A 220 V DC shunt motor runs at 630 rpm when the armature current is 50 A. Find the speed if the torque is doubled. Assume armature resistance = 0.2 Ω.

Concept:

The EMF equation of the DC shunt motor is:

E = V - I_aR_a

The speed of a DC motor is given by:

E = kϕω 

The torque produced by DC motor is:

T = kϕI_a

where,  E = Back EMF

V = Terminal voltage

I_a = Armature current

R_a = Armature resistance

ω = Speed in rpm

T = Torque

Calculation:

Given, V = 220 volt

Ia = 50 A

Ra = 0.2Ω 

E₁ = 220 - ( 50 × 0.2)

E1 = 210 V

If torque gets doubled, then armature current also doubles.

E₂ = 220 - ( 100 × 0.2)

E₂ = 200 V

\({E_1 \over E_2}= {N_1 \over N_2}\)

\({210 \over 200}= {630 \over N_2}\)

N₂ = 600 rpm