enter image description here

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 ?

  • \$\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, 2018 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\$ Jan 12, 2018 at 2:15
  • \$\begingroup\$ I would try it in a simulator to see what I got. \$\endgroup\$
    – Andy aka
    Jan 12, 2018 at 11:22
  • \$\begingroup\$ @Andyaka, can you suggest something? \$\endgroup\$ Jan 16, 2018 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, 2018 at 15:02

2 Answers 2


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.

  • \$\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\$ Jan 17, 2018 at 4:51
  • \$\begingroup\$ @HeathRaftery Thanks for the notice. I made some fixes. \$\endgroup\$
    – user136077
    Jan 17, 2018 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\$ Jan 17, 2018 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\$
    – user136077
    Jan 17, 2018 at 12:21
  • \$\begingroup\$ you took a and c to maintain NI(mmf) balance right ? And in b and c, it is 0. \$\endgroup\$ Jan 17, 2018 at 13:40

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.

  • \$\begingroup\$ I can't get that 3*1.732 thing. Please explain using power balance as I was trying that way. \$\endgroup\$ Jan 16, 2018 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.