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Vg � Ip(Rp � j XLp ) �� Xm�90° Thus, the input impedance is Zi � �Rp�j XLp � or, defining Zp � Rp � j XLp Zs � Rs � j XLs and Xm �90° ��j qM we have (�j qM) Zi � Zp � 2 (Xm �90°) � Zs � ZL 2 �IpXm �90° �� Rs � j XLs � ZL Vg � �� Ip Rs � j XLs � ZL ( qM) and Zi � Zp � �� (21.36) Z � Z The term (qM) 2 /(Z s � Z L) is called the coupled impedance, and it is independent of the sign of M since it is squared in the equation. Consider also that since (qM) 2 is a constant with 0° phase angle, if the load Z L is resistive, the resulting coupled impedance term will appear capacitive due to division of (Z s � R L) into (qM) 2 . This resulting capacitive reactance will oppose the series primary inductance L p, causing a reduction in Z i. Including the effect of the mutual term, the input impedance to the network will appear as shown in Fig. 21.32. EXAMPLE 21.10 Determine the input impedance to the air-core transformer in Fig. 21.33. Z i Solution: q = 400 R p 3 � L p = 6 H M = 0.9 H s 2 L s = 1 H FIG. 21.33 Example 21.10. L R s 0.5 � R L 40 � Zi � Zp � � Rp � j XLp � � 3 ��j 2.4 k�� j 2.4 k�� 129.6 � 103 ((400 rad/s)(0.9 H)) � �� 40.5 � j 400 2 (qM) �� 0.5 ��j 400 ��40 � 2 (qM) �� Rs � j XLs � RL 2 � Zs � ZL AIR-CORE TRANSFORMER ⏐⏐⏐ 957 Z i R p L p Coupled impedence 2M2 –––––– ω Zs + ZL FIG. 21.32 Input characteristics for the air-core transformer.
958 ⏐⏐⏐ TRANSFORMERS Iron core Secondary IL = –––––– 2000 V = 4 A > 2.5 A (rated) 500 Ω + 2000 V XC = 500 Ω – FIG. 21.34 Demonstrating why transformers are rated in kVA rather than kW. � j 2.4 k��322.4 � ��84.22° ⎪ ⎪ ⎪ ⎭ ⎫ ⎪ ⎪ ⎪ ⎬ capacitive � j 2.4 k��(0.0325 k��j 0.3208 k�) � 0.0325 k��j (2.40 � 0.3208) k� and Zi � Ri � j XLi � 32.5 � � j 2079 � � 2079.25 � �89.10° 21.10 NAMEPLATE DATA A typical iron-core power transformer rating, included in the nameplate data for the transformer, might be the following: 5 kVA 2000/100 V 60 Hz The 2000 V or the 100 V can be either the primary or the secondary voltage; that is, if 2000 V is the primary voltage, then 100 V is the secondary voltage, and vice versa. The 5 kVA is the apparent power (S � VI ) rating of the transformer. If the secondary voltage is 100 V, then the maximum load current is S 5000 VA IL ��50 A VL 100 V and if the secondary voltage is 2000 V, then the maximum load current is S 5000 VA IL ��2.5 A VL 2000 V The transformer is rated in terms of the apparent power rather than the average, or real, power for the reason demonstrated by the circuit of Fig. 21.34. Since the current through the load is greater than that determined by the apparent power rating, the transformer may be permanently damaged. Note, however, that since the load is purely capacitive, the average power to the load is zero. The wattage rating would therefore be meaningless regarding the ability of this load to damage the transformer. The transformation ratio of the transformer under discussion can be either of two values. If the secondary voltage is 2000 V, the transformation ratio is a � Np/Ns � Vg /VL � 100 V/2000 V � 1/20, and the transformer is a step-up transformer. If the secondary voltage is 100 V, the transformation ratio is a � Np/Ns � Vg /VL � 2000 V/100 V � 20, and the transformer is a step-down transformer. The rated primary current can be determined simply by applying Eq. (21.18): Is Ip � � a which is equal to [2.5A/(1/20)] � 50A if the secondary voltage is 2000 V, and (50 A/20) � 2.5 A if the secondary voltage is 100 V. To explain the necessity for including the frequency in the nameplate data, consider Eq. (21.12): Ep � 4.44fpNp�m and the B-H curve for the iron core of the transformer (Fig. 21.35). The point of operation on the B-H curve for most transformers is at the knee of the curve. If the frequency of the applied signal should
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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
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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
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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
Page 808 and 809:
Th Method 3: See Fig. 18.49. Eg Ig
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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 ��
Page 826 and 827:
Th For 100 W, Is � � Np �Ip
Page 828 and 829:
Th Redrawing the network as shown i
Page 830 and 831:
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
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,
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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 ��
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
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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
Page 878 and 879:
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
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
Page 949 and 950: 948 ⏐⏐⏐ TRANSFORMERS + v x -
Page 951 and 952: 950 ⏐⏐⏐ TRANSFORMERS + V g -
Page 953 and 954: 952 ⏐⏐⏐ TRANSFORMERS + V g -
Page 955 and 956: 954 ⏐⏐⏐ TRANSFORMERS (a) (b)
Page 957: 956 ⏐⏐⏐ TRANSFORMERS polariti
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
Page 971 and 972: 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 �
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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
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GLOSSARY �-connected ac generator
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1018 ⏐⏐⏐ DECIBELS, FILTERS, A
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1094 ⏐⏐⏐ PULSE WAVEFORMS AND
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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.
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
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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
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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
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1200 ⏐⏐⏐ APPENDIXES 2 3 �3
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1202 ⏐⏐⏐ APPENDIXES x � War
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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
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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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