By using Kircho adt's current law. 5MA -> -> 8MA 式 -> 1.5MA Appy KCL at node X, らmA =エュ+4mA Tュ= ちmA-4mA エ221MA In=I2+4MA = IMA+4MA =5MA KLL aty u 8MA = I +Iz 8MA= 5MA+ Iz Iz=3MA

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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Please correct this one I’m stuck
By using
Kircho adt's
current law.
5MA
->
エ
->
8MA
式
->
1.5MA
Appy KCL at node X,
らmA =エュ+4mA
Tュ= ちmA-4MA
Gu1にてエ
In= I2+4MA = IMA+4MA =5MA
%3D
KLL aty u
SMA = In t+Iz
8MA= 5MA+IZ
Iz=3MA
Transcribed Image Text:By using Kircho adt's current law. 5MA -> エ -> 8MA 式 -> 1.5MA Appy KCL at node X, らmA =エュ+4mA Tュ= ちmA-4MA Gu1にてエ In= I2+4MA = IMA+4MA =5MA %3D KLL aty u SMA = In t+Iz 8MA= 5MA+IZ Iz=3MA
T3 tl.5MA = 3MA
I3 = 3MA-I.sMA
ニ
Ig=105MA
The actual
Circit is
8MA
L 3MA .
とらmA
S- や
(MA
lov
R, =
= 6.667kr
1-SMA
= 6.667KL
lo
R3=
= Iokr
IMA
R4= l0
= 2.5ド人
Transcribed Image Text:T3 tl.5MA = 3MA I3 = 3MA-I.sMA ニ Ig=105MA The actual Circit is 8MA L 3MA . とらmA S- や (MA lov R, = = 6.667kr 1-SMA = 6.667KL lo R3= = Iokr IMA R4= l0 = 2.5ド人
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