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Question

The main and auxiliary winding impedances of a 50 Hz, capacitor start single phase induction motor are: Main winding →Zlm=3+j2.7Ω; Auxiliary winding →Zla=7+j3Ω The value of capacitor in μF to be connected in series with auxiliary winding to achieve a phase difference of α=67∘ between the currents of two winding at start is

A
147.75
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B
508.145
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C
295.5
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D
259.07
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Solution

The correct option is B 508.145Phase angle of main winding current ∠Im=−∠Zm=−∠tan−12.73=−42∘ Phase angle of auxiliary winding current with capacitor in series, ∠Ia=−∠tan−1(3−1ωC)7 Now phase difference, α=67∘ (Given) 67∘=∠Ia−∠Im 67∘=−tan−1(3−1ωC)7+42∘ tan−1⎛⎝3−1ωC7⎞⎠=−25∘ 3−1ωC7=−0.466 3−1ωC=−3.2641 1ωC=3+3.2641 1ωC=6.2641 C=12π×50×6.2641=508.145μF

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