Calculation of Fault Current Using Zbus Matrix
Consider the circuit of Fig. 3.3 which is redrawn as shown in Fig. 6.10.
Fig. 6.10 Network depicting a symmetrical fault at bus-4.
We assume that a symmetrical fault has occurred in bus-4 such that it is now connected to the reference bus. Let us assume that the pre-fault voltage at this bus is Vf . To denote that bus-4 is short circuit, we add two voltage sources Vf and - Vf together in series between bus-4 and the reference bus. Also note that the subtransient fault current I²f flows from bus-4 to the reference bus. This implies that a current that is equal to - I²f is injected into bus-4. This current, which is due to the source - Vf will flow through the various branches of the network and will cause a change in the bus voltages. Assuming that the two sources and Vf are short circuited. Then - Vf is the only source left in the network that injects a current - I²f into bus-4. The voltages of the different nodes that are caused by the voltage - Vf and the current - I²f are then given by
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(6.14) |
where the prefix Δ indicates the changes in the bus voltages due to the current - I²f .
From the fourth row of (6.14) we can write
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(6.15) |
Combining (6.14) and (6.15) we get
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(6.16) |
We further assume that the system is unloaded before the fault occurs and that the magnitude and phase angles of all the generator internal emfs are the same. Then there will be no current circulating anywhere in the network and the bus voltages of all the nodes before the fault will be same and equal to Vf . Then the new altered bus voltages due to the fault will be given from (6.16) by
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(6.17) |
Example 6.3
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