Question

A 200 V, 1000 rpm dc shunt motor drives a load that requires a constant torque regardless of the speed of operation. The armature circuit resistance is 0.03$\Omega$. When this motor takes armature current of 75A. If the flux is reduced to 75% of its original value, then new speed will be ______ rpm.

• A. 1328.28

Answer 1

The correct option is A 1328.28
$E=200-(75\times 0.03)$

$E=197.75V$

$E\propto \varphi {\omega }_m$


$\text{Load torque is constant,}$

$\text{T = constant}$

$T\propto \varphi I_a$

$\text{so,}{\varphi }_2{I}_{a2}={\varphi }_1{I}_{a1}$

$I_{a2}=\frac{{\varphi }_1}{0.75{\varphi }_1}{I}_{a1}=100A$

$\text{now,}\frac{E_{a2}}{E_{a1}}=\frac{{\varphi }_2}{{\varphi }_1}\times \frac{N_2}{N_1}$

$\frac{200-100\times 0.03}{197.75}=\frac{0.75\times N_2}{1000}$

$N_2=1328.28\text{rpm}$