Question

In the circuit shown, the phase currents are
\[ I_A=572.812+j50.115\text{ A} \]
\[ I_B=-254.525-j459.175\text{ A} \]
\[ I_C=-207.083+j444.091\text{ A} \]

Given that the CTs are ideal with no saturation, and the turns ratio of the Main CT is \(300:5\) and that of the Auxiliary Transformer (\(Yn\Delta\)) is \(2:1\) on every phase, the value of \(I_{AR}\), rounded off to three decimal places, is:

• A. \(0\) A
• B. \(0.653\angle17.556^{\circ}\) A
• C. \(537.240\angle4.105^{\circ}\) A
• D. \(8.954\angle4.105^{\circ}\) A

Hint:

First scale the phase currents down through the 300:5 Main CT, then subtract the zero sequence average (IA+IB+IC)/3 from the phase A current to get IAR.

Answer 1

The Correct Option is D

Step 1: Reduce the phase currents through the Main CT.
The Main CT has a ratio of \(300:5\), the same as \(60:1\). Dividing each primary phase current by \(60\) gives the current available on the CT secondary side:
\[ I_{A,sec}=\frac{572.812+j50.115}{60}=9.547+j0.835\text{ A} \]
\[ I_{B,sec}=\frac{-254.525-j459.175}{60}=-4.242-j7.653\text{ A} \]
\[ I_{C,sec}=\frac{-207.083+j444.091}{60}=-3.451+j7.402\text{ A} \]

Step 2: Recognize what the Auxiliary Transformer does.
The Auxiliary Transformer is connected \(Yn\Delta\), meaning its three primary windings are star connected with the star point grounded, fed from the three CT secondary currents. A grounded star primary sums the three phase currents, and this sum drives a circulating current in the delta secondary. This is the standard way to extract the zero sequence part of a set of currents:
\[ I_{0,sec}=\frac{I_{A,sec}+I_{B,sec}+I_{C,sec}}{3} \]

Step 3: Compute the zero sequence current.
\[ I_{A,sec}+I_{B,sec}+I_{C,sec}=(9.547-4.242-3.451)+j(0.835-7.653+7.402)=1.854+j0.584\text{ A} \]
\[ I_{0,sec}=\frac{1.854+j0.584}{3}=0.618+j0.195\text{ A}=0.653\angle17.556^{\circ}\text{ A} \]

Step 4: Identify how \(I_{AR}\) is formed.
From the circuit, the resistor carrying \(I_{AR}\) is fed directly from the phase A CT secondary line, while the Auxiliary Transformer's delta output subtracts the zero sequence current \(I_{0,sec}\) out of that line (the \(2:1\) ratio of the Auxiliary Transformer is exactly the turns ratio needed for its delta output to remove this average, \(I_{0,sec}\), from the direct phase A current). So
\[ I_{AR}=I_{A,sec}-I_{0,sec} \]

Step 5: Substitute the numbers.
\[ I_{AR}=(9.547+j0.835)-(0.618+j0.195)=8.929+j0.640\text{ A} \]

Step 6: Convert to polar form.
\[ |I_{AR}|=\sqrt{8.929^2+0.640^2}=8.954\text{ A} \]
\[ \angle I_{AR}=\tan^{-1}\left(\frac{0.640}{8.929}\right)=4.105^{\circ} \]

Step 7: Analyze the options.

(A) 0 A: Would only hold if the three currents were perfectly balanced with no zero sequence at all, which is not the case here, and it is also not what \(I_{AR}\) physically represents. Incorrect.

(B) \(0.653\angle17.556^{\circ}\) A: This is actually \(I_{0,sec}\), the zero sequence current itself, not \(I_{AR}\). Incorrect.

(C) \(537.240\angle4.105^{\circ}\) A: This is \(60\) times the correct answer, the value obtained if the \(300:5\) Main CT ratio is never applied. Incorrect.

(D) \(8.954\angle4.105^{\circ}\) A: Matches the value found above. Correct.

Final Answer:
\[ \boxed{8.954\angle4.105^{\circ}\text{ A}} \]