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A DC motor takes an armature current of 130 A at 460 V. The armature circuit resistance is 0.4 Ω. The machine has 6 poles and armature is lap-connected with 864 conductors, the flux per pole is 0.07 wb. Calculate the gross torque developed by the armature.
1. 1250 N-m
2. 1450 N-m
3. 1350 N-m
4. 1500 N-m

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Correct Answer - Option 1 : 1250 N-m


Speed of the motor is given as

\({ω} = \frac{{{N} \times 2\pi }}{{60}}\)

Torque developed by the motor is

T = P / ω 


Power developed P = E Ia

\(E = \frac{ϕ ZNp}{60A}\)

∴ \(T= \frac{\frac{ϕ ZNp}{60A}I_a}{\frac{{{N_s} \times 2\pi }}{{60}}}=\frac{ϕ ZpI_a}{2\pi A}\)



Armature current Ia = 130 A

Number of poles P = 6

Lap wound A = P = 6

Number of conductors Z = 864

The flux per pole ϕ = 0.07 Wb

\(T=\frac{ϕ ZpI_a}{2\pi A}=\frac{0.07\times 864\times 6 \times130}{2\pi\times 6 }=1251.97\;Nm\)

T ≈ 1250 N-m

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