For an induction motor, operating at a slip ‘s’, the ratio of gross power output to the air gap power is equal to

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ISRO Scientist Electrical 2019 Paper

Option 2 : (1 - s)

NTPC Diploma Trainee 2020: Full Mock Test

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120 Questions
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120 Mins

Concept:

Power flow in the Induction motor is as shown below.

Rotor input or air gap power \({{P}_{in}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}\)

Rotor copper losses \({{P}_{cu}}=s\times {{P}_{in}}=3I_{2}^{2}{{R}_{2}}\)

Gross mechanical power output \({{P}_{g}}={{P}_{in}}-{{P}_{cu}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}-3I_{2}^{2}{{R}_{2}}=3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

The relation between the rotor air gap power, rotor copper losses and gross mechanical power output is,

\({{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}:3I_{2}^{2}{{R}_{2}}:3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

\(\Rightarrow {{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{1}{s}:1:\left( \frac{1-s}{s} \right)=1:s:\left( 1-s \right)\)