= -46.667 V-18Practice 5.4Find v1,v2, and v3 in the circuit in the figure below using nodal analysis.6210 V5iSupevNodelSuperModle2 24Ω32Answer:V9S02y = 3.043 V, v2 = -6.956 V. vz = 0.6522 V.

Here V1,V2 and V3 form super node.hence we did super node analysis and kcl.

Answered: ww 5i 10 V 12 Supe Modle 22 42 32 ,v2…

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ww 5i 10 V 12 Supe Modle 22 42 32 ,v2 =-6.956 V, v3 = 0.6522 V. %3D ww

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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Current Division Method

The generalized current divider formula is given by the following relation below,

Question

= -46.667 V
-18
Practice 5.4
Find v1,v2, and v3 in the circuit in the figure below using nodal analysis.
62
10 V
5i
Supev
Nodel
Super
Modle
2 2
4Ω
32
Answer:
V9S0
2y = 3.043 V, v2 = -6.956 V. vz = 0.6522 V.

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