Question

The armature of a 6 pole dc shunt motor has a wave connected winding accommodated in 60 slot each containing 40 conductors. If the useful flux per pole is 25 mWb and armature current is 40 A, then value of torque developed in motor will be

• A. 2044.2 N-m
• B. 1488 N-m
• C. 1145.4 N-m
• D. 756 N-m

Answer 1

The correct option is C 1145.4 N-m

$\text{Torque developed}=\frac{\text{Power developed in armature}}{\left(\frac{2\pi N}{60}\right)}$
$=\frac{E_a{I}_a\times 60}{2\pi \times N}..\left(i\right)$

$Also,E_a=\frac{P\varphi NZ}{60A}..\left(ii\right)$

$\text{using equation (i) and (ii)}$

$T=\frac{P\varphi NZ}{60\times A}\times \frac{I_a\times 60}{2\pi \times N}$

$\text{For wave connection, A = 2}$
$T=\frac{6\times 25\times {10}^{-3}\times 40\times 40\times 60\times 60}{60\times 2\times 2\pi }$

$=\frac{150\times 40\times 40\times {10}^{-3}}{120\times 2\pi }\times 60\times 60=1145.4N-m$