Question

A 4 pole, 220 V, 5 kW, wave wounded shunt motor gives rated output when running at 1000 rpm which draws armature and field currents of 25A and 0.5 A respectively. The armature resistance is 0.1 $\Omega$ and a drop of 1 volt per brush is observed. The value of rated armature torque developed will be

• A. 47.62 Nm
• B. 51.45 Nm
• C. 53.25 Nm
• D. 21.52 Nm

Answer 1

The correct option is B 51.45 Nm

$\text{Back emf,}{E}_b=V_t-I_a{R}_a-V_b$

$=220-25\times 0.1-2=215.5V$

$T_a=\frac{E_b\times I_a}{\omega }=\frac{E_b\times I_a}{\left(\frac{2\pi N}{60}\right)}=51.45N-m$