Inorder to decide on the selection of CT for protection application, we first need to determine the maximum value of fault current available at the location, then we have to take this maximum value and divide by the rating of the CT the result will determine the multiplication factor, then we have to choose the accuracy depending on our application
eg. if we have maximum fault current available as 15 kA, and our CT current rating is 1000/5 then 15000/1000 =15.
therefore a CT XP20 is sufficient, X is determined based on the accuracy requirement eg 5 , 10 etc.