Question

An 8 kW, 250 V, 1200 rpm dc shunt motor has $R_a=0.7\Omega$. The field current is adjusted, until on no-load with a supply of 250 V, the motor runs at 1250 rpm and draws armature current of 1.6 A. A load torque is then applied to the motor shaft, which causes the $I_a$ to rise to 40 A and the speed falls to 1150 rpm. The reduction in flux per pole due to armature reaction is___________%.

• A. 3.01

Answer 1

The correct option is A 3.01
$\text{Speed,}$ $N=k_a^{\prime }\left(\frac{E_a}{\varphi }\right)=k_a^{\prime }\left(\frac{V-I_a{R}_a}{\varphi }\right)$

$\text{or, Flux,}$ $\varphi =k_a^{\prime }\left(\frac{V-I_a{R}_a}{N}\right)$

$\varphi (no-load)=k_a^{\prime }\left(\frac{250-40\times 0.7}{1250}\right)=0.199k_a^{\prime }$

$\varphi \left(load\right)=k_a^{\prime }\left(\frac{250-40\times 0.7}{1150}\right)=0.193k_a^{\prime }$

$\text{Reduction in flux per pole due to armature reaction is,}$
$\Delta \varphi =\left(\frac{0.199-0.193}{0.199}\right)\times 100$

$=3.01\%$