Discussion

To limit that machine's starting current to 100 A, the resistance to be inserted in series is

Answer: C

Correct answer: 2.25 Ω

The total resistance required is

\[ R_{total} = \frac{250}{100} = 2.5 \ \Omega \]

and the armature already provides 0.25, so the external resistance is

\[ 2.5 - 0.25 = 2.25 \ \Omega \]

Answering 2.5 forgets that the armature is part of the circuit - a small error here, but a serious one in a machine of very low armature resistance.

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