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GATE EE 2014 Official Paper: Shift 3

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

**Concept:**

The torque of the motor is, \(T = \frac{P}{\omega } = \frac{{V{I_a}}}{\omega }\)

Where \(\omega = \frac{{2\pi {N_s}}}{{60}}\)

I_{a} = Armature current, V = Excitation voltage

**Calculation:**

No load speed, N = 1400 rpm

I_{a} = 8A, V = 230 V

\(T = \frac{{V{I_a}}}{\omega } = \frac{{V{I_a}}}{{\frac{{2\pi\; \times \;1400}}{{60}}}} = \frac{{230\; \times\; 8}}{{\frac{{2\; \times \;\pi \; \times \;1400}}{{60}}}}\)

\(T = \frac{{9.55\; \times \;230\; \times\; 8}}{{1400}}\)

T = 12.554 Nm