Example 3.1
Consider the impedance diagram of Fig. 3.2 in which the system parameters are given in per unit by
| Z11 = Z22 = j0.25, Z12 = j0.2, Z13 = j0.25, Z23 = Z34 = j0.4 and Z24 = j0.5 |
|
The system admittance can then be written in per unit as
| Y11 = Y22 = - j4, Y12 = - j5, Y13 = - j4, Y23 = Y34 = - j2.5 and Y24 = - j2 |
|
The Ybus is then given from (3.10) as
 per unit |
|
Consequently the bus impedance matrix is given by
 per unit |
|
It can be seen that like the Ybus matrix the Zbus matrix is also symmetric.
Let us now assume that the voltages EG1 and EG2 are given by
The current sources I1 and I2 are then given by
We then get the node voltages from (3.4) as
 per unit |
|
Solving the above equation we get the node voltages as
 per unit |
|
|
