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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