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ƒ r The total impedance of this network at any frequency is determined by ZT � R � j XL � j XC � R � j (XL � XC) The resonant conditions described in the introduction will occur when (20.2) removing the reactive component from the total impedance equation. The total impedance at resonance is then simply (20.3) representing the minimum value of ZT at any frequency. The subscript s will be employed to indicate series resonant conditions. The resonant frequency can be determined in terms of the inductance and capacitance by examining the defining equation for resonance [Eq. (20.2)]: XL � XC Substituting yields X L � X C Z Ts � R qL � and q 2 1 1 � � � qC LC 1 and qs � � (20.4) �L�C� � 1 f � hertz (Hz) or fs �� L � henries (H) (20.5) 2p�L�C� C � farads (F) The current through the circuit at resonance is E �0° E I �� 0° R �0° R which you will note is the maximum current for the circuit of Fig. 20.2 for an applied voltage E since ZT is a minimum value. Consider also that the input voltage and current are in phase at resonance. Since the current is the same through the capacitor and inductor, the voltage across each is equal in magnitude but 180° out of phase at resonance: VL � (I �0°)(XL �90°) � IXL �90° ⎫ 180° ⎬ out of VC � (I �0°)(XC ��90°) � IXC ��90° ⎭ phase and, since X L � X C, the magnitude of V L equals V C at resonance; that is, V Ls � V Cs (20.6) Figure 20.3, a phasor diagram of the voltages and current, clearly indicates that the voltage across the resistor at resonance is the input voltage, and E, and I, and V R are in phase at resonance. SERIES RESONANT CIRCUIT ⏐⏐⏐ 889 I E V R V L V C FIG. 20.3 Phasor diagram for the series resonant circuit at resonance.
890 ⏐⏐⏐ RESONANCE Q L = I 2 X L S = EI P = I 2 R = EI Q C = I 2 X C FIG. 20.4 Power triangle for the series resonant circuit at resonance. The average power to the resistor at resonance is equal to I 2 R, and the reactive power to the capacitor and inductor are I 2 X C and I 2 X L, respectively. The power triangle at resonance (Fig. 20.4) shows that the total apparent power is equal to the average power dissipated by the resistor since Q L � Q C. The power factor of the circuit at resonance is F p � cos v � and Fps � 1 (20.7) Plotting the power curves of each element on the same set of axes (Fig. 20.5), we note that, even though the total reactive power at any instant is equal to zero (note that t � t ′), energy is still being absorbed and released by the inductor and capacitor at resonance. Power supplied to element 0 Power returned by element p L p C p R p C P � S t 1 t2 t 3 t 4 t 5 p L FIG. 20.5 Power curves at resonance for the series resonant circuit. ƒ r p′ L t p′ C = p′ L A closer examination reveals that the energy absorbed by the inductor from time 0 to t 1 is the same as the energy released by the capacitor from 0 to t 1. The reverse occurs from t 1 to t 2, and so on. Therefore, the total apparent power continues to be equal to the average power, even though the inductor and capacitor are absorbing and releasing energy. This condition occurs only at resonance. The slightest change in frequency introduces a reactive component into the power triangle, which will increase the apparent power of the system above the average power dissipation, and resonance will no longer exist. 20.3 THE QUALITY FACTOR (Q ) The quality factor Q of a series resonant circuit is defined as the ratio of the reactive power of either the inductor or the capacitor to the average power of the resistor at resonance; that is, Qs �� reactive power � average power (20.8) The quality factor is also an indication of how much energy is placed in storage (continual transfer from one reactive element to the other) compared to that dissipated. The lower the level of dissipation for the same
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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
Page 800 and 801:
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
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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
Page 834 and 835:
Th FIG. 18.101 Using PSpice to dete
Page 836 and 837:
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
Page 854 and 855: P q s The reason for rating some el
Page 856 and 857: 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
Page 866 and 867: P q s + E = E ∠0° - I L Solving
Page 868 and 869: P q s The equivalent parallel load
Page 870 and 871: P q s As the name implies, power-fa
Page 872 and 873: P q s duced, the eddy current loss
Page 874 and 875: P q s The vast majority of generato
Page 876 and 877: 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
Page 884 and 885: P q s *14. For the circuit of Fig.
Page 886: P q s SECTION 19.10 Effective Resis
Page 889: 888 ⏐⏐⏐ RESONANCE E s + - Sou
Page 893 and 894: 892 ⏐⏐⏐ RESONANCE R R( f ) 0
Page 895 and 896: 894 ⏐⏐⏐ RESONANCE f < f s: ne
Page 897 and 898: 896 ⏐⏐⏐ RESONANCE Solving the
Page 899 and 900: 898 ⏐⏐⏐ RESONANCE As Q s of t
Page 901 and 902: 900 ⏐⏐⏐ RESONANCE b. Since Qs
Page 903 and 904: 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)
Page 957 and 958:
956 ⏐⏐⏐ TRANSFORMERS polariti
Page 959 and 960:
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
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 �
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
Page 1021 and 1022:
Page 1023 and 1024:
Page 1025 and 1026:
Page 1027 and 1028:
Page 1029 and 1030:
Page 1031 and 1032:
Page 1033 and 1034:
Page 1035 and 1036:
Page 1037 and 1038:
Page 1039 and 1040:
Page 1041 and 1042:
Page 1043 and 1044:
Page 1045 and 1046:
Page 1047 and 1048:
Page 1049 and 1050:
Page 1051 and 1052:
Page 1053 and 1054:
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Page 1063 and 1064:
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Page 1077 and 1078:
Page 1079 and 1080:
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Page 1085 and 1086:
Page 1087 and 1088:
Page 1089 and 1090:
Page 1091 and 1092:
Page 1093 and 1094:
Page 1095 and 1096:
1094 ⏐⏐⏐ PULSE WAVEFORMS AND
Page 1097 and 1098:
Page 1099 and 1100:
Page 1101 and 1102:
Page 1103 and 1104:
Page 1105 and 1106:
Page 1107 and 1108:
Page 1109 and 1110:
Page 1111 and 1112:
Page 1113 and 1114:
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Page 1117 and 1118:
Page 1119 and 1120:
Page 1121 and 1122:
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
Page 1153 and 1154:
Page 1155 and 1156:
Page 1157 and 1158:
Page 1159 and 1160:
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Page 1163 and 1164:
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Page 1177 and 1178:
Page 1179 and 1180:
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Page 1189 and 1190:
Page 1191 and 1192:
Page 1193 and 1194:
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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