Problem #2 A multi-ratio current transformer with the following ratios (600:5, 4

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 connected 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 CT Secondary ato res stance 600 5t ? 061 50 5 1005 0082 15050.104 2005 0125 7505 0 146 3005 0 168 400 50211 4505 0 230 5005 0242 6005 0 296 500.5 10 450 5 400 5 300 5 50.5 1005 1505 2005 250 5 0 001 001 10 100 Secordary exctng amps-

Explanation / 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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