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A 3-ϕ induction motor drives a constant torque load. It operates at a slip of 4% at rated voltage. If the voltage is reduced by 10% the slip will (approximately)
1. Increase by 25%
2. Decrease by 20%
3. Decrease by 67%
4. Increase by 33%

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Correct Answer - Option 1 : Increase by 25%

At constant torque load slip is inversely proportional to square of the voltage.

\(\begin{array}{l} s \propto \frac{1}{{{V^2}}}\\ \Rightarrow \frac{{{s_1}}}{{{s_2}}} = {\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^2}\\ \Rightarrow \frac{{{s_1}}}{{{s_2}}} = {\left( {\frac{{0.9\;{V_1}}}{{{V_1}}}} \right)^2}\\ \Rightarrow {s_2} = 1.23\;{s_1} \end{array}\)

Hence the slip will increase by 25% approximately.

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