Discussion
A dc shunt motor is driving a constant torque load with rated excitation. If the field current is halved, then the speed of the motor __________.
Answer: A
The speed becomes slightly less than doubled.
Motor torque is \(T = k\phi I_{a}\). With a constant load torque, halving the field (and so, assuming a linear magnetic circuit, the flux) forces the armature current to double. The speed is \(N \propto \frac{V - I_{a}R_{a}}{\phi}\).
\[ \begin{aligned} \frac{N_{2}}{N_{1}} &= \frac{V - 2I_{a}R_{a}}{V - I_{a}R_{a}} \times \frac{\phi}{\phi/2} \\ &= 2 \times \frac{V - 2I_{a}R_{a}}{V - I_{a}R_{a}} \end{aligned} \]
The second factor is a little less than 1, because the doubled current causes a larger armature drop, so the speed rises to slightly less than double. Exactly double would ignore the armature drop, and the 'half' options reverse the effect: weakening the field raises a dc motor's speed.
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