Discussion
A dc shunt motor's field is weakened to 80 per cent while the load torque is unchanged. The armature current will
Answer: A
Correct answer: rise to about 125 per cent of its former value
The torque required has not changed, and \( T = k\Phi I_a \), so with the flux at 80 per cent the current must rise to compensate:
\[ I_a \propto \frac{1}{\Phi} = \frac{1}{0.8} = 1.25 \]Field weakening therefore raises the speed by 25 per cent and the armature current by 25 per cent at the same torque, which is why the constant-power region above base speed is limited by armature heating as much as by anything else.
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