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

I_{a} = 100 A

R_{a} = 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**