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GATE 2015 EE (EE2) – Question 52

Power Systems · Principles of over-current, differential, directional and distance protection; Circuit breakers · 2 marks · Multiple choice

The 33/11 kV delta–star transformer and CT connections are shown. Both-side CT ratios are 500/5. For the indicated 300 A fault, the currents i1 and i2 are

Diagram for GATE 2015 EE (EE2) question 52
  1. $i_1=1/\sqrt3\ \mathrm A,\ i_2=0$
  2. $i_1=0,\ i_2=0$
  3. $i_1=0,\ i_2=1/\sqrt3\ \mathrm A$
  4. $i_1=1/\sqrt3\ \mathrm A,\ i_2=1/\sqrt3\ \mathrm A$

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Correct answer: (A) $i_1=1/\sqrt3\ \mathrm A,\ i_2=0$

Explanation

The per-winding HV/LV turns ratio is 33000/(11000/√3)=3√3. A 300 A current in the faulted LV winding is therefore balanced by 300/(3√3)=100/√3 A in the associated delta winding.

The indicated HV-side CT lead consequently carries (100/√3)/100=1/√3 A, using the 500/5=100 ratio. The indicated LV CT lead i2 belongs to the unfaulted phase path in the drawing and carries zero for this single-phase fault.

Thus the pair is **1/√3 A and 0 A, A**. Current directions and the specific lead labels follow the supplied CT connection diagram.

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