Question

An 8 kW, 230 V, 1200 rpm dc shunt motor has armature resistance as $0.7\Omega$. The field current is adjusted until on no load with a supply voltage 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 armature current to rise to 40 A and the speed fall to 1150 rpm. The reduction in flux per pole due to the armature reaction is____%

• A. 3.06

Answer 1

The correct option is A 3.06
$E_a=K_a\varphi N$ $\Rightarrow \varphi =\frac{E_a}{K_{a}N}$

$\varphi =\frac{V-I_a{R}_a}{K_{a}N}$

${\varphi }_{(no-load)}=\left[\frac{250-1.6\times 0.7}{K_a\times 1250}\right]=\frac{0.1991}{K_a}$

${\varphi }_{load}=\left[\frac{250-40\times 0.7}{1150\times K_a}\right]=\frac{0.193}{K_a}$

$\therefore \text{Reduction in}\varphi \text{due to armature redrawn}$
$=\left(\frac{0.1991-0.193}{0.1991}\right)\times 100\%=3.06\%$