4. A CT with an excitation curve given in the figure has a rated current ratio of 500:5 A and a secondary leakage impedance of 0.1 +j0.5 ohms. 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 ohms and the primary CT load current is 400 A. b.) The impedance of the terminating device is 4.9 +j0.5 ohms and the primary CT load current is 1200 A c.) The impedance of the terminating device is 14.9 +j 1.5 ohms and the primary CT fault current is 400 A. d.) The impedance of the terminating device is 14.9 +j 1.5 ohms and the primary CT fault current is 1200 A w 100 80 60 40 20 10 Secondary exciting volts-E'

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Author:Robert L. Boylestad
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4. A CT with an excitation curve given in the
figure has a rated current ratio of 500:5 A and a
secondary leakage impedance of 0.1 +j0.5
ohms. 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 ohms and the primary CT load
current is 400 A. b.) The impedance of the
terminating device is 4.9 +j0.5 ohms and the
primary CT load current is 1200A c.) The
impedance of the terminating device is 14.9 + j
1.5 ohms and the primary CT fault current is
400 A. d.) The impedance of the terminating
device is 14.9 + j 1.5 ohms and the primary CT
fault current is 1200 A
w 100
80
60
40
20
2 4
6
10
Secondary exciting amps-,
Secondary exciting volts-E'
Transcribed Image Text:4. A CT with an excitation curve given in the figure has a rated current ratio of 500:5 A and a secondary leakage impedance of 0.1 +j0.5 ohms. 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 ohms and the primary CT load current is 400 A. b.) The impedance of the terminating device is 4.9 +j0.5 ohms and the primary CT load current is 1200A c.) The impedance of the terminating device is 14.9 + j 1.5 ohms and the primary CT fault current is 400 A. d.) The impedance of the terminating device is 14.9 + j 1.5 ohms and the primary CT fault current is 1200 A w 100 80 60 40 20 2 4 6 10 Secondary exciting amps-, Secondary exciting volts-E'
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