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Use the node voltage method to find the phasor current lg in the circuit shown below: NODAL ANALYSIS: I = V R SLOA Is -34-2 -;8-2 1222 V3 1 542 121 २०८१०० v V₁ NODE V, '> en-'n V₁ -j8 Va -38 = Is NODE V2 : 0 = V₂-V, + √2 - V3 = S j 1 V₁ = 5 + j l Va 8 V₁ = V₂ - j40 0 C₁₁ = j ел 3 - 14, + 8 ↓ 121 V₁ + 뚜). √2 - V3 12 - -j 1 V3 12 Va = - 5 1 (√ ₂ - j 40) + ( 1 − j 1 ) V2 - 1 V3 - 5 + 1 Va 12 1 V₂ 8 - 1 V3 12 8 5 = √₂ (± - ; +) - 1 V₁. 12

Use the node voltage method to find the phasor current lg in the circuit shown below: NODAL ANALYSIS: I = V R SLOA Is -34-2 -;8-2 1222 V3 1 542 121 २०८१०० v V₁ NODE V, '> en-'n V₁ -j8 Va -38 = Is NODE V2 : 0 = V₂-V, + √2 - V3 = S j 1 V₁ = 5 + j l Va 8 V₁ = V₂ - j40 0 C₁₁ = j ел 3 - 14, + 8 ↓ 121 V₁ + 뚜). √2 - V3 12 - -j 1 V3 12 Va = - 5 1 (√ ₂ - j 40) + ( 1 − j 1 ) V2 - 1 V3 - 5 + 1 Va 12 1 V₂ 8 - 1 V3 12 8 5 = √₂ (± - ; +) - 1 V₁. 12

Power System Analysis and Design (MindTap Course List)
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter6: Power Flows
Section: Chapter Questions
Problem 6.61P
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Use the node voltage method to find the phasor current lg in the circuit shown below: NODAL ANALYSIS: I = V R SLOA Is -34-2 -;8-2 1222 V3 1 542 121 २०८१०० v V₁ NODE V, '> en-'n V₁ -j8 Va -38 = Is NODE V2 : 0 = V₂-V, + √2 - V3 = S j 1 V₁ = 5 + j l Va 8 V₁ = V₂ - j40 0 C₁₁ = j ел 3 - 14, + 8 ↓ 121 V₁ + 뚜). √2 - V3 12 - -j 1 V3 12 Va = - 5 1 (√ ₂ - j 40) + ( 1 − j 1 ) V2 - 1 V3 - 5 + 1 Va 12 1 V₂ 8 - 1 V3 12 8 5 = √₂ (± - ; +) - 1 V₁. 12
Transcribed Image Text:Use the node voltage method to find the phasor current lg in the circuit shown below: NODAL ANALYSIS: I = V R SLOA Is -34-2 -;8-2 1222 V3 1 542 121 २०८१०० v V₁ NODE V, '> en-'n V₁ -j8 Va -38 = Is NODE V2 : 0 = V₂-V, + √2 - V3 = S j 1 V₁ = 5 + j l Va 8 V₁ = V₂ - j40 0 C₁₁ = j ел 3 - 14, + 8 ↓ 121 V₁ + 뚜). √2 - V3 12 - -j 1 V3 12 Va = - 5 1 (√ ₂ - j 40) + ( 1 − j 1 ) V2 - 1 V3 - 5 + 1 Va 12 1 V₂ 8 - 1 V3 12 8 5 = √₂ (± - ; +) - 1 V₁. 12
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