A three-phase, three winding Δ/Δ/Y (1.1 kV/6.6 kV/400 V) transformer is energized from AC mains at the 1.1 kV side. It supplies 900 kVA load at 0.8 power factor lag from the 6.6 kV winding and 300 kVA load at 0.6 power factor lag from the 400 V winding. The RMS line current in ampere drawn by the 1.1 kV winding from the mains is ___________. (Give the answer up to one decimal place.)
Concept:
Power balance equation,
input kVA = output kVA
Explanation:
Equivalent circuit is given as,

using power balance equation,
input kVA = output kVA
kVA1 = kVA2 + kVA3
kVA1 = 900 ∠-36.87° + 300 ∠-53.13°
kVA1 = 720 - j 540 + 160 - j 240
kvA1 = 900 + j 780
kvA1 = 1190.9 ∠-40.9° kVA
We know that,
kVA = √3 Vl Il
where,
Vl = line voltage
Il = line current
Now,
\(\rm kVA=\sqrt{3}V_{l_1}I_{l_1}=1190.9\angle-40.9^\circ\)
⇒ \(I_{l_1}=\frac{1190.9\angle-40.9\ \rm kVA}{\sqrt{3}\times1.1\ \rm kV}\)
⇒ \(\rm I_{l_1}=625.09\angle-40.9^\circ\ \rm A\)
Hence,
RMS line current drawn by the 1.1 kV winding from the mains is 625.09 A
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