Question
Download Solution PDFA 400 V, 50 Hz, three-phase induction motor runs at a slip of 0.06 when it is driving a constant load torque. With what slip will it run driving the same load if the voltage applied is reduced to 200 V, neglecting stator resistance and the equivalent reactance of the motor?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept
For a constant load torque, the slip varies inversely with the square of the voltage.
s α V
\({s_2\over s_1}=({V_1\over V_2})^2\)
Calculation
Given, s1 = 0.06
V1 = 400 V
V2 = 200 V
\({s_2\over 0.06}=({400\over 200})^2\)
s2 = 0.06 × 4
s2 = 0.24
Last updated on May 29, 2025
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