GATE 2017 EE – Question 57
A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 $\Omega$ and 0.1 $\Omega$, respectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50%, the resistance in ohms that should be added in series with the armature is ________. (Give the answer up to two decimal places.)
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Correct answer: 9.5 to 12
Explanation
In a series motor the torque is proportional to $I_a^2$, and the load torque is proportional to $N^2$, so $I_a \propto N$. At half speed the current is 15 A. At first, $E_{b1} = 220 - 30 \times 0.5 = 205$ V. The back emf is proportional to flux times speed, so $E_{b2} = 205 \times \frac{15}{30} \times 0.5 = 51.25$ V. Then $220 = 51.25 + 15(0.5 + R)$, so $0.5 + R = 11.25$ and $R = 10.75\ \Omega$.