fundamentals of electric circuits solutions manual chapter 9
fundamentals of electric circuits
... 190 Problems 191 Comprehensive Problems 200Contentsxi Chapter 2 Basic Laws 27 Chapter 3 Methods of Analysis 75PART 1 DC CIRCUITS 1 Chapter 1 Basic Concepts 3 Chapter 4 Circuit Theorems 1 19 Chapter ... Odd-Numbered Problems 893 Selected Bibliography 92 9Index 93 3xivCONTENTS Chapter 16 The Fourier Series 707 Chapter 17 Fourier Transform 7 59 Chapter 18 Two-Port Networks 795 | | | |▲▲e-Text ... MenuTextbook Table of ContentsProblem Solving Workbook Contents 1DC CIRCUITS PART 1 Chapter 1Basic Concepts Chapter 2Basic Laws Chapter 3Methods of Analysis Chapter 4Circuit Theorems Chapter 5Operational...
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FUNDAMENTALS OF ELECTRIC DRIVERS doc
... reserved.Any use is subject to the Terms of Use as given at the website. FUNDAMENTALS OF ELECTRIC SYSTEMS FUNDAMENTALS OF ELECTRIC SYSTEMS 1 .9 FIGURE 1 .9 Iron filings around a wire carrying a ... 2.15 55,000 80 10.7Supermendur 49% Co, V 2.4 15 ,90 0 80,000 8 26Vanadium Permendur 49% Co, V 2.3 12,700 4 ,90 0 92 40 93 2Hyperco 27 27% Co 2.36 70,000 2,800 198 19 925Flake iron Carbonal power ... subject to the Terms of Use as given at the website. FUNDAMENTALS OF ELECTRIC SYSTEMS FUNDAMENTALS OF ELECTRIC SYSTEMSCAPACITORSFigure 1.1 illustrates a capacitor. It consists of two insulated...
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Fundamentals of Corporate Finance 8th edition: Solutions Manual
... = .5 294 Now we can use the sustainable growth rate equation to get: Sustainable growth rate = (ROE ì b) / [1 (ROE ì b)] Sustainable growth rate = [.1 899 (.5 294 )] / [1 – .1 899 (.5 294 )] ... $10,157 / (1 – 0.34) = $15,3 89. 39 Now, we can add interest to EBT to get EBIT as follows: EBIT = EBT + Interest paid = $15,3 89. 39 + 3,405 = $18, 794 . 39 CHAPTER 4 B-35 10. Below is ... $41,7 69 + 85,000 = $126,7 69 Now we can calculate Tobin’s Q, which is: Tobin’s Q = (Market value of equity + Book value of debt) / Book value of assets Tobin’s Q = ($860,000 + 126,7 69) ...
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Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf
... limit of the capability curve would be drawn from an origin of 3V 2 3( )1328 V 2 Q = ⎞ = = 4810 kVAR SX 1.1 & The radius of the ... generate the power by means of a motor-generator set consisting of a synchronous motor driving a synchronous generator. How many poles should each of the two machines have in ... voltage of 200 V, and the maximum I F is 10 A. The resistance of the field circuit is adjustable over the range from 20 to 200 &. The OCC of this generator...
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Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt
... speed of 1800 r/min and a full-load speed of 1785 r/min. The loads supplied by the two generators consist of 100 kW at 0.85 PF lagging. (a) Calculate the speed droops of generator ... advantage of this fact. % M-file: prob5_4e.m % M-file to calculate and plot the terminal voltage % of a synchronous generator as a function of load % for power factors of 0.8 ... magnitude of E A is 12,040 V. (b) The torque angle of the generator at rated conditions is ™ = 17.6°. (c) Ignoring RA , the maximum output power of the generator...
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Electric Machinery Fundamentals (Solutions Manual) Part 3 docx
... of Gen A fnlB = 61.5; % No-load freq of Gen B fnlC = 60.5; % No-load freq of Gen C spA = 1.5; % Slope of Gen A (MW/Hz) spB = 1.676; % Slope of Gen B (MW/Hz) spC = 1 .96 1; ... /Hz ( ) sys 90 00 kW+ 0.15 MW /Hz sysf f 0.25 MW/Hz fsys 6050 kW 90 00 kW= + 100 kW fsys 14, 95 0 kW 59. 8 Hz= = 0.25 MW/Hz ... 61.5 sys 1 .96 1 60.5f f sysf + + 7 MW 91 .5 1.= 5.137 fsys = 306.2 fsys = 59. 61 Hz 5 sys 103.07 1.676 sys 118.64 1 .96 1 sysf...
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Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx
... P sinA 3( )277 V (5 09 V ) ( ) sin 471 kW sin™ ™ ™ = = = SXG 0. 899 & The power supplied as a function of the torque angle ™ may be ... % M-file: prob5_ 19. m % M-file to plot the power output of a % synchronous generator as a function of % the torque angle. % Calculate Xs delta = (0:1 :90 ); % Torque ... reactance of this generator at the rated conditions? (b) What is the unsaturated synchronous reactance of this generator? (c) Plot the saturated synchronous reactance of this...
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Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx
... )46 19 36.87 A (1.716 )( )46 19 36.87 A EA = ° + & ° j+ & ° EA 13, 590 27.6 V= ° Therefore, the magnitude of ... I A A⎞ A S AR jX 7621 0 ( )0 .9 ( 52 49 36.87 A ) E A j= ° + & ° EA 11,120 19. 9 V= ° The resulting voltage regulation ... 1 39 When this program is executed, the plot of reactive power versus flux is (d) The program in part (c) of this program calculated I A as a function of ...
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Electric Machinery Fundamentals (Solutions Manual) Part 6 ppt
... j= ° + & ° EA 92 64 19. 9= V° The resulting voltage regulation is 92 64 6351 RV 100% 45 .9% = ⋅ = 6351 Because voltage, ... MW/Hz f nl1 = 61 .91 Hz fnl1 60 Hz (d) If the swing generator trips off the line, the other three generators would have to supply all 290 MW of the load. Therefore, ... 290 MW 34 62.21 = 8.5 29 186.63 3 f sys fsys = 59. 37 Hz sys 34 62.21 sys 34 62.21f f sysf Each generator will supply 96 .7...
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Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf
... 9, 92 3 V , and sin 1 1E 1A sin sin 13, 230™ ™ V sin= = 27 .9 38.6° = ° 2 2 EA2 Therefore, the new armature current is 9, 923 ... 298 .4 kW= = 4 49 A= L 3 PF 3 ( )480 V (0.8) TV Because the motor is -connected, the corresponding phase current is 4 49 / 3 2 59 AI ... ⎞ 9, 92 3 38.6 70 ° 44 0 762 6.6 A E V AI = = jX S j8.18 ° = ° (f) A MATLAB program to plot the magnitude of the armature...
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Electric Machinery Fundamentals (Solutions Manual) Part 8 pps
... change? (Hint: Think about the derivation of X S .) 157 198 & 0 A ° 14 39 = 22. 7 V° These values of E A and ™ at unity ... (141.5 31.8 A) E A = ° j & ° EA 4 29 24= .9 V° So EA = 4 29 V at rated conditions. The resulting plot is shown below 260 ... voltage of would require a field current of 4.6 A. (b) The motor’s efficiency at full load and unity power factor is ⎜ OUTP 100%= ⋅ 746 k= W 100%⋅ 89. 3%=...
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Electric Machinery Fundamentals (Solutions Manual) Part 9 ppt
... ° EA 124 14 .9 = V° The torque angle ™ of this machine is –14 .9 °. (b) A phasor diagram of the motor operating at a power factor of 0.78 leading is shown ... generated voltage of the motor is 161 = E V I ,m ⎞ ,m jX ,mA S A ( )( ) E ,mA E ,mA 277 0 V= ° j 1.1 297 20.= 9 V° 96 & ... the magnitudes and angles of EA for both machines. (b) If the flux of the motor is increased by 10 percent, what happens to the terminal voltage of the power system? What is...
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Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt
... are 168 reactance of 0 .90 and a per-unit resistance of 0.02. (a) What is the rated input power of this motor? (b) What is the magnitude of EA at rated conditions? ... What is the speed of rotation of this motor? (b) What is the output torque of this motor at the rated conditions? (c) What is the internal generated voltage of this motor ... hp ( )1200 r/min 91 , 91 0 lb ft = = = ⊕ mn 6-13. A 440-V three-phase Y-connected synchronous motor has a synchronous reactance of 1.5 & per phase. ...
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Electric Machinery Fundamentals (Solutions Manual) Part 11 ppsx
... 3 A AP I 3R 79. 4 A 0.22 4.16 kW (f) coP nv at rated conditions is = = = conv IN CP P U 83P .8 kW 4.16 kW 79. 6 kW (g) If EA ... If EA is decreased by 10%, the new value if EA = (0 .9) (603 V) = 543 V. To simplify this part of the problem, we will ignore RA . Then the quantity EA sin ... EA changes. Therefore, sin E 1A sin sin 603 V™ ™ sin ( )1 1= = 19. 5 21.8 ° = ° 2 1 Therefore, I EA2 ⎞ AV...
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