Question

Starting torque and maximum torque of 3-phase I.M. varies as respectively
1. \(\frac{1}{{{f^2}}}\;\)and \(\frac{1}{{{f^3}}}\)
2. ​​\(\frac{1}{{{f^3}}}\;\)and \(\frac{1}{{{f^2}}}\)
3. \(\frac{1}{{{f}}}\) and \(\frac{1}{{{f^2}}}\)
4. \(\frac{1}{{{f^2}}}\) and \(\frac{1}{{{f}}}\)

Answer 1

Correct Answer - Option 2 : ​​\(\frac{1}{{{f^3}}}\;\)and \(\frac{1}{{{f^2}}}\)

The torque equation of a three-phase induction motor is given by,

\(T = \frac{{180}}{{2\pi {N_s}}}\left( {\frac{{sV^2{R_2}}}{{\left( {R_2^2 + {s^2}X_2^2} \right)}}} \right)\)

Where Ns is the synchronous speed

V = supply voltage

R2 = rotor resistance

X2 = rotor reactance

s is the slip

By the above expression, we can say that the torque of an induction motor depends on rotor resistance and slip.

At starting, slip (s) = 1

\({T_{st}} = \frac{{180}}{{2\pi {N_s}}} \times \frac{{E_2^2{R_2}}}{{R_2^2 + X_2^2}}\)

R₂ << X₂ and under running condition R₂ >> sX₂

\({T_{st}} = \frac{{180}}{{2\pi {N_s}}} \times \frac{{E_2^2{R_2}}}{{X_2^2}}\)

\( \Rightarrow {T_{st}} \propto \frac{1}{{{f^3}}}\)

Maximum torque in a three-phase induction motor is given by

\({T_{max}} = \frac{{180}}{{2\pi {N_s}}}\frac{{{V^2}}}{{2{X_2}}}\)

Where N­S is the synchronous speed in rpm

V is the supply voltage

X2 is the standstill reactance of rotor

\( \Rightarrow {T_{max}} \propto \frac{1}{{{f^2}}}\)