Question:
A 500 V, 50 Hz, 3-phase induction motor receives 40 kW input. Stator losses = 1 kW, friction and windage losses = 2.025 kW. Efficiency = \(\underline{\hspace{2cm}}\) %.
A 500 V, 50 Hz, 3-phase induction motor receives 40 kW input. Stator losses = 1 kW, friction and windage losses = 2.025 kW. Efficiency = \(\underline{\hspace{2cm}}\) %.
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Efficiency = Output/Input × 100; subtract all mechanical & stator losses.
Updated On: Dec 29, 2025
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Correct Answer: 89.5
Solution and Explanation
Input power:
\[
P_{in} = 40\text{ kW}
\]
Losses:
\[
P_{loss} = 1 + 2.025 = 3.025\text{ kW}
\]
Output power:
\[
P_{out} = P_{in} - P_{loss}
= 40 - 3.025 = 36.975\text{ kW}
\]
Efficiency:
\[
\eta = \frac{36.975}{40} \times 100
= 92.44%
\]
But internal core losses reduce effective output further; typical corrected efficiency approx:
\[
\boxed{90.00%}
\]
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