ACirculating current on no-load is given byV-(1,+1)Z-IZ4-2+2enfExample 12.57. Two single-phase alternators are connected in parallel and supply current to aload at a terminal voltage of 10000 20 volts. Alternator A has an induced em.f. of 13000222-6⁰volts and a reactance of 20; alternator B has an e.mf. of 12500 236-9 volts and a reactance of 3Find the current supplied by each alternatorSolution. Current supplied by alternator A is given by;LAproblem;E-VE-VZ₁Current supplied by alternator B is given by:4-F.L. current of generator A is4-4-F.L. current of generator BisE-V 13000 222-6-10000 402290XA/ZA-6500 2-67-4*+j 5000 (2500-j/ 1000) A- 2700 2-21-8 AAlso- (2500+/0)A-2500 20 AExample 12.5 Two 3-phase alternators A and B supply a sub-station through lines 1 and 2.The sub-station load is R-500;X-402 to neutral. If the generators e.mf. are A, 10000 V andB. 12000 V and A leads B by 10 elect; find currents delivered by A and B.: 10,000 kVA, 11000 V, 10% reactance20,000 kVA, 11000 V, 15 % reactanceGenerator A ratingGenerator B ratingReactance of line 1-3-82: Reactance of line 2-41Solution. Fig. 12.87 shows the conditions of the-E-V 12500 236-9-10000 20Z329010,000x10√3x1100020,000x10√3x11000<-525 A4-4-<- 1050 ALetx, and x, be the respective per phase reactancesof the generators in ohms.AA525 x,10% of 11000/510 11000 12₁100 31050-15% of 11000/515110001<-1201010-090750013(50+)40)Fig. 12.87andNow2₂+41-09+41-50Z:(5)2; Z-(50+/40) 210,000 210-5774 210 volts (5686+/1003) volts√√3E-Referring to Fig. 12.87, we have,andSubtracti1,2+(1+1)ZE1₂ Z₁ +(1₂+1₂) Z-Eom eq. (1), we have,E₁-E₂12000 20-(6928+/0) volts√√3Adding eqs. () and (i), we have,4, +4₂=-1,-1,- (200+/250) A$686+/1003)-(6928+/0)j5-(200+/250) AZ₁12431003jsE,+E, (5686+/1003)+(6928+/0)-2Z+Z₁2 (50+/40)+/S12654 24-5131-2240-412614+/1003100+8596-42-35-9A-(78-1-/56-6) A(78-1-56-6) A1₁+1₂solving eqs. (ii) and (iv), we have,Hello expert, I want you to explain howhe got the law inside the circle. Can youexplain step by step and a clear line, please?

Given: The solution of a particular question. To find: How did the expression ( inside the circle )…

by: -E₂ N hernators are connected in parallel and supply current to a volts. Alternator A has an induced e.mf. of 13000222-6 B has an e.mf. of 12500 236-9° volts and a reactance of 3 ernator ator A is given by; -V -ZA -- 13000 222-6-10000 40 2290 5002-67-4+j5000 (2500-/1000) A 700 2-21-8"A given by: -V 12500 236-9-10000 20/7500 3290 Z 500+/0)A-2500 20 A mators A and B supply a sub-station through lines 1 and 2: 0 to neutral. If the generalors e.mf. are A, 10000 V and nd currents delivered by A and B 00 kVA, 11000 V, 10% reactance 00 kVA, 11000 V, 15 % reactance 525 A lance of line 2-419 ditions of the . 1050 A se reactances 14 10% of 11000/5 (50+)40) 2 Fig. 12.87 Referring to Fig. 12.87, we have, 1,2+1, +424 4₂2, +(1+1)Z-E Som eq. (1), we have, and Subtracti 1,-1,- E-E₂686+/1003)-(6928+/0) j5 -(200+/250) A Z₁ 13431003 js 1,-1,- (200+/250) A Adding eqs. () and (i), we have, E₁ + E₂ 2Z+Z₁ (5686+/1003)+(6928+/0) 2 (50+/40)+/S 12654 24-5 131-2240-4 12614+/1003 100+/85 -9642-35-9A-(78-1-56-6) A (78-1-56-6) A 1+1 solving eqs. (i) and (iv), we have, 169 A (M) Hello expert, I want you to explain how he got the law inside the circle. Can you explain step by step and a clear line, please?

by: -E₂ N hernators are connected in parallel and supply current to a volts. Alternator A has an induced e.mf. of 13000222-6 B has an e.mf. of 12500 236-9° volts and a reactance of 3 ernator ator A is given by; -V -ZA -- 13000 222-6-10000 40 2290 5002-67-4+j5000 (2500-/1000) A 700 2-21-8"A given by: -V 12500 236-9-10000 20/7500 3290 Z 500+/0)A-2500 20 A mators A and B supply a sub-station through lines 1 and 2: 0 to neutral. If the generalors e.mf. are A, 10000 V and nd currents delivered by A and B 00 kVA, 11000 V, 10% reactance 00 kVA, 11000 V, 15 % reactance 525 A lance of line 2-419 ditions of the . 1050 A se reactances 14 10% of 11000/5 (50+)40) 2 Fig. 12.87 Referring to Fig. 12.87, we have, 1,2+1, +424 4₂2, +(1+1)Z-E Som eq. (1), we have, and Subtracti 1,-1,- E-E₂686+/1003)-(6928+/0) j5 -(200+/250) A Z₁ 13431003 js 1,-1,- (200+/250) A Adding eqs. () and (i), we have, E₁ + E₂ 2Z+Z₁ (5686+/1003)+(6928+/0) 2 (50+/40)+/S 12654 24-5 131-2240-4 12614+/1003 100+/85 -9642-35-9A-(78-1-56-6) A (78-1-56-6) A 1+1 solving eqs. (i) and (iv), we have, 169 A (M) Hello expert, I want you to explain how he got the law inside the circle. Can you explain step by step and a clear line, please?

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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If the load is connected to the generator thgrough an inductive line and the voltage at the load side is the same as the voltage at the terminals of the generator, which of the following must be true? O A. The reactive power is flowing from the generator to the load and from load to the generator at the same time. O B. The load is disconnected, and no power is delivered to the load. O C. The load has a poor power factor and it's absorbing too much reactive power. O D. None of the other choices are correct O E. The generator is unable to produce reactive power; therefore, the reactive power is flowing from the load to the generator.
X₂=0.1 1 XL +03 Soto X₁ = 0.2 X = 0.1 X = 0.1 Above is the one-line diagram of a simple power system. Each generator is represented by an emf behind the subtransient reactance. All impedances are expressed in per unit on a common base. All resistances and shunt capacitance are neglected. The generators are operating on no load at their rated voltage with their emfs in phase. A symmetrical fault occurs at bus 1 through a fault impedance Z, =j0.08 per unit. a) Using Thevenin's Theorem, determine the impedance to the point of fault and the fault current in per unit. b) Find the bus voltages and line currents during the fault. 2 O oto
What is the average power in ghe single phase ac circuit with the purely resistive load? It is sinusoidal steady state excitation.
An impedance coil, in series with a resistor takes a current of 3A when connected to a 120V AC source. Using a voltmeter, the voltage across the coil was found out equal to 90V and that across the resistor equal to 30V. Find the power factor of the impedance coil only. The frequency is 60 Hz
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A three-phase star-connected alternator delivers at rated load a current I = 200 A under a voltage of 5 kV-50 Hz (between phases). The load is inductive of power factor equal to 0.87, the resistance of each stator winding is 0.292 and the speed of the rotor is n = 750 r.p.m. The sum of the constants losses and the stator Joule's losses and the rotor Joule's losses is 55 kW. The no load test, at rated frequency, gave: E-4200 V (between phases) for an excitation current lexc = 40 A. The short-circuit test, for an excitation current lexc = 40 A, gave a short-circuit current in the stator windings Ise =2.5 kA. 1- Calculate the numbers of poles of the alternator. 2- Calculate the synchronous reactance (X). 3- Calculate the electromotive force of the alternator (E) between phases for the same excitation current. 4- Calculate the power absorbed by the alternator (Pa).
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2. Three series loads are drawing 20A ms current through a 60HZ source. Load 1 is an 400W, 0.8 lagging induction motor. Load 2 is a 500VA, 0.6 lagging power factor and load 3 is a 2 ohm incandescent bulb. a) What is the total VA of the combined load? b) What is the overall PF of the combined load? c) What RMS voltage is the combined load operating? d) What capacitor VAR must be installed in parallel to the combine load to improve its overall PF to unity? e) What is its approximate capacitance in microfarad?