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(1)= 0.4 p.u. V₁ = 1.040⁰ #12 = 0.12 + j0.16 100 MW + 50 Mvar

(1)= 0.4 p.u. V₁ = 1.040⁰ #12 = 0.12 + j0.16 100 MW + 50 Mvar

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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In the figure, bus 1 is the reference bus, V1 = 1.06 p.u and the angle is 0°. A load with active power= 100 MW and reactive power = 50 Mvar is connected to bus 2. The line impedance is z12 = 0.12 +j0.16 p.u on a base of 100 MVA. Start with an initial estimate of |V2|0) = 1.0 p.u and 8200) = 0°. Find |V2|(¹) (one iteration). 2 100 MW H O *12= 0.12 + j0.16 + 50 Mvar V₁ = 1.040⁰ O a. |V2|(¹) = 0.4 p.u. O b. |V2|(¹) = 0.5 p.u. OC. |V2|(¹) = 0.7 p.u. O d. |V2|(¹) = 0.8 p.u. O e. |V2|(¹) = 0.6 p.u.
Transcribed Image Text:In the figure, bus 1 is the reference bus, V1 = 1.06 p.u and the angle is 0°. A load with active power= 100 MW and reactive power = 50 Mvar is connected to bus 2. The line impedance is z12 = 0.12 +j0.16 p.u on a base of 100 MVA. Start with an initial estimate of |V2|0) = 1.0 p.u and 8200) = 0°. Find |V2|(¹) (one iteration). 2 100 MW H O *12= 0.12 + j0.16 + 50 Mvar V₁ = 1.040⁰ O a. |V2|(¹) = 0.4 p.u. O b. |V2|(¹) = 0.5 p.u. OC. |V2|(¹) = 0.7 p.u. O d. |V2|(¹) = 0.8 p.u. O e. |V2|(¹) = 0.6 p.u.
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