AC Fault Location
Two single-phase AC voltage sources supply a faulted transmission line. The fault resistance R F divides the line into two impedances: m Z L on the left, and ( 1 − m ) Z L on the right, where 0 < m < 1 . Source voltages V S and V R are known, as well as source currents I S and I R . Real-valued quantities m and R F are unknown.
Determine the value of m .
Details and Assumptions:
1)
V
S
=
9
+
j
5
2)
V
R
=
1
0
+
j
0
3)
Z
L
=
0
+
j
1
0
4)
I
S
=
1
.
4
6
6
4
7
4
2
4
2
9
5
+
j
0
.
0
5
5
2
7
4
7
5
6
6
1
5
8
5)
I
R
=
0
.
0
6
7
6
1
1
8
5
6
9
6
7
8
−
j
0
.
1
2
6
2
6
7
2
0
6
4
3
2
6)
j
=
−
1
The answer is 0.37.
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2 solutions
A fault is an unintended short circuit
To make this one easy to understand, I will write out the equations, which can then be put into matrix form. The first equates two expression for the voltage at the fault point. The second makes a loop from the left side through the fault resistance.
V S − m Z L I S = V R − ( 1 − m ) Z L I R V S − m Z L I S − R F ( I S + I R ) = 0
Solve these two equations for m and R F
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Applying Kirchhoff's laws for this circuit leads to the equations cast into a matrix form as such:
[ Z L I S − Z L I R I S + I R I S + I R ] [ m R F ] = [ V S V R − I R Z L ] ⟹ m = [ 1 0 ] [ Z L I S − Z L I R I S + I R I S + I R ] − 1 [ V S V R − I R Z L ] = 0 . 3 7 ⟹ R F = [ 0 1 ] [ Z L I S − Z L I R I S + I R I S + I R ] − 1 [ V S V R − I R Z L ] = 6
From a mathematical point of view, this circuit is easy to solve, but I do not quite grasp the intuition behind a 'fault'. Some explanation of that would be helpful.