CALCULATE THE FOLLOWING THREE CURRENTS RMS SYMMETRICAL FAULT CURRENT RMS ASYMMETRICAL FAULT CURRENT

In Figure 7.1 (p.323 of your textbook), assume the following values:
= = =
V 4 kV, X 3 n, R 1 n, and ID= 211:60 rad/sec. Calculate the following three
currents: the rms symmetrical fault current, the rms asymmetrical fault current at the
instant the switch closes (assuming maximum dc offset), and the rms asymmetrical fault
current 5 cycles after the switch closes (assuming maximum dc offset).



a.
I~ =1,065.4, 1 (0)=1,191.4, I (5)=1,065.4
b.
1~ = 1,365.4, I"". (0)= 2,291.4, I (5) = 1,365 A.
c.
I=-.=1,565.4, 1 (0)=3,191.4, I.". (5) =1,565.4
d.
I=c=1,265.4, I (0)=2,191.4, 1 (5) =1,265.4

SECTION 7.1 SERIES
R-L CIRCUIT TRANSIENTS 323

FIGURE 7.1




E:3
in a series R-L R L
with ac voltage
source

~" = V2V"-I.' + .,- - - ,.", ,= 0
i(t)
i,cW
. /1'
, ...
IdC{t) /
\
-+
/
cut
I-a \
"" /
'--/

iae(t) = ~V sin + a -
(cot 0) A (7.1.3)

. V2V.
Ide(t ) = -- Z sm(a - 0)e-liT A (7.1.4)

Z = VR2 + (coL)2= VR2 + X2 Q (7.1.5)

0= tan-l coL= tan-l X (7.1.6)
R R
L X X
(7.1.7)
T = R = coR = 2nfR s
The total fault current in (7.1.2), called the asymmetrical fault current, is
plotted in Figure 7.1 along with its two components. The ac fault current
(also called symmetrical or steady-state fault current), given by (7.1.3), is a
sinusoid. The dc offset current, given by (7.1.4), decays exponentially with
time constant T = L/R.
The rms ac fault current is lac = VIZ. The magnitude of the dc offset,
which depends on a, varies from 0 when a = 0 to V2lae when a = (0 I nI2).
Note that a short circuit may occur at any instant during a cycle of the ac
source; that is, a can have any value. Since we are primarily interested in the
largest fault current, we choose a = (0 - nI2). Then (7.1.2) becomes
i(t) = V2lae[sin(cot - n12) + e-I/T] A (7.1.8)
where
V
I -- A (7.1.9)
ae - Z

The rms value of i(t) is of interest. Since i(t) in (7.1.8) is not strictly periodic,
its rms value is not strictly defined. However, treating the exponential term as