Question B.29: he armature resistance of a 55-hp, 525-V, DC shunt wound mot...

 The armature resistance of a 55-hp, 525-V, DC shunt wound motor is 0.4  \Omega. The full-load armature current of this motor is 80 \mathrm{~A}. If the initial starting current is 175 percent of the full-load value, the resistance of the starting coil in \Omega should be nearest to
a. 6.6125      b. 3.75      c. 3.35      d. 4.15      e. 2.75
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I_{\text {start }}(\text { starting current })=1.75 \times 80=140 \mathrm{~A}

Therefore, the total resistance is

R=\frac{V}{I_{S}}=\frac{525}{140}=3.75  \Omega

Thus, the resistance of the starter is 3.75-0.4=3.35  \Omega. The correct answer is c.

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