Correct Answer - Option 4 : 1 Ω and 0.2 p u
Concept:
The slip at maximum torque is given by \({s_m} = \dfrac{{{R_m}}}{{{X_m}}}\)
Where Rm = Motor resistance per phase
Xm = Motor reactance per phase
n a three-phase induction motor, synchronous speed is
\({N_s} = \dfrac{{120f}}{P}\)
Where,
Ns = Synchronous Speed in rpm
f = supply frequency in Hz
P = number of poles
Rotor speed Nr = Ns (1 – s)
Per Unit rotor speed(asynchronous) is given by
\(N_{r\ (pu)}=\dfrac{N_r}{N_s}=(1-s)\)
Per Unit Synchronous speed Ns(pu) = Ns/Ns = 1 PU
Application:
Given:
Number of Poles (P) = 4
f = 50 HZ implies \({N_s} = \dfrac{{120f}}{P}=1500 \ rpm\)
Rotor resistance at Maximum Torque = 0.2 Ω
Nr = 1200 rpm at Maximum Torque
Slip at maximum Torque is given by
\({s_m} = \dfrac{N_s-N_r}{N_s}=\dfrac{1500-1200}{1500}=0.2=\dfrac{R_m}{X_m}\)
Therefore,
Rotor reactance at Maximum Torque is
\(Xm = \dfrac{0.2}{0.2}\)
= 1 Ω
The difference between synchronous and asynchronous speeds is
Ns(pu) - Nr(pu) = 1 - (1 - s) = s
= 0.2 PU