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Answer :
(i) The operating time of the feeder relay for a fault current of 5000A is approximately 0.17 seconds.
(ii) The minimum plug setting of the transformer relay is 6.8.
(i) To determine the operating time of the feeder relay, we need to refer to the operating characteristic curve in Fig.Q.2. Since the fault current is 5000A, we find the corresponding time on the curve, which is approximately 0.17 seconds.
(ii) The minimum plug setting of the transformer relay is determined by multiplying the rated current (20MVA) by the overload percentage (30%). This gives us 6MVA. Since the transformer is operating at 11kV, we can use the formula P = V x I to find the current: 6MVA / 11kV = 545.45A. Therefore, the minimum plug setting of the transformer relay is 6.8 (125% of 545.45A).
Assuming a discriminative time margin of 0.5 seconds means that the time setting of the transformer relay should be at least 0.5 seconds higher than the operating time of the feeder relay. Therefore, the time setting for the transformer relay would be approximately 0.67 seconds.
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