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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