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779 @ ۳:۳۲ م O Z 4.5G+ Iı. تم الأس س الكهربائية الكورس الثاني_ا. . . Thus, the frequency domain equivalent circuit is as shown in Fig. 10.2. 10Ω V j4 2 V2 ww ll 20/0° V -j2.5 2 21, J2Ω How solve the step?! F Frequency domain equivalent of the circuit in Fig. 10.1. Applying KCL at node 1, V, - V2 -j2.5 V- 20 V1 10 j4 or : نطبق طريقة النود تا في )DC( بالضبط رض أن النود هو الذي بارات. والنود الذي معادلة هو الأكبر ودات تطرح منه . التيار ) اذا كان داخل جب واذا خارج من تبة سالب( بعد (1 + jl.5)V, + j2.5V2 = 20 At node 2, V2 V2- V 21, j2 j4 But I, = V/-j2.5. Substituting this gives %3D V2 V, - V2 2V1 j2 j4 -j2.5 By simplifying, we 15V ll

779 @ ۳:۳۲ م O Z 4.5G+ Iı. تم الأس س الكهربائية الكورس الثاني_ا. . . Thus, the frequency domain equivalent circuit is as shown in Fig. 10.2. 10Ω V j4 2 V2 ww ll 20/0° V -j2.5 2 21, J2Ω How solve the step?! F Frequency domain equivalent of the circuit in Fig. 10.1. Applying KCL at node 1, V, - V2 -j2.5 V- 20 V1 10 j4 or : نطبق طريقة النود تا في )DC( بالضبط رض أن النود هو الذي بارات. والنود الذي معادلة هو الأكبر ودات تطرح منه . التيار ) اذا كان داخل جب واذا خارج من تبة سالب( بعد (1 + jl.5)V, + j2.5V2 = 20 At node 2, V2 V2- V 21, j2 j4 But I, = V/-j2.5. Substituting this gives %3D V2 V, - V2 2V1 j2 j4 -j2.5 By simplifying, we 15V ll

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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779 @ ۳:۳۲ م O Z 4.5G+ liı. تم الأس س_ الكهربائية الكورس الثاني _ا. . . Thus, the frequency domain equivalent circuit is as shown in Fig. 10.2. 10 Ω j4 2 ll V V2 ww 20/0° V -j2.5 2 21, J2 Ω How solve the step?! F Frequency domain equivalent of the circuit in Fig. 10.1. Applying KCL at node 1, V- 20 V1 V, - V. 2 = 0 j4 10 -j2.5 or : نطبق طريقة النود تا في )DC( بالضبط رض أن النود هو الذي بارات. والنود الذي معادلة هو الأكبر ودات تطرح منه . التيار ) اذا كان داخل جب واذا خارج من تبة سالب( بعد (1 + jl.5)V, + j2.5V2 = 20 At node 2, V2 V2 - V = 21, j2 j4 But I, = V/-j2.5. Substituting this gives V2 V, - V2 2V1 j2 j4 -j2.5 By simplifying, we 11V, + 15V2 = 0 Equations (10.1.1) a(10.1.2) can be nutm matrix form as 1+ jl.5 j2.5 M-P 11 15 We obtain the determinants as | |1 + jl.5 j2.5| = 15 - j5 15 A = 11 |20 j2.5| A, = |1 + jl.5 20| A2 300, =-220 15 11 300 18.97/18.43° V %3D %3D %3D 15 - j5 A2 -220 = 13.91/198.3° V 15 - j5
Transcribed Image Text:779 @ ۳:۳۲ م O Z 4.5G+ liı. تم الأس س_ الكهربائية الكورس الثاني _ا. . . Thus, the frequency domain equivalent circuit is as shown in Fig. 10.2. 10 Ω j4 2 ll V V2 ww 20/0° V -j2.5 2 21, J2 Ω How solve the step?! F Frequency domain equivalent of the circuit in Fig. 10.1. Applying KCL at node 1, V- 20 V1 V, - V. 2 = 0 j4 10 -j2.5 or : نطبق طريقة النود تا في )DC( بالضبط رض أن النود هو الذي بارات. والنود الذي معادلة هو الأكبر ودات تطرح منه . التيار ) اذا كان داخل جب واذا خارج من تبة سالب( بعد (1 + jl.5)V, + j2.5V2 = 20 At node 2, V2 V2 - V = 21, j2 j4 But I, = V/-j2.5. Substituting this gives V2 V, - V2 2V1 j2 j4 -j2.5 By simplifying, we 11V, + 15V2 = 0 Equations (10.1.1) a(10.1.2) can be nutm matrix form as 1+ jl.5 j2.5 M-P 11 15 We obtain the determinants as | |1 + jl.5 j2.5| = 15 - j5 15 A = 11 |20 j2.5| A, = |1 + jl.5 20| A2 300, =-220 15 11 300 18.97/18.43° V %3D %3D %3D 15 - j5 A2 -220 = 13.91/198.3° V 15 - j5
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