Discussion
The flux of a dc motor is reduced to 80 per cent of its rated value at constant armature voltage and negligible resistive drop. The no-load speed becomes
Answer: D
Correct answer: 125 per cent of its former value
With the resistive drop neglected, \( V \approx E = k\Phi N \), so at constant voltage the speed varies inversely with the flux:
\[ N \propto \frac{1}{\Phi} = \frac{1}{0.8} = 1.25 \]Weakening the field to 80 per cent therefore raises the no-load speed by 25 per cent. This is field weakening, the standard way of running above base speed - and the reason a shunt machine that loses its field runs away.
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