This question was previously asked in

GATE EE 2014 Official Paper: Shift 1

Option 3 : 2.125

ST 1: Logical reasoning

5265

20 Questions
20 Marks
20 Mins

**Concept:**

For the induction motor, we have \({T_f}\left( {full\;load\;torque} \right) \propto \frac{{{s_f}{R_2}}}{{R_2^2 + {{\left( {{s_f}{X_2}} \right)}^2}}}\) or \({T_{max}}\left( {\mathop {\max }\ torque} \right) \propto \frac{1}{{2{X_2}}}\)

So, we get,

\(\frac{{{T_f}}}{{{T_{max}}}} = \frac{{2{s_f}{R_2}{X_2}}}{{R_2^2 + {{\left( {{s_f}{X_2}} \right)}^2}}}\) or \(\frac{{{T_f}}}{{{T_{max}}}} = \frac{2}{{\frac{{{S_{max}}}}{{{s_f}}} + \frac{{{S_f}}}{{{S_{max}}}}}}\) ……..(i)

\(\frac{{{T_f}}}{{{T_{max}}}} = \frac{2s_{max} s_f}{{{{{S_{max}^2}}} + {{{S_f}^2}}{{}}}}\)

Where,

s_{f} = Full load slip

s_{max }= (R/X) = Maximum slip

**Calculation:**

Now, the slip at full load is 3%, i.e s_{f} = 0.05 and

\({s_{max}} = \frac{{{R_2}}}{{{X_2}}} = \frac{{0.1}}{{0.5}} = 0.2\)

Substituting these values in equation (i), we get

\(\frac{{{T_f}}}{{{T_{max}}}} = \frac{2}{{\frac{{0.2}}{{0.05}} + \frac{{0.05}}{{0.2}}}}\) or

\(\frac{{{T_{max}}}}{{{T_f}}} = \frac{{4 + 0.25}}{2}\)

\(\frac{{{T_{max}}}}{{{T_f}}} = 2.125\)

Hence, the ratio of maximum to full load torque is **2.125**