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Question

# A 20 MVA, 3-phase star connected alternator with an impedance of 5 Ω and a resistance of 0.5 Ω is operating in parallel with constant voltage 11 kV bus-bars. If its field current is adjusted to give an excitation voltage of 12 kV. The value of armature current and power factor under maximum power conditions will be respectively

A
1784 A, 0.705 lagging
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B
1879.57 A, 0.669 lagging
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C
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D
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Solution

## The correct option is D 1784 A, 0.705 leadingAt maximum power conditions, δ=θs δ=cos−1[0.55]=84.260∘ Ef∠δ=Vt∠0+IaZs Ef per phase =12000√3=6928.20 V Vt per phase =11000√3=6350.852 V Efcosδ+jEfsinδ=Vt+Ia∠−ϕ.Zs∠θs =Vt+IaZs∠θs−ϕ Efcosδ+jEfsinδ=Vt+IaZscos(θs−ϕ)+jIaZssin(θs−ϕ) Efcosδ=Vt+IaZscos(θs−ϕ) 692.98=6350+5Iacos(θs−ϕ) 5Iacos(θs−ϕ)=−5657.015 Iacos(θs−ϕ)=−1131.403 ...(i) Again, 5Iasin(θs−ϕ)=6893.46193 ...(ii) From (i) and (ii) squaring and adding we get, Ia=1783.5 A cos(θs−ϕ)=−0.6343 θs−ϕ=129.373 ϕ=−45.11 cosϕ=0.705 leading Alternate Solution: For Pmax, θs=δ=84.26∘ and leading pf Pmax=115[12−11cos84.26∘]=√3×11×Icosϕ I cosϕ=1.2586 ....(i) Q=115(−11sin84.26)=√3×11×Isinϕ Isinϕ=−1.2638 ....(iii) From equation (i) and (iii), ϕ=45.11∘(leading) pf=cos45.11∘=0.705 (leading) I=1.25860.705=1.7836 kA=1783.6 A

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