ACirculating current on no-load is given byV-(1,+1)Z-IZ4-Z₁ + Z₂enfExample 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;JAproblem;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 12₁10031050-15% of 11000/515 x 11000x1<-1201010-0907500J3(50+)40)Fig. 12.87andNow2₂+41-09+41-50Z:(5)2; Z-(50+/40) 210,000 210-5774 210 volts (5686+/1003) volts√√312000 20-(6928+/0) voltsE-Referring to Fig. 12.87, we have,andSubtracti1,2+1,+1)ZE1₂ Z₁+(1₁+1₂) Z-ESom eq. (1), we have,E₁-E₂Adding eqs. () and (i), we have,4+4=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 here a data and asked to explain one equation.

E Ę Referring to Fig. 12.87, we have, and Subtracting 10,000 210-5774 210 volts (5686+/1003) volts √√3 12000 20- (6928+10) volts 1, 2+1, +1)ZE - 1₂ Z+(1+1)ZĘ eq. (), we have, 1₁-1₂-₂686+/1003)-(6928+ / 0) Z₁ 13431003 js 1,-1,- (200+/250) A Adding eqs. () and (i), we have, E₁ + E₂ 2Z+Z₁ 12614+/1003 100+/85 (5686+/1003)+(6928+/0) 2 (50+/40)+/S 12654 24-5 131-2240-4 1+1 solving eqs. (ii) and (iv), we have, j5 -(200+/250) A -96-42-35-9A-(78-1-/566) A (78-1-56-6)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?

E Ę Referring to Fig. 12.87, we have, and Subtracting 10,000 210-5774 210 volts (5686+/1003) volts √√3 12000 20- (6928+10) volts 1, 2+1, +1)ZE - 1₂ Z+(1+1)ZĘ eq. (), we have, 1₁-1₂-₂686+/1003)-(6928+ / 0) Z₁ 13431003 js 1,-1,- (200+/250) A Adding eqs. () and (i), we have, E₁ + E₂ 2Z+Z₁ 12614+/1003 100+/85 (5686+/1003)+(6928+/0) 2 (50+/40)+/S 12654 24-5 131-2240-4 1+1 solving eqs. (ii) and (iv), we have, j5 -(200+/250) A -96-42-35-9A-(78-1-/566) A (78-1-56-6)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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