SSC JE Electrical Engineering 2013 Paper

Question : 162 of 200

Marks: +1, -0

A dc series motor has an armature resistance of 0.06  and series field resistance of 0.08 Ω. The motor is connected to a 400 V supply. The line current is 20 A when the speed of the machine is 1100 rpm. when the line current is 50 A and the excitation is increased by 30%, speed of the machine in rpm is

1100
1003
837
938

Validate

Solution:

V = 400 V,

I_1 = 20A

N_1 = 1100

rpm

R_a = 0.6Ω, R_\se = .08 Ω

I_2 = 50 A, N_2 = ?

ϕ_2 = ϕ_1(1.3)

While drawing line current of 200 A

E_{b1} = V – I_1 (R_a + R_{\se})

= 400 – 20(0.06 + 0.08) = 397.2 volt
while drawing line current 50 A.

E_{b2}

= 400 – 50(.06 + .08) = 393 V

N_2/N_1=E_{b2}/E_{b1}×ϕ_1/ϕ_2

N_2/1100=393/397.2×ϕ_1/{ϕ_1×1.3}

N_2=837\RPM

Go to Question:

Prev Question

Next Question