For the short circuit test of a 3 phase transformer (star-star connected) I am given ( or have already calculated from previous questions) $$I=9720 A\\cosφ=\sqrt 2/2\\P=300kW\\R=0.001 (phase)$$I am asked to find the leakage reactance of the transformer. I started by finding the apparent power from cosφ $$P/S=cosφ=>S=\sqrt 2P$$Then S for any phase is $$\frac{S}{3}=\frac{V}{\sqrt 3}I=>V=\frac{\sqrt 3S}{3I}$$ And now I can find the total impedance Z $$Z=\frac{V}{\sqrt 3 I}=0.001496$$ Finally for the leakage reactance $$X=\sqrt{Z^2-R^2}=0.00111$$

The solution manual did just one calculation for all this. $$\tanφ=\frac{X}{R}=>X=0.001$$

How is this true and what have I assumed wrong in my calculations and the result is different?


Both the answers are right.

There are many instances where my answer and the solution manual's answer are different by < 0.1% as in this case.

Coming to the reasoning part, the cos(phi) term given is the R(phase) / Z(phase).

where Z(phase) = sqrt((R(phase))^2 + (X(phase))^2).

As cos(phi) = sqrt(2) / 2, tan(phi) = 1 = R(phase) / X(phase). Hence X(phase) = R(phase).

So your answer and the manual's answer are correct, and your answer is a bit more accurate.


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