Question is ⇒ A 220 V. DC shunt motor is operating at a speed of 1,440 rpm. The armature resistance is 1.0 Ω and armature current is 10 A. If the excitation of the machine is reduced by 10%, the extra resistance to be put in the armature circuit to maintain the same speed and torque will be, Options are ⇒ (A) 1.79Ω, (B) 2.1Ω, (C) 3.1Ω, (D) 18.9Ω, (E) , Leave your comments or Download question paper.

Question is ⇒ A 220 V. DC shunt motor is operating at a speed of 1,440 rpm. The armature resistance is 1.0 Ω and armature current is 10 A. If the excitation of the machine is reduced by 10%, the extra resistance to be put in the armature circuit to maintain the same speed and torque will be, Options are ⇒ (A) 1.79Ω, (B) 2.1Ω, (C) 3.1Ω, (D) 18.9Ω, (E) , Leave your comments or Download question paper. Previous question Next question

Q1.

A 220 V. DC shunt motor is operating at a speed of 1,440 rpm. The armature resistance is 1.0 Ω and armature current is 10 A. If the excitation of the machine is reduced by 10%, the extra resistance to be put in the armature circuit to maintain the same speed and torque will be

1.79Ω

2.1Ω

3.1Ω

18.9Ω

View Answer 19 -29

Explanation:-
Answer : A
Discuss it below :

Abdul Ghafoor Said on 2021-04-09 13:28:56

Ia1 = 10
Now flux is decreased by 10%, so φ2 = 0.9 φ1
Torque is constant, & T = kT φ Ia
Then, Ia1 φ1 = I a2 φ2

⇒ Ia2 =	10	= 11.11 A
	0.9

Now , N ∝	Eb
	φ

⇒	N1	=	Eb1	×	φ2
	N2		Eb2		φ1

=	220 - Ia1 r1	×	0.9φ1
	220 - Ia2 (r1 + R)		φ1

∴ 1 =	210 × 0.9
	220 - 11.11 (1 + R)

⇒ 1 + R = 2.79
⇒ R = 1.79 Ω

Shahryar Muhammad Said on 2019-11-02 01:32:42

how to solve it?

Hussain mohammed Said on 2019-03-19 23:21:53

What was the answer how to do this problem.

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