Question

400-V dc shunt motor takes a current of 5.6 A on no-load and 68.3 A on full-load. Armature reaction weakens the field by 3%. What is the ratio of full-load speed to no-load speed? Given: Ra = 0.18 Ω, brush voltage drop= 2 V, Rf = 200 Ω.

(a) 1.2

(b) 0.8

(c) 1.4

(d) 1

The question was posed to me in an internship interview.

This key question is from Speed Control Using Field Control of Shunt Motor in chapter DC Motors of DC Machines

Answer 1

Right option is (d) 1

The explanation is: If = 400/200= 2 A

No-load:

Ia0 = 5.6 – 2 = 3.6 A

Ea0 = 400 – 0.18 \ 3.6 – 2 = 397.4 V

Full-load:

Ib>a (fl) = 68.3 – 2 = 66.3 A

Ea (fl) = 400 – 0.18 / 66.3 – 2 = 386.1 V

n (fl)/n (nl) = [386.1/397.4] [1/0.97] = 1.