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Question
A 3-phase, 400 V synchronous motor takes 52.5 A at a power factor of 0.8 leading. The motor impedance per phase is (0.25+j3.2)Ω. Then the induced emf (line to line) of motor is _______V.
- 608.5
- 608.5
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Solution
The correct option is A 608.5
For leading power factor, (Ef)2=(V cos ϕ−IaRa)2+(V sin ϕ+IaXs)2
=(400√3×0.8−52.5×0.25)2+(400√3×0.6+52.5×3.2)2
=(171.6)2+(306.57)2
Ef=351.3 V
Ef(line to line)=351.3×√3=608.5 Volts
For leading power factor, (Ef)2=(V cos ϕ−IaRa)2+(V sin ϕ+IaXs)2
=(400√3×0.8−52.5×0.25)2+(400√3×0.6+52.5×3.2)2
=(171.6)2+(306.57)2
Ef=351.3 V
Ef(line to line)=351.3×√3=608.5 Volts
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