Chapter 10 Symmetrical Faults

V bus (F) = V bus (0) + Z bus I bus (F) (6)
Equation (6) can be written in matrix form as
= + (7)
Equation (7) can be written as a system of n equations as follows:
V 1 (F) = V 1 (0) – Z 1K I k (F)
V 2 (F) = V 2 (0) – Z 2K I k (F)
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V k (F) = V k (0) – Z kK I k (F)
(7A)
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V n (F) = V n (0) – Z nK I k (F)
The voltage at bus k during the fault is V k (F) which is the k th equation in the set of n
equation (indicated in italics). So
V k (F) = V k (0) – Z kK I k (F) (8)
But the voltage at bus k during the fault is the product of fault impedance Z f and fault
current I k (F) i.e. V k (F) = Z f I k (F), therefore equation (8) becomes
Z f I k (F) = V k (0) – Z kK I k (F)
Solving for I k (F) gives
I k (F) = (9)
Thus the fault current I k (F) for a fault at bus k depends on the pre-fault voltage V k (0) at
bus k, the diagonal element Z kk of the bus impedance matrix Z bus and the fault
impedance Z f . Remember that for a bolted fault or a solid fault Z f = 0.
From equation (7A), for any bus i the bus voltage during the fault is
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