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The answer given in my workbook to the above problem is A.

The things I was able to point out

  1. secondary c-phase current = 300A and other secondary phase currents = 0A since the entire current flows through the fault.

  2. Total kVA of primary = 1.732 * 33kV * IL and primary kVA = secondary kVA

  3. i1 = (IL/3) and i2 = 0

I am confused on how to calculate secondary kVA

and why it wouldn't be {(33kV/1.732) * 300A} + 0 + 0 ?

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  • \$\begingroup\$ If I1 is non-zero then some current has to pass into the node where I2 is indicated irrespective of that node being part of a secondary CT. This means it can't be answer A unless there are some assumptions you know of that haven't been given above. \$\endgroup\$ – Andy aka Jan 11 '18 at 16:50
  • \$\begingroup\$ This is the complete problem. No other information was provided. Can you think of some assumption which supports option A ? \$\endgroup\$ – Nikhil Kashyap Jan 12 '18 at 2:15
  • \$\begingroup\$ I would try it in a simulator to see what I got. \$\endgroup\$ – Andy aka Jan 12 '18 at 11:22
  • \$\begingroup\$ @Andyaka, can you suggest something? \$\endgroup\$ – Nikhil Kashyap Jan 16 '18 at 12:45
  • \$\begingroup\$ @NikhilKashyap Do you want a justification of why A is the answer? or you want to calculate the KVA of secondary? \$\endgroup\$ – Hazem Jan 16 '18 at 15:02
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We can exclude options (B), (C) and (D) easily.

(C) and (D) are impossible, because the ground fault in the secondary causes current only to an unmonitored phase

=> only (A) and (B) are left

DY configuration distributes one phase secondary loading to all primary phase lines or at least to a and c in idealized case. => (A) is left, but is it OK, checking it needs proper calculations. Let's do them.

The voltage rating 33kV/11kV tells phase to phase voltages. The voltage between the faulty line and GND is 11kV/sqrt(3).

Let's write: the current in primary lines a and c = Ia.

Powers in primary and secondary must be equal: Ia * 33kV = 300A * 11kV/sqrt(3). This equation gives Ia=100A/sqrt(3)

The current transformer spec: I1=Ia/100

=> I1=1A/sqrt(3) => option (A) is ok.

NOTE: Power equations are useful to check what should be multiplied or divided by sqrt(3) in 3-phase systems.

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  • \$\begingroup\$ I can't do much better than this. Just fix "options (C) and (D) easily" and maybe make formula clearer: "Primary power = secondary power \$ \implies Ia * 33kV = 300A * 11kV/\sqrt{3} \$" \$\endgroup\$ – Heath Raftery Jan 17 '18 at 4:51
  • \$\begingroup\$ @HeathRaftery Thanks for the notice. I made some fixes. \$\endgroup\$ – user287001 Jan 17 '18 at 8:13
  • \$\begingroup\$ @user287001 Can you show me currents in primary ? I'm still confused with Ppri=Ia*33kV. Shouldn't it be 3*33kV*(Ia/1.732) ? \$\endgroup\$ – Nikhil Kashyap Jan 17 '18 at 10:51
  • \$\begingroup\$ @NikhilKashyap only winding between a and c in primary takes current if the transformer is ideal. That winding operates with the loaded secondary winding. The others are out of the game. The voltage between a and c is 33kV. The current Ia from line a through the active winding to line c is unknown, but the power is Ia*33kV. It's useless to try to use the equations of symmetric 3-phase system because the load is highly non-symmetric, only one phase in secondary and between phase a and c in primary. \$\endgroup\$ – user287001 Jan 17 '18 at 12:21
  • \$\begingroup\$ you took a and c to maintain NI(mmf) balance right ? And in b and c, it is 0. \$\endgroup\$ – Nikhil Kashyap Jan 17 '18 at 13:40
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Because of the wye (star) configuration of the secondary, the sum of the secondary currents must equal zero. Therefore, i2 is always zero. The 300 amp imbalance is reflected back to the primary. For a 3:1 voltage on a delta-star transformer, the turns ratio must be 3√3:1. So the current imbalance is reflected as 300A/(3√3) or 100A/√3. Since the current transformer ratio is 100:1, i1 is 1√3.

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  • \$\begingroup\$ I can't get that 3*1.732 thing. Please explain using power balance as I was trying that way. \$\endgroup\$ – Nikhil Kashyap Jan 16 '18 at 17:48

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