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S wiring, you will find that since the bulbs are connected in series, there is one wire to each bulb with additional wires from plug to plug. Why would they need two additional wires if the bulbs are connected in series? The answer lies in the fact that when each set is connected together, they will actually be in parallel (to be discussed in the next chapter) by a unique wiring arrangement shown in Fig. 5.63(b) and redrawn in Fig. 5.63(c) to clearly show the parallel arrangement. Note that the top line is the hot line to all the connected sets, and the bottom line is the return, neutral, or ground line for all the sets. Inside the plug of Fig. 5.63(d) the hot line and return are connected to each set, with the connections to the metal spades of the plug as shown in Fig. 5.63(b). We will find in the next chapter that the current drawn from the wall outlet for parallel loads is the sum of the current to each branch. The result, as shown in Fig. 5.63(c), is that the current drawn from the supply is 6 � 200 mA � 1.2 A, and the total wattage for all six sets is the product of the applied voltage and the source current or (120 V)(1.2 A) � 144 W with 144 W/6 � 24 W per set. Microwave Oven Series circuits can be very effective in the design of safety equipment. Although we all recognize the usefulness of the microwave oven, it can be quite dangerous if the door is not closed or sealed properly. It is not enough to test the closure at only one point around the door because the door may be bent or distorted from continual use, and leakage could result at some point distant from the test point. One common safety arrangement appears in Fig. 5.64. Note that magnetic switches are located all around the door, with the magnet in the door itself and the magnetic door switch in the main frame. Magnetic switches are simply switches where the magnet draws a magnetic conducting bar between two contacts to complete the circuit—somewhat revealed by the symbol Magnets Magnetic door switches Magnets FIG. 5.64 Series safety switches in a microwave oven. Series safety switches Microwave power unit APPLICATIONS ⏐⏐⏐ 153
154 ⏐⏐⏐ SERIES CIRCUITS S Door switch Window foil Magnetic switch E I s VR1 R1 + – 6 Ω 54 V Sensors ≅ 5 mA Sensing relay FIG. 5.65 Series alarm circuit. + V 1 – + VR2 – R2 7 Ω To bell circuit 5 V @ 5mA 1 kΩ 5 V Power supply FIG. 5.66 Series dc network to be analyzed using PSpice. + V 2 – + VR3 – R3 5 Ω for the device in the circuit diagram of Fig. 5.64. Since the magnetic switches are all in series, they must all be closed to complete the circuit and turn on the power unit. If the door is sufficiently out of shape to prevent a single magnet from getting close enough to the switching mechanism, the circuit will not be complete, and the power cannot be turned on. Within the control unit of the power supply, either the series circuit completes a circuit for operation or a sensing current is established and monitored that controls the system operation. Series Alarm Circuit The circuit of Fig. 5.65 is a simple alarm circuit. Note that every element of the design is in a series configuration. The power supply is a 5-V dc supply that can be provided through a design similar to that of Fig. 2.31, a dc battery, or a combination of an ac and a dc supply that ensures that the battery will always be at full charge. If all the sensors are closed, a current of 5 mA will result because of the terminal load of the relay of about 1 k�. That current energizes the relay and maintains an off position for the alarm. However, if any of the sensors are opened, the current will be interrupted, the relay will let go, and the alarm circuit will be energized. With relatively short wires and a few sensors, the system should work well since the voltage drop across each will be minimal. However, since the alarm wire is usually relatively thin, resulting in a measurable resistance level, if the wire to the sensors is too long, a sufficient voltage drop could occur across the line, reducing the voltage across the relay to a point where the alarm fails to operate properly. Thus, wire length is a factor that must be considered if a series configuration is used. Proper sensitivity to the length of the line should remove any concerns about its operation. An improved design will be described in Chapter 8. 5.12 COMPUTER ANALYSIS PSpice In Section 4.9, the basic procedure for setting up the PSpice folder and running the program were presented. Because of the detail provided in that section, you should review it before proceeding with this example. Because this is only the second example using PSpice, some detail will be provided, but not at the level of Section 4.9. The circuit to be investigated appears in Fig. 5.66. Since the PSpice folder was established in Section 4.9, there is no need to repeat the process here—it is immediately available. Double-clicking on the Orcad Lite Edition icon will generate the Orcad Capture-Lite Edition window. A new project is then initiated by selecting the Create document key at the top left of the screen (it looks like a page with a star in the upper left corner). The result is the New Project dialog box in which SeriesDC is inserted as the Name. The Analog or Mixed A/D is already selected, and C:\PSpice appears as the Location—only the Name had to be entered! Click OK, and the Create PSpice Project dialog box will appear. Select Create a blank project, click OK, and the working windows will appear. Grabbing the left edge of the SCHEMATIC1:PAGE1 window will allow you to move it to the right so that you can see both screens. Clicking the + sign in the Project Manager window will allow you to set the sequence down to PAGE1. If you prefer to change the name of the SCHEMATIC1, just select it
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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
Page 103 and 104: 102 ⏐⏐⏐ OHM’S LAW, POWER, A
Page 130 and 131: 5 Series Circuits 5.1 INTRODUCTION
Page 132 and 133: S the same through series elements
Page 134 and 135: S Solution: RT � R1 � R2 � R3
Page 136 and 137: S type of element. In other words,
Page 138 and 139: S c. V1 � IR1 � (2 A)(4 �)
Page 140 and 141: S In the above discussion the curre
Page 142 and 143: S Voltage Sources and Ground Except
Page 144 and 145: S EXAMPLE 5.14 Find the voltage Vab
Page 146 and 147: S EXAMPLE 5.19 Using the voltage di
Page 148 and 149: S ∆V L 120 V 100 V 0 V L voltage
Page 150 and 151: S EXAMPLE 5.24 Determine the voltag
Page 152 and 153: S 5.11 APPLICATIONS Holiday Lights
Page 156 and 157: S and right-click on the mouse. A l
Page 158 and 159: S FIG. 5.68 Applying Electronics Wo
Page 160 and 161: S also be used to remind the progra
Page 162 and 163: S 3. Find the applied voltage E nec
Page 164 and 165: S 9. Determine the current I and th
Page 166 and 167: S 40 V I + R 3 R 2 R 1 V 3 - 30 �
Page 168 and 169: S *28. For the network of Fig. 5.97
Page 170 and 171: 6 Parallel Circuits 6.1 INTRODUCTIO
Page 172 and 173: P rent level. In other words, as th
Page 174 and 175: P EXAMPLE 6.4 a. Find the total res
Page 176 and 177: P R T R 6 � R′ T � ��
Page 178 and 179: P d. RT � 30 � � 30 � � 0
Page 180 and 181: P 0.25 S � 0.1 S � 0.05 S � 0
Page 182 and 183: P EXAMPLE 6.13 Determine the curren
Page 184 and 185: P I2 � I4 � I5 12 A � I4 �
Page 186 and 187: P For the particular case of two pa
Page 188 and 189: P and (R 1 � R 2)I 1 � R 2I R 1
Page 190 and 191: P A short circuit is a very low res
Page 192 and 193: P must therefore be zero volts, as
Page 194 and 195: P + E - A + - I s + - ensure a posi
Page 196 and 197: P 12-gage fuse link Filter capacito
Page 198 and 199: P car such as the lights, air condi
Page 200 and 201: P careful, you can work with one li
Page 202 and 203: 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
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Speakers and Microphones Electromag
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4,000,000,000,000 bits of informati
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ingly enough one that perhaps will
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Hall Effect Sensor The Hall effect
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ciently close to establish contact
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SECTION 11.8 Hysteresis 9. For the
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*18. For the series-parallel magnet
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12 Inductors 12.1 INTRODUCTION We h
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Inductors are coils of various dime
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(a) (d) (b) FIG. 12.10 Various type
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ciated with the applied ac signal m
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the voltage across the coil is not
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y 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0
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12.8 INITIAL VALUES This section wi
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Let us now test the validity of the
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The mathematical expression for the
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v R1 R1 Defined polarity + vL - v L
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iL � (1 � e �t/t ) t � �
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12.12 INDUCTORS IN SERIES AND PARAL
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Applying the current divider rule,
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EXAMPLE 12.12 Find the energy store
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+ Feed 120 V ac - Return ��
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would need for 15 W, not to mention
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will scatter to all sides of the mo
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ing trace appears in the bottom of
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FIG. 12.61 Using PSpice to determin
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Parameters, use 0 s as the Start ti
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*11. Find the waveform for the curr
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*19. For the network of Fig. 12.76:
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*29. The switch of Fig. 12.83 has b
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37. Find the voltage V 1 and the cu
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522 ⏐⏐⏐ 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
Page 796 and 797:
Th EXAMPLE 18.4 For the network of
Page 798 and 799:
Th pendent sources. The solution ob
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Th is the replacement of the term r
Page 802 and 803:
Th Z1 � R1 � j XL1 � 6 ��
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Th will behave like the actual tran
Page 806 and 807:
Th obtained by applying a source of
Page 808 and 809:
Th Method 3: See Fig. 18.49. Eg Ig
Page 810 and 811:
Th or Eoc�1 � � � k1k2R2
Page 812 and 813:
Th Independent Sources The procedur
Page 814 and 815:
Th Solution: Steps 1 and 2 (Fig. 18
Page 816 and 817:
Th The Norton equivalent circuit ap
Page 818 and 819:
Th so Eg(1 � h) � Ig[R1 � (1
Page 820 and 821:
Th + E = 9 V ∠ 0° - Z1Z 2 (10
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Th (8.54 V) 72.93 and Pmax � �
Page 824 and 825:
Th ��
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Th For 100 W, Is � � Np �Ip
Page 828 and 829:
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
Page 834 and 835:
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
Page 840 and 841:
Th *5. Using superposition, find th
Page 842 and 843:
Th PROBLEMS ⏐⏐⏐ 841 11. Find
Page 844 and 845:
Th PROBLEMS ⏐⏐⏐ 843 20. Find
Page 846 and 847:
Th PROBLEMS ⏐⏐⏐ 845 *40. Find
Page 848 and 849:
Th GLOSSARY ⏐⏐⏐ 847 49. Using
Page 850 and 851:
19 Power (ac) 19.1 INTRODUCTION The
Page 852 and 853:
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,
Page 858 and 859:
P q s 2 V and QC � �� (VAR) (
Page 860 and 861:
P q s j I 2 X L = Q L I 2 X C = Q C
Page 862 and 863:
P q s Thus, PT 600 W Fp ��
Page 864 and 865:
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
Page 870 and 871:
P q s As the name implies, power-fa
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P q s duced, the eddy current loss
Page 874 and 875:
P q s The vast majority of generato
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P q s 349.2 kW � 240 kW � 109.2
Page 878 and 879:
P q s Peak icon to the right of the
Page 880 and 881:
P q s small region below the axis i
Page 882 and 883:
P q s 6. For the circuit of Fig. 19
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P q s *14. For the circuit of Fig.
Page 886:
P q s SECTION 19.10 Effective Resis
Page 889 and 890:
888 ⏐⏐⏐ RESONANCE E s + - Sou
Page 891 and 892:
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
Page 897 and 898:
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
Page 905 and 906:
904 ⏐⏐⏐ RESONANCE R l Z Tm Z
Page 907 and 908:
906 ⏐⏐⏐ RESONANCE Setting the
Page 909 and 910:
908 ⏐⏐⏐ RESONANCE Inductive R
Page 911 and 912:
910 ⏐⏐⏐ RESONANCE For an idea
Page 913 and 914:
912 ⏐⏐⏐ RESONANCE TABLE 20.1
Page 915 and 916:
914 ⏐⏐⏐ RESONANCE Example 20.
Page 917 and 918:
916 ⏐⏐⏐ RESONANCE I e. Ql ≥
Page 919 and 920:
918 ⏐⏐⏐ RESONANCE Therefore,
Page 921 and 922:
920 ⏐⏐⏐ RESONANCE Volume Full
Page 923 and 924:
922 ⏐⏐⏐ RESONANCE bandwidth
Page 925 and 926:
924 ⏐⏐⏐ RESONANCE FIG. 20.43
Page 927 and 928:
926 ⏐⏐⏐ RESONANCE cursor esta
Page 929 and 930:
928 ⏐⏐⏐ RESONANCE FIG. 20.48
Page 931 and 932:
930 ⏐⏐⏐ RESONANCE I 2 mA IL I
Page 933 and 934:
932 ⏐⏐⏐ RESONANCE ZT I = 5 mA
Page 935 and 936:
934 ⏐⏐⏐ RESONANCE Z Tp GLOSSA
Page 937 and 938:
936 ⏐⏐⏐ TRANSFORMERS + e p -
Page 939 and 940:
938 ⏐⏐⏐ TRANSFORMERS L p = 20
Page 941 and 942:
940 ⏐⏐⏐ TRANSFORMERS revealin
Page 943 and 944:
942 ⏐⏐⏐ TRANSFORMERS Since th
Page 945 and 946:
944 ⏐⏐⏐ TRANSFORMERS Solution
Page 947 and 948:
946 ⏐⏐⏐ TRANSFORMERS Public a
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948 ⏐⏐⏐ TRANSFORMERS + v x -
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950 ⏐⏐⏐ TRANSFORMERS + V g -
Page 953 and 954:
952 ⏐⏐⏐ TRANSFORMERS + V g -
Page 955 and 956:
954 ⏐⏐⏐ TRANSFORMERS (a) (b)
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956 ⏐⏐⏐ TRANSFORMERS polariti
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958 ⏐⏐⏐ TRANSFORMERS Iron cor
Page 961 and 962:
960 ⏐⏐⏐ TRANSFORMERS Primary
Page 963 and 964:
962 ⏐⏐⏐ TRANSFORMERS Z i + E
Page 965 and 966:
964 ⏐⏐⏐ TRANSFORMERS Source +
Page 967 and 968:
966 ⏐⏐⏐ TRANSFORMERS ��
Page 969 and 970:
968 ⏐⏐⏐ TRANSFORMERS insertio
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970 ⏐⏐⏐ TRANSFORMERS FIG. 21.
Page 973 and 974:
972 ⏐⏐⏐ TRANSFORMERS + Vg = 2
Page 975 and 976:
974 ⏐⏐⏐ TRANSFORMERS q = 1000
Page 978 and 979:
22 Polyphase Systems 22.1 INTRODUCT
Page 980 and 981:
0.866 E m(CN) 0.866 E m(BN) This is
Page 982 and 983:
E BC The length x is C B + - E CN E
Page 984 and 985:
B C E BC E CA (a) ⎪ ⎬ ⎪ ⎭ E
Page 986 and 987:
. 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
Page 992 and 993:
The phase voltages are Van � IanZ
Page 994 and 995:
Power Factor The power factor of th
Page 996 and 997:
Power Factor S T � 3S f � �3
Page 998 and 999:
A + EAN - or EAN � IfZline � Vf
Page 1000 and 1001:
methods of determining whether the
Page 1002 and 1003:
IBb � Ibc � Iab � 8.32 A �
Page 1004 and 1005:
or multiphase, result in a total lo
Page 1006 and 1007:
40,000 V �120° � 33,200 V �
Page 1008 and 1009:
7. For the system of Fig. 22.39, fi
Page 1010 and 1011:
C 14. Repeat Problem 13 if the phas
Page 1012 and 1013:
3-phase ∆-connected generator Pha
Page 1014 and 1015:
*43. The Y-Y system of Fig. 22.48 h
Page 1016:
GLOSSARY �-connected ac generator
Page 1019 and 1020:
1018 ⏐⏐⏐ DECIBELS, FILTERS, A
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1094 ⏐⏐⏐ PULSE WAVEFORMS AND
Page 1097 and 1098:
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Page 1124 and 1125:
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.
Page 1134 and 1135:
NON Vm 0 v 1 (a) and v � Vm�1
Page 1136 and 1137:
NON v(a) � V 0 � V m1 sin a �
Page 1138 and 1139:
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,
Page 1195 and 1196:
1194 ⏐⏐⏐ APPENDIXES A.3 Mathc
Page 1197 and 1198:
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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