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Example 4.2: Use the the circuit 201 4.2: Use the node-voltage power dissipated рошет + shown in Fig. 4.10. 2 Node voltage at V₁ - 20 2 Node Voltage V₂ - V₁ 5 + + node + 1 V₁ 20 V₁ + 20 at method to find in the 5-2 resister in the V₁ + V₂ 10 Problem: We have Solution: Express is, the the voltages! 51 www - V₁ - V/₂ 5 + V₂ 10₂ V₂ 12-810 2 2 = 24 www - O two equations, but con bralling 10 = V₁ V₂ but three un/knowns. in terms of Big curmat using passive sign convention

Example 4.2: Use the the circuit 201 4.2: Use the node-voltage power dissipated рошет + shown in Fig. 4.10. 2 Node voltage at V₁ - 20 2 Node Voltage V₂ - V₁ 5 + + node + 1 V₁ 20 V₁ + 20 at method to find in the 5-2 resister in the V₁ + V₂ 10 Problem: We have Solution: Express is, the the voltages! 51 www - V₁ - V/₂ 5 + V₂ 10₂ V₂ 12-810 2 2 = 24 www - O two equations, but con bralling 10 = V₁ V₂ but three un/knowns. in terms of Big curmat using passive sign convention

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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I need help solving for i-phi
ZVk₂ хо 201 + √₂ + 5 V/2 TO TO Tie 10 -201-8V₁ 10 - IOU, 10 10 2 V2 10 - + Vz 10 84₁ To +5Vz 8 V₂ 10 -8V2 10
Transcribed Image Text:ZVk₂ хо 201 + √₂ + 5 V/2 TO TO Tie 10 -201-8V₁ 10 - IOU, 10 10 2 V2 10 - + Vz 10 84₁ To +5Vz 8 V₂ 10 -8V2 10
Example 4.2: Use the the circuit 201 power dissipated 2 shown in Fig. 4.10. گرام 21 www Node voltage at Vi V₁ - 20 V₁ Node Voltage √₂-V₁ 5 V₁ 5 十 + + + V₂ 10 20 V₂ 10 1/2 10 1 20 + + + at + Problem: We have Solution: Express is the controlling the поле voltages! node - Voltage in the 5-1 52 www V₂ + N V₂ ≥ 10 V₁ - V/₂ 5 Substitute this relationship in to equation 2 V₂ - V₁ V₂ 8 1/( 5 2 1/₂ 12₂-810 2 2 two equations, but 24 www = 0 8v, 10 method resister in the систnt using passive sign convention 10 = V₁-1₂ 5 V₁ – U₂ 5 8V₂ 10 to find three unlenowns. in terms of Big =0
Transcribed Image Text:Example 4.2: Use the the circuit 201 power dissipated 2 shown in Fig. 4.10. گرام 21 www Node voltage at Vi V₁ - 20 V₁ Node Voltage √₂-V₁ 5 V₁ 5 十 + + + V₂ 10 20 V₂ 10 1/2 10 1 20 + + + at + Problem: We have Solution: Express is the controlling the поле voltages! node - Voltage in the 5-1 52 www V₂ + N V₂ ≥ 10 V₁ - V/₂ 5 Substitute this relationship in to equation 2 V₂ - V₁ V₂ 8 1/( 5 2 1/₂ 12₂-810 2 2 two equations, but 24 www = 0 8v, 10 method resister in the систnt using passive sign convention 10 = V₁-1₂ 5 V₁ – U₂ 5 8V₂ 10 to find three unlenowns. in terms of Big =0
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