The CT of Problem 10.5 is utilized in conjunction with a currentsensitive device that will operate at current levels of \(8 \mathrm{~A}\) or above. Check whether the device will detect the 1300-A fault current for cases (b) and (d) in Problem 10.5.

Problem 10.5

A CT with an excitation curve given in Figure 10.39 has a rated current ratio of 500:5 A and a secondary leakage impedance of \(0.1+j 0.5 \Omega\). Calculate the CT secondary output current and the CT error for the following cases:

(a) The impedance of the terminating device is \(4.9+j 0.5 \Omega\) and the primary CT load current is \(400 \mathrm{~A}\).

(b) The impedance of the terminating device is \(4.9+j 0.5 \Omega\) and the primary CT fault current is \(1200 \mathrm{~A}\).

(c) The impedance of the terminating device is \(14.9+j 1.5 \Omega\) and the primary CT load current is \(400 \mathrm{~A}\).

(d) The impedance of the terminating device is \(14.9+j 1.5 \Omega\) and the primary CT fault current is \(1200 \mathrm{~A}\).

Transcribed Image Text:

Secondary exciting volts-E' 100 80 60 40 20 2 4 6 8 10 Secondary exciting amps-le 0