Question

A 6-pole, 3-phase, 50 Hz induction motor runs on full load with a slip of 4%. Given the rotor stand still impedance/phase as (0.01 + j0.05) $\Omega$, calculate the available maximum torque in terms of full load torque

• A. $0.384T_{fL}$
• B. $2.6T_{fL}$
• C. $0.079T_{fL}$
• D. $12.52T_{fL}$

Answer 1

The correct option is B

$2.6T_{fL}$

$\text{s = 0.04}$

$s_m=\frac{R_2}{X_2}=\frac{0.01}{0.05}=0.2$

$\frac{T_{max}}{T_{fL}}=\frac{s^2+s_m^2}{2s{s}_m}=\frac{{0.04}^2+{0.2}^2}{2\times 0.04\times 0.2}$

$\frac{T_{max}}{T_{fL}}=2.6$

$T_{max}=2.6T_{fL}$