ql ≥
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918 ⏐⏐⏐ RESONANCE Therefore,
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920 ⏐⏐⏐ RESONANCE Volume Full
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922 ⏐⏐⏐ RESONANCE bandwidth
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924 ⏐⏐⏐ RESONANCE FIG. 20.43
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926 ⏐⏐⏐ RESONANCE cursor esta
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928 ⏐⏐⏐ RESONANCE FIG. 20.48
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930 ⏐⏐⏐ RESONANCE I 2 mA IL I
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932 ⏐⏐⏐ RESONANCE ZT I = 5 mA
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934 ⏐⏐⏐ RESONANCE Z Tp GLOSSA
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936 ⏐⏐⏐ TRANSFORMERS + e p -
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938 ⏐⏐⏐ TRANSFORMERS L p = 20
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940 ⏐⏐⏐ TRANSFORMERS revealin
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942 ⏐⏐⏐ TRANSFORMERS Since th
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944 ⏐⏐⏐ TRANSFORMERS Solution
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946 ⏐⏐⏐ TRANSFORMERS Public a
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948 ⏐⏐⏐ TRANSFORMERS + v x -
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950 ⏐⏐⏐ TRANSFORMERS + V g -
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952 ⏐⏐⏐ TRANSFORMERS + V g -
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954 ⏐⏐⏐ TRANSFORMERS (a) (b)
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956 ⏐⏐⏐ TRANSFORMERS polariti
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958 ⏐⏐⏐ TRANSFORMERS Iron cor
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960 ⏐⏐⏐ TRANSFORMERS Primary
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962 ⏐⏐⏐ TRANSFORMERS Z i + E
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964 ⏐⏐⏐ TRANSFORMERS Source +
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966 ⏐⏐⏐ TRANSFORMERS ��
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968 ⏐⏐⏐ TRANSFORMERS insertio
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970 ⏐⏐⏐ TRANSFORMERS FIG. 21.
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972 ⏐⏐⏐ TRANSFORMERS + Vg = 2
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974 ⏐⏐⏐ TRANSFORMERS q = 1000
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22 Polyphase Systems 22.1 INTRODUCT
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0.866 E m(CN) 0.866 E m(BN) This is
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E BC The length x is C B + - E CN E
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B C E BC E CA (a) ⎪ ⎬ ⎪ ⎭ E
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. EL � �3�Ef � (1.73)(120 V
Page 988 and 989:
E CA = 150 V ∠ v 3 b. Vf � EL.T
Page 990 and 991:
I Bb I AC 30° 120° It can be show
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The phase voltages are Van � IanZ
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Power Factor The power factor of th
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Power Factor S T � 3S f � �3
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A + EAN - or EAN � IfZline � Vf
Page 1000 and 1001:
methods of determining whether the
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IBb � Ibc � Iab � 8.32 A �
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or multiphase, result in a total lo
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40,000 V �120° � 33,200 V �
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7. For the system of Fig. 22.39, fi
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C 14. Repeat Problem 13 if the phas
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3-phase ∆-connected generator Pha
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*43. The Y-Y system of Fig. 22.48 h
Page 1016:
GLOSSARY �-connected ac generator
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1018 ⏐⏐⏐ DECIBELS, FILTERS, A
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Page 1023 and 1024:
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1094 ⏐⏐⏐ PULSE WAVEFORMS AND
Page 1097 and 1098:
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25 Nonsinusoidal Circuits 25.1 INTR
Page 1126 and 1127:
NON Average Value: A 0 The dc term
Page 1128 and 1129:
NON f(t) � �f � t � � T 2
Page 1130 and 1131:
NON EXAMPLE 25.1 Determine which co
Page 1132 and 1133:
NON 1 sin qt i t 1 (i = 0) FIG. 25.
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NON Vm 0 v 1 (a) and v � Vm�1
Page 1136 and 1137:
NON v(a) � V 0 � V m1 sin a �
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NON EXAMPLE 25.7 The input to the c
Page 1140 and 1141:
NON ZT1 � 6 ��j 37.7 ��38
Page 1142 and 1143:
NON FIG. 25.28 Using PSpice to appl
Page 1144 and 1145:
NON you can change the range to 0 H
Page 1146 and 1147:
NON 9. Find the total average power
Page 1148:
NON 24. Given any nonsinusoidal fun
Page 1151 and 1152:
1150 ⏐⏐⏐ SYSTEM ANALYSIS: AN
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1192 Appendixes APPENDIX A PSpice,
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1194 ⏐⏐⏐ APPENDIXES A.3 Mathc
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1196 ⏐⏐⏐ APPENDIXES To Conver
Page 1199 and 1200:
1198 Appendix C DETERMINANTS Determ
Page 1201 and 1202:
1200 ⏐⏐⏐ APPENDIXES 2 3 �3
Page 1203 and 1204:
1202 ⏐⏐⏐ APPENDIXES x � War
Page 1205 and 1206:
1204 ⏐⏐⏐ APPENDIXES D Note th
Page 1207 and 1208:
1206 Appendix D COLOR CODING OF MOL
Page 1209 and 1210:
1208 Appendix F MAGNETIC PARAMETER
Page 1211 and 1212:
1210 ⏐⏐⏐ APPENDIXES and the p
Page 1213 and 1214:
1212 ⏐⏐⏐ APPENDIXES 3. (a) 16
Page 1215 and 1216:
1214 ⏐⏐⏐ APPENDIXES (b) 50(1
Page 1217 and 1218:
1216 ⏐⏐⏐ APPENDIXES (f) 300 W
Page 1219 and 1220:
1218 ⏐⏐⏐ APPENDIXES 9. (a) E
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