Induction Motor in Running Conditions MCQ Quiz in मराठी - Objective Question with Answer for Induction Motor in Running Conditions - मोफत PDF डाउनलोड करा

The equivalent circuit for the machine has been depicted in the figure.

\(Rated\ slip = \frac{{1500 - 1410}}{{1500}}\)

which is equal to 0.06

Rated current at a slip of 0.06 through the machine is given by

\({I_{rated}} = \frac{{\frac{{440}}{{\sqrt 3 }}}}{{\sqrt {{{\left( {2 + \frac{2}{s}} \right)}^2} + {{\left( 7 \right)}^2}} }}\)

\({I_{rated}} = 7.052A\)

Similarly rated torque of the machine is given by

\({T_{rated}} = \frac{3}{{{\omega _s}}} \times I_{rated}^2 \times \frac{{{r_2}}}{s}\)

Substituting the data here we get T_rated as 31.664 Nm.

According to the question the current at 300 V is same as 7.052A hence

\(7.052 = \frac{{\frac{{300}}{{\sqrt 3 }}}}{{\sqrt {{{\left( {2 + \frac{2}{s}} \right)}^2} + {{\left( 7 \right)}^2}} }}\)

From this value of slip can be extracted out as 0.0928 Putting I_rated as 7.052 and s as 0.0928 in torque equation we get new torque as 20.469

Hence

\(\% Reduction\ in\ Torque = \frac{{31.664 - 20.469}}{{31.664}}\)

which is equal to 35.35% ≈ 35%