Correct Answer - Option 1 : 1250 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 = 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