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Question

# A 2000 V, 3-phase star-connected synchronous motor has synchronous impedance of (0.5 + j5) Ω per phase. For an excitation voltage of 3000 V, the motor takes an input of 900 kW at rated voltage. The power angle of motor would be around

A
53
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B
49
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C
60
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D
40
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Solution

## The correct option is B 49∘Synchronous impedance, Zs=(0.5+j5)Ω=5.025∠84.29∘ Ω Ia=V∠0∘−E∠−δZs∠θ S=VIa=V∠0[V∠0−E∠−δZs∠θ] S=V2Zs∠θ−EVZs∠θ+δ So from above equation, P=V2Zscosθ−EVZscos(θ+δ) 900×103=200025.025(cos84.29∘)−2000×30005.025cos(84.29∘+δ) 84.29∘+δ=133.426∘ Power angle, δ=49.13∘

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