# 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-connecte

61 views
in General
closed
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

by (110k points)
selected

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