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Unformatted text preview: ’0.952âˆ âˆ’ 18.2Â°
g Solving for the (2) network we get
(2)
Ig
(2)
Im I (2)
f Ef
1.05âˆ 0Â°
=
=
= âˆ’ j7
j0.15
j0.15
E f 1.05âˆ 0Â°
=
=
= âˆ’ j 2.1
j0.5
j0.5
= âˆ’ j 7 âˆ’ j 2.1 = âˆ’ j 9.1 I g = 0.952âˆ âˆ’ 18.2Â° âˆ’ j 7 = 7.35âˆ âˆ’ 82.9Â° This matches
what we
calculated
earlier
11 ! Extension to Larger Systems
The superposition approach can be easily extended
to larger systems. Using the Ybus we have
Ybus V = I
For the second (2) system there is only one voltage
source so I is all zeros except at the fault location
! M"
#0$
#
$
I = #âˆ’ I f $
#
$
#0$
# M$
&
' However to use this
approach we need to
first determine If
12 ! Determination of Fault Current
Define the bus impedance matrix Z bus as
âˆ’1
Z bus @ Ybus " Z11 L
Then $ M O
$
$ Z n1 L
& V = Z busI
"V1(2) #
" M#
$ 0 % $V (2) %
Z1n #
$
% $2 %
M % $âˆ’ I f % = $ M %
%$
% $ (2) %
Z nn % $ 0 % $V %
'
n âˆ’1
$ M % $ (2) %
&
' $Vn %
&
' For a fault a b...
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This note was uploaded on 02/09/2014 for the course ECE 5042 taught by Professor Degroat during the Fall '13 term at Ohio State.
 Fall '13
 Degroat

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