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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)
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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A Circulating current on no-load is given by V-(1,+1)Z-IZ 4- Z₁ + Z₂ enf Example 12.57. Two single-phase alternators are connected in parallel and supply current to a load at a terminal voltage of 10000 20 volts. Alternator A has an induced em.f. of 13000222-6⁰ volts and a reactance of 20; alternator B has an e.mf. of 12500 236-9 volts and a reactance of 3 Find the current supplied by each alternator Solution. Current supplied by alternator A is given by; JA problem ; E-VE-V Z₁ Current supplied by alternator B is given by: 4- F.L. current of generator A is 4-4- F.L. current of generator Bis E-V 13000 222-6-10000 40 2290 XA/ZA -6500 2-67-4*+j 5000 (2500-j/ 1000) A - 2700 2-21-8 A Also - (2500+/0)A-2500 20 A Example 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 and B. 12000 V and A leads B by 10 elect; find currents delivered by A and B. : 10,000 kVA, 11000 V, 10% reactance 20,000 kVA, 11000 V, 15 % reactance Generator A rating Generator B rating Reactance of line 1-3-82: Reactance of line 2-41 Solution. Fig. 12.87 shows the conditions of the - E-V 12500 236-9-10000 20 Z 3290 10,000x10 √3x11000 20,000x10 √3x11000 <-525 A 4-4- <- 1050 A Letx, and x, be the respective per phase reactances of the generators in ohms. A A 525 x, 10% of 11000/5 10 11000 1 2₁ 1003 1050-15% of 11000/5 15 x 11000x1 <-120 1010-090 7500 J3 (50+)40) Fig. 12.87 and Now 2₂+41-09+41-50 Z:(5)2; Z-(50+/40) 2 10,000 210-5774 210 volts (5686+/1003) volts √√3 12000 20-(6928+/0) volts E- Referring to Fig. 12.87, we have, and Subtracti 1,2+1,+1)ZE 1₂ Z₁+(1₁+1₂) Z-E Som eq. (1), we have, E₁-E₂ Adding eqs. () and (i), we have, 4+4= 1,-1,- (200+/250) A - $686+/1003)-(6928+/0) j5 -(200+/250) A Z₁ 12431003 js E,+E, (5686+/1003)+(6928+/0) - 2Z+Z₁ 2 (50+/40)+/S 12654 24-5 131-2240-4 12614+/1003 100+85 96-42-35-9A-(78-1-/56-6) A (78-1-56-6) A 1₁+1₂ solving eqs. (ii) and (iv), we have, 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?
Transcribed Image Text:A Circulating current on no-load is given by V-(1,+1)Z-IZ 4- Z₁ + Z₂ enf Example 12.57. Two single-phase alternators are connected in parallel and supply current to a load at a terminal voltage of 10000 20 volts. Alternator A has an induced em.f. of 13000222-6⁰ volts and a reactance of 20; alternator B has an e.mf. of 12500 236-9 volts and a reactance of 3 Find the current supplied by each alternator Solution. Current supplied by alternator A is given by; JA problem ; E-VE-V Z₁ Current supplied by alternator B is given by: 4- F.L. current of generator A is 4-4- F.L. current of generator Bis E-V 13000 222-6-10000 40 2290 XA/ZA -6500 2-67-4*+j 5000 (2500-j/ 1000) A - 2700 2-21-8 A Also - (2500+/0)A-2500 20 A Example 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 and B. 12000 V and A leads B by 10 elect; find currents delivered by A and B. : 10,000 kVA, 11000 V, 10% reactance 20,000 kVA, 11000 V, 15 % reactance Generator A rating Generator B rating Reactance of line 1-3-82: Reactance of line 2-41 Solution. Fig. 12.87 shows the conditions of the - E-V 12500 236-9-10000 20 Z 3290 10,000x10 √3x11000 20,000x10 √3x11000 <-525 A 4-4- <- 1050 A Letx, and x, be the respective per phase reactances of the generators in ohms. A A 525 x, 10% of 11000/5 10 11000 1 2₁ 1003 1050-15% of 11000/5 15 x 11000x1 <-120 1010-090 7500 J3 (50+)40) Fig. 12.87 and Now 2₂+41-09+41-50 Z:(5)2; Z-(50+/40) 2 10,000 210-5774 210 volts (5686+/1003) volts √√3 12000 20-(6928+/0) volts E- Referring to Fig. 12.87, we have, and Subtracti 1,2+1,+1)ZE 1₂ Z₁+(1₁+1₂) Z-E Som eq. (1), we have, E₁-E₂ Adding eqs. () and (i), we have, 4+4= 1,-1,- (200+/250) A - $686+/1003)-(6928+/0) j5 -(200+/250) A Z₁ 12431003 js E,+E, (5686+/1003)+(6928+/0) - 2Z+Z₁ 2 (50+/40)+/S 12654 24-5 131-2240-4 12614+/1003 100+85 96-42-35-9A-(78-1-/56-6) A (78-1-56-6) A 1₁+1₂ solving eqs. (ii) and (iv), we have, 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?
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