33For example, in an application where the maximum fault current is expected to be 40,000 amps, we would choose a CT with a ratio of at least 2000:5 to drive the protective relay, because 40,000 amps is twenty times this CT’s primary current rating of 2000 amps. We could also select a CT with a larger ratio such as 3000:5. The point is to have the CT be able to faithfully transform any reasonable fault current into a proportionately lower value for the protective relay(s) to sense.