Example 1:
A , C400 CT with excitation curves shown on above fig 5.7, is connected to a 2.0 burden. Based on the accuracy classification, what is the maximum symmetrical fault current that may be applied to this CT without exceeding a 10% ratio error?

Answer:
CT ratio = 1200/5
Secondary resistance = 0.61
Relay burden = 2
For 20 times rated secondary current, i.e., 100A
Secondary voltage = 100 x (2 + 0.61) = 261 Volts which is less than knee point of the CT. Since this voltage is less than 400V, from electrical perspective, linearity will not be lost at even higher currents. Approximate limit on secondary current is given by . Hence maximum symmetrical fault current = 36720A.
More exact calculation requires usage of fig 5.7. This would involve 'cut and try' approach.

Example 2:
A , C400 CT is connected on the tap. What is the maximum secondary burden that can be used and we can maintain rated accuracy at 20 times rated symmetrical secondary current?

Answer:
The secondary voltage corresponding to the tap 1000/5, = 333V
Secondary current

, from calibration curve.
333 = 100(0.51+R_B)
Secondary burden = 3.33 - 0.51 = 2.72