Problem #2 A multi-ratio current transformer with the following ratios (600:5, 4
ID: 2293117 • Letter: P
Question
Problem #2 A multi-ratio current transformer with the following ratios (600:5, 400/5, and 300:5) has a secondary resistance of 0.61 Ohms (same for all ratios). If the transformer is conneted to a relay with a burden of 2.2 Ohms with a 15 meter lead with a resistance of 3.5 Ohms/km. Determine if the for each turns ratios the CT will saturate for a fault current of 2.0kA CT's using the CT curves provided? 500 100 C Secordary rato res stance 600 5 50 5 1005 0082 150 50104 2005 7505 0 146 3005 0 168 400 5 4505 0 230 500 5 600 5 0 296 ? 061 500 5 10 450 5 0 125 400 5 300 5 0 211 50.5 100 5 0 242 150 5 2005 250 5 0 001 0 01 0 1 10 100 Secordary excing ampsExplanation / Answer
Multiturn CT has three turns ratios
600/5
400/5
300/5
Secondary resistance 0.61 Ohm
Burden 2.2 Ohm
Lead resistance = 3.5*0.015=0.05 Ohm
Total resistance= 2.86 Ohm
Maximum operating current 2 kA
With 600/5 ratio, secondary current = 2000 A *(5/600)= 16.66 A
Secondary voltage Vs =16.66*(2.86) = 47.647 Volts
This operating point is well below the sturation point, so CT will not saturate.
With 400/5 ratio, secondary current = 2000 A *(5/400)= 25 A
Secondary voltage Vs =25*(2.86) = 71.5 Volts
This operating point is very close to the sturation point. So CT may go into saturation.
With 600/5 ratio, secondary current = 2000 A *(5/300)= 33.33 A
Secondary voltage Vs =33.33 *(2.86) = 95.32 Volts.
This operating point is well above the sturation point, so CT will definitely saturate.
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