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10.30 The three loads in the circuit seen in Fig. P10.30 S₁ = 6+j3 kVA, S₂ =7.5-j4.5 kVA, S3 = 12 + 19 kVA. are a) Calculate the complex power associated with each voltage source, Vg1 and V2. b) Verify that the total real and reactive power delivered by the sources equals the total real and reactive power absorbed by the network. Figure P10.30 0.10 0.20 www 0.1Ω ww S₁150 /0° V (rms) S₂ 1500 V (rms) S3

10.30 The three loads in the circuit seen in Fig. P10.30 S₁ = 6+j3 kVA, S₂ =7.5-j4.5 kVA, S3 = 12 + 19 kVA. are a) Calculate the complex power associated with each voltage source, Vg1 and V2. b) Verify that the total real and reactive power delivered by the sources equals the total real and reactive power absorbed by the network. Figure P10.30 0.10 0.20 www 0.1Ω ww S₁150 /0° V (rms) S₂ 1500 V (rms) S3

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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a. S=VI* Z₁ = (150) ² 6+j3kVA = 3-j1.5-2 I₁ = 150 10.30 The three loads in the circuit seen in Fig. P10.30 S₁ = 6+j3 kVA, S₂= 7.5- j4.5 kVA, are S3 = 12 + 19 kVA. I= \/2 3-11-5 =40+120 a) Calculate the complex power associated with each voltage source, Vg₁ and V₂2- b) Verify that the total real and reactive power delivered by the sources equals the total real and reactive power absorbed by the network. Figure P10.30 0.1 Ω www I₂ = 0.22 Z₂ = (150) ² 7.5-4.5KVA S₁ = 150 (40+)zon). = 6+;3KVA 0.12 ww =2.21+j1.322 150 2.21+j1.32 = 50- j30 A S₁ 150 0º V (rms) S₂ 150/0 V (rms) 5₂=150(50+130) 23= (150) ² = 7.5+ 4.5kVA I₂ = S3 12+;9 KVA = 1.2-jo.92 150 1.2-jo.9 =80+160 A 53= 150(80-360) = 12-19KVA
Transcribed Image Text:a. S=VI* Z₁ = (150) ² 6+j3kVA = 3-j1.5-2 I₁ = 150 10.30 The three loads in the circuit seen in Fig. P10.30 S₁ = 6+j3 kVA, S₂= 7.5- j4.5 kVA, are S3 = 12 + 19 kVA. I= \/2 3-11-5 =40+120 a) Calculate the complex power associated with each voltage source, Vg₁ and V₂2- b) Verify that the total real and reactive power delivered by the sources equals the total real and reactive power absorbed by the network. Figure P10.30 0.1 Ω www I₂ = 0.22 Z₂ = (150) ² 7.5-4.5KVA S₁ = 150 (40+)zon). = 6+;3KVA 0.12 ww =2.21+j1.322 150 2.21+j1.32 = 50- j30 A S₁ 150 0º V (rms) S₂ 150/0 V (rms) 5₂=150(50+130) 23= (150) ² = 7.5+ 4.5kVA I₂ = S3 12+;9 KVA = 1.2-jo.92 150 1.2-jo.9 =80+160 A 53= 150(80-360) = 12-19KVA
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