| The electrical network shown can be viewed as con- sisting of three loops. Applying Kirchoff's law (Evoltage drops = Evoltage sources) to each loop yields the following equations for the loop cur- rents i1, i2, iz: 20 Ω 5Ω mn-220 V ww i, iz -OV www 102 5i1 + 15(i1 – i3) = 220 V R(i2 – i3) + 5i2 + 10i2 = 0 20i3 + R(i3 – i2) + 15(i3 – i1) = 0 %3D Compute the three loop currents in terms of R. 152 wwww www
| The electrical network shown can be viewed as con- sisting of three loops. Applying Kirchoff's law (Evoltage drops = Evoltage sources) to each loop yields the following equations for the loop cur- rents i1, i2, iz: 20 Ω 5Ω mn-220 V ww i, iz -OV www 102 5i1 + 15(i1 – i3) = 220 V R(i2 – i3) + 5i2 + 10i2 = 0 20i3 + R(i3 – i2) + 15(i3 – i1) = 0 %3D Compute the three loop currents in terms of R. 152 wwww www
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
Related questions
Question
Subject: Circuits
show your solutions
![| The electrical network shown can be viewed as con-
sisting of three loops. Applying Kirchoff's law
(Evoltage drops = Evoltage sources)
to each loop yields the following equations for the loop cur-
rents i1, i2, iz:
20 Ω
5Ω
mn-220 V
ww
i,
iz
-OV
www
102
5i1 + 15(i1 – i3)
= 220 V
R(i2 – i3) + 5i2 + 10i2
= 0
20i3 + R(i3 – i2) + 15(i3 – i1)
= 0
%3D
Compute the three loop currents in terms of R.
152
wwww
www](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4729a6c1-71f7-4999-8507-bfb005f7d790%2Fcbd2d82d-2de3-4110-9278-96051008fda6%2Fl1xg14r_processed.png&w=3840&q=75)
Transcribed Image Text:| The electrical network shown can be viewed as con-
sisting of three loops. Applying Kirchoff's law
(Evoltage drops = Evoltage sources)
to each loop yields the following equations for the loop cur-
rents i1, i2, iz:
20 Ω
5Ω
mn-220 V
ww
i,
iz
-OV
www
102
5i1 + 15(i1 – i3)
= 220 V
R(i2 – i3) + 5i2 + 10i2
= 0
20i3 + R(i3 – i2) + 15(i3 – i1)
= 0
%3D
Compute the three loop currents in terms of R.
152
wwww
www
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