GATE 2025 EE – Question 44
A DC series motor with negligible series resistance is running at a certain speed driving a load, where the load torque varies as cube of the speed. The motor is fed from a 400 V DC source and draws 40 A armature current. Assume linear magnetic circuit. The external resistance, in Ω, that must be connected in series with the armature to reduce the speed of the motor by half, is closest to
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Correct answer: (A) 23.28
Explanation
For a series motor with a linear magnetic circuit, $\phi\propto I$, so $T\propto I^2$ and $E\propto\omega I$. The load torque is $T_L\propto\omega^3$, so $I\propto\omega^{3/2}$. At half speed, $I'=40\times0.5^{1.5}=14.14$ A and, since the resistance is negligible at the start ($E=400$ V), $E'=400\times0.5\times0.3536=70.71$ V. Then $400=70.71+14.14R_{ext}$ gives $R_{ext}=23.28\ \Omega$.