If we take the low voltage current of phase Y as basis then the other currents-low and high voltage, arrow and amplitude-it seems to me to be correct[as per attached image].
Since the only supply it is low voltage short-circuit currents all the currents are in the same phase or opposite. The currents in secondary of CT are opposite to the primary.
Sum of currents leaving the point A are:
IRY-IBR+0=0
Sum of currents leaving the point B are:
-IYB-IRY+4800=0
Sum of currents leaving the point C are:
IYB+IBR-4800=0
Let's put IRY=X;IYB=Y;IBR=Z
Then X=Z
X+Y=4800
The general equation X-Y+Z=0 [current circulation around in ∆]
Then Y=2X
3X=4800 ;X=1600 A;Y=2X=3200 A