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A 4-pole dc motor takes a 50 A armature current. The armature has lap connected 480 conductors. The flux per pole is 20 mWb. Calculate the gross torque developed by the armature of the motor.
1. 52.6 N-m
2. 39.5 N-m
3. 76.4 N-m
4. 15.2 N-m

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Correct Answer - Option 3 : 76.4 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 = 50 A

Number of poles p = 4

Lap wound A = p = 4

Number of conductors Z = 480

The flux per pole ϕ = 20 mWb

\(T=\frac{ϕ ZpI_a}{2\pi A}=\frac{20\times 10^{-3}\times 480\times 50 \times4}{4\times 2\pi }\)

T = 76.4 N-m

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