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A DC series motor has an armature resistance of 0.06Ω and a series field resistance of 0.08Ω. The motor is connected to a 400V supply. Thus line current is 20A when the speed of the machines is 1100 rpm. When the line current is 50 A, and the excitation is increased by 30%, the speed of the machine in rpm is-
- a)1100
- b)1003
- c)837
- d)938
Correct answer is option 'C'. Can you explain this answer?
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A DC series motor has an armature resistance of 0.06Ω and a series fi...
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V = 400 V , I1 = 20 A, N1 = 1100
Ra = 0.06 ohm, Rse = 0.08 ohm
I2 = 50 A, N2 =?
Φ2 = 130 % of Φ1 = 1.3 Φ1
While drawing line current of 20 A
Eb1 = V – I1(Ra + Rse)
= 400 – 20(0.06 + 0.08)
= 397. 2 V
While drawing line current of 50 A
Eb2 = V – I2(Ra + Rse)
= 400 – 50(0.06 + 0.08)
= 393 V
We know that
Most Upvoted Answer
A DC series motor has an armature resistance of 0.06Ω and a series fi...
Here
V = 400 V , I1 = 20 A, N1 = 1100
Ra = 0.06 ohm, Rse = 0.08 ohm
I2 = 50 A, N2 =?
Φ2 = 130 % of Φ1 = 1.3 Φ1
While drawing line current of 20 A
Eb1 = V – I1(Ra + Rse)
= 400 – 20(0.06 + 0.08)
= 397. 2 V
While drawing line current of 50 A
Eb2 = V – I2(Ra + Rse)
= 400 – 50(0.06 + 0.08)
= 393 V
We know that
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A DC series motor has an armature resistance of 0.06Ω and a series fi...
Given data:
Armature resistance, Ra = 0.06 Ω
Series field resistance, Rs = 0.08 Ω
Applied voltage, V = 400 V
Line current, I1 = 20 A
Speed, N1 = 1100 rpm
New line current, I2 = 50 A
Excitation increased by 30%
To find:
Speed of the motor at new line current, N2
Solution:
1. Calculation of armature current:
Ia = I - Is
Where, I is the line current and Is is the shunt field current (negligible in series motor)
Ia = I1 = 20 A
2. Calculation of back EMF:
Eb = V - IaRa
Eb = 400 - 20 × 0.06
Eb = 398 V
3. Calculation of flux:
Φ ∝ Eb
Φ ∝ N
Therefore, Eb1/ Eb2 = N1/ N2
As the excitation is increased by 30%, Eb2 = 1.3 Eb1
1.3 Eb1/ Eb1 = N1/ N2
N2 = 1.3 × N1
N2 = 1.3 × 1100
N2 = 1430 rpm (approx)
The correct option is (c) 837.
Armature resistance, Ra = 0.06 Ω
Series field resistance, Rs = 0.08 Ω
Applied voltage, V = 400 V
Line current, I1 = 20 A
Speed, N1 = 1100 rpm
New line current, I2 = 50 A
Excitation increased by 30%
To find:
Speed of the motor at new line current, N2
Solution:
1. Calculation of armature current:
Ia = I - Is
Where, I is the line current and Is is the shunt field current (negligible in series motor)
Ia = I1 = 20 A
2. Calculation of back EMF:
Eb = V - IaRa
Eb = 400 - 20 × 0.06
Eb = 398 V
3. Calculation of flux:
Φ ∝ Eb
Φ ∝ N
Therefore, Eb1/ Eb2 = N1/ N2
As the excitation is increased by 30%, Eb2 = 1.3 Eb1
1.3 Eb1/ Eb1 = N1/ N2
N2 = 1.3 × N1
N2 = 1.3 × 1100
N2 = 1430 rpm (approx)
The correct option is (c) 837.
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A DC series motor has an armature resistance of 0.06Ω and a series field resistance of 0.08Ω. The motor is connected to a 400V supply. Thus line current is 20A when the speed of the machines is 1100 rpm. When the line current is 50 A, and the excitation is increased by 30%, the speed of the machine in rpm is-a)1100b)1003c)837d)938Correct answer is option 'C'. Can you explain this answer?
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A DC series motor has an armature resistance of 0.06Ω and a series field resistance of 0.08Ω. The motor is connected to a 400V supply. Thus line current is 20A when the speed of the machines is 1100 rpm. When the line current is 50 A, and the excitation is increased by 30%, the speed of the machine in rpm is-a)1100b)1003c)837d)938Correct answer is option 'C'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A DC series motor has an armature resistance of 0.06Ω and a series field resistance of 0.08Ω. The motor is connected to a 400V supply. Thus line current is 20A when the speed of the machines is 1100 rpm. When the line current is 50 A, and the excitation is increased by 30%, the speed of the machine in rpm is-a)1100b)1003c)837d)938Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A DC series motor has an armature resistance of 0.06Ω and a series field resistance of 0.08Ω. The motor is connected to a 400V supply. Thus line current is 20A when the speed of the machines is 1100 rpm. When the line current is 50 A, and the excitation is increased by 30%, the speed of the machine in rpm is-a)1100b)1003c)837d)938Correct answer is option 'C'. Can you explain this answer?.
A DC series motor has an armature resistance of 0.06Ω and a series field resistance of 0.08Ω. The motor is connected to a 400V supply. Thus line current is 20A when the speed of the machines is 1100 rpm. When the line current is 50 A, and the excitation is increased by 30%, the speed of the machine in rpm is-a)1100b)1003c)837d)938Correct answer is option 'C'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A DC series motor has an armature resistance of 0.06Ω and a series field resistance of 0.08Ω. The motor is connected to a 400V supply. Thus line current is 20A when the speed of the machines is 1100 rpm. When the line current is 50 A, and the excitation is increased by 30%, the speed of the machine in rpm is-a)1100b)1003c)837d)938Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A DC series motor has an armature resistance of 0.06Ω and a series field resistance of 0.08Ω. The motor is connected to a 400V supply. Thus line current is 20A when the speed of the machines is 1100 rpm. When the line current is 50 A, and the excitation is increased by 30%, the speed of the machine in rpm is-a)1100b)1003c)837d)938Correct answer is option 'C'. Can you explain this answer?.
Solutions for A DC series motor has an armature resistance of 0.06Ω and a series field resistance of 0.08Ω. The motor is connected to a 400V supply. Thus line current is 20A when the speed of the machines is 1100 rpm. When the line current is 50 A, and the excitation is increased by 30%, the speed of the machine in rpm is-a)1100b)1003c)837d)938Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for SSC.
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A DC series motor has an armature resistance of 0.06Ω and a series field resistance of 0.08Ω. The motor is connected to a 400V supply. Thus line current is 20A when the speed of the machines is 1100 rpm. When the line current is 50 A, and the excitation is increased by 30%, the speed of the machine in rpm is-a)1100b)1003c)837d)938Correct answer is option 'C'. Can you explain this answer?, a detailed solution for A DC series motor has an armature resistance of 0.06Ω and a series field resistance of 0.08Ω. The motor is connected to a 400V supply. Thus line current is 20A when the speed of the machines is 1100 rpm. When the line current is 50 A, and the excitation is increased by 30%, the speed of the machine in rpm is-a)1100b)1003c)837d)938Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of A DC series motor has an armature resistance of 0.06Ω and a series field resistance of 0.08Ω. The motor is connected to a 400V supply. Thus line current is 20A when the speed of the machines is 1100 rpm. When the line current is 50 A, and the excitation is increased by 30%, the speed of the machine in rpm is-a)1100b)1003c)837d)938Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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