Correct Answer - Option 3 : 76.4 N-m
Concept:
Speed of the motor is given as
\({ω} = \frac{{{N} \times 2\pi }}{{60}}\)
Torque developed by the motor is
T = P / ω
where,
Power developed P = E Ia
\(E = \frac{ϕ ZNp}{60A}\)
∴ \(T= \frac{\frac{ϕ ZNp}{60A}I_a}{\frac{{{N_s} \times 2\pi }}{{60}}}=\frac{ϕ ZpI_a}{2\pi A}\)
Calculation:
Given,
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