GATE 2020 EE – Question 38
A 250 V dc shunt motor has an armature resistance of $0.2\ \Omega$ and a field resistance of $100\ \Omega$. When the motor is operated on no-load at rated voltage, it draws an armature current of 5 A and runs at 1200 rpm. When a load is coupled to the motor, it draws total line current of 50 A at rated voltage, with a 5 % reduction in the air-gap flux due to armature reaction. Voltage drop across the brushes can be taken as 1 V per brush under all operating conditions. The speed of the motor, in rpm, under this loaded condition, is closest to:
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: (C) 1220
Explanation
At no load $E_b=250-5\times0.2-2=247$ V at 1200 rpm. The field current is $2.5$ A, so the loaded armature current is $50-2.5=47.5$ A and $E_b=250-47.5\times0.2-2=238.5$ V. With 95% of the flux, $N=1200\times\frac{238.5}{247}\times\frac{1}{0.95}=1220$ rpm.