# A DC motor takes an armature current of 100 A at 440 V. The armature circuit resistance is 0.3 Ω. The machine has 6 poles and armature is lap-connecte

116 views
in General
closed
A DC motor takes an armature current of 100 A at 440 V. The armature circuit resistance is 0.3 Ω. The machine has 6 poles and armature is lap-connected with 822 conductors, the flux per pole is 0.05 wb. Calculate the speed.
1. 598.5 rpm
2. 610 rpm
3. 698 rpm
4. 710 rpm

by (110k points)
selected

Correct Answer - Option 1 : 598.5 rpm

Concept:

The EMF equation of a DC Machine is

${E_b} = \frac{{NPϕ Z}}{{60A}}$

In a dc motor, back emf is

Eb = V – IaRa

Where,

N is the speed in rpm

ϕ is the flux per pole

P is the number of poles

Z is the number of conductors

V is the terminal voltage

Ia is the armature current

Ra is the armature resistance

Calculation:

Given V = 440 V

Ia = 100 A

Ra = 0.3 Ω

P = 6

A = P = 6 (LAP winding)

Z = 822

ϕ = 0.05 wb

Eb = V – IaRa

Eb = 440 - 100 × 0.3 = 410 V

${E_b} = \frac{{NPϕ Z}}{{60A}}\Rightarrow N =\frac{E_b\times 60\times A}{Pϕ Z}$

$N = \frac{410 \times 60 \times6 }{6\times 0.05\times 822}=598.54\;rpm$

∴ The speed is 598.5 rpm