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1 Answer

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by (41.2k points)

The given load is star connected load.

L = 0.02 H

Inductive reactance X = 2 × π × 50 × 0.02 ≈ 6.28Ω

R = 10 Ω

Impedance, Zph = 10 + j6.28 = 11.81∠32.13°

Phase voltage, \(V_{ph} = \frac{440}{\sqrt 3} = 254.03 V\)

Line current

\(I_{ph} = I_L = \frac {V_ph}{z_ph} = \frac {254.03}{11.81} = 21.51 A\)

Power in three-phase is given by

\(P = \sqrt 3 \times 440 \times \frac{440}{\sqrt 3 \times 11.8} \times \cos 32.12 = 13 .89 k W\) 

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