A 50 Hz, induction motor wound for pole-amplitude modulation has 20 initial poles and the modulating function has 8 poles. At what two speeds will the motor run?
Question 9.16: A 50 Hz, induction motor wound for pole-amplitude modulation...
P=20, \quad P_{M}=8
P_{1}=20-8=12
P_{2}=20+8=28
From Eq. (9.81) P_{M} / P=\left(1-\frac{3 n}{r}\right)
P_{M} / P=\frac{8}{20}=\frac{2}{5}=\left(1-\frac{3 n}{r}\right) ; \quad n=1, \quad r=5
Hence P_{1} poles are suppressed
Speed (20 poles) =\frac{20 \times 50}{20}=300 rpm
Speed (28 poles) =\frac{20 \times 50}{28}=214.286 rpm
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