Electrical Power Systems

Symmetrical ault 217
Solution: Inject unit current at bus 1 and keeping bus 2 open circuited as shown in
ig. 8.21(a).
ig: 8.21(a)
I 1 = 1.0 pu, I 2 = 0.0
Calculating voltages at bus 1 and bus 2,
we have,
Z11 = V1 = 11.0
Z21 = V2 = 6.0
Now inject unit current at bus 2 and keeping bus 1 open circuited as shown in ig. 8.21(b)
Therefore, I1 = 0, I2 = 1.0 pu.
rom ig. 8.21(b), similarly we have,
Z22 = V2 = 10
Z12 = V1 = 6
Therefore,
Z BUS =
L
NM
Z Z
Z Z
11 12
21 22
ig. 8.21(b)
O
QP
=L NM
11 6
6 10
The Z BUS matrix also referred to or the “open circuit impedance matrix”.
8.6 ALGORITHM OR BUILDING Z BUS MATRIX
Z BUS building algorithm is a step-by-step procedure which proceeds branch by branch. Main
advantage of this method is that, any modification of the network elements does not require
complete rebuilding of Z BUS–matrix . Details of Z BUS formulation is given below:
O
QP