Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > A 500 V, 40 kW, 1000 rpm d.c. shunt motor has...
Start Learning for Free
A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______
- a)1.125 Ω
- b)1.0 Ω
- c)2.9 Ω
- d)3.0 Ω
Correct answer is option 'D'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency ...
Most Upvoted Answer
A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency ...
Given data:
Voltage, V = 500 V
Power, P = 40 kW
Speed, N = 1000 rpm
Efficiency, η = 80%
Armature circuit resistance, Ra = 0.125 ohm
Field current, If = 5 A
To find: Resistance to be inserted in the armature circuit to limit starting current to 1.65 times the full load current.
Solution:
1. Calculation of full load current (Ifl):
Power, P = VIfl
Ifl = P/V = 40000/500 = 80 A
2. Calculation of starting current (Ist):
Ist = 1.65 x Ifl = 1.65 x 80 = 132 A
3. Calculation of armature circuit resistance required to limit starting current:
Let R be the resistance to be inserted in the armature circuit.
Voltage drop across armature circuit, Vra = Ist x R
At starting, the voltage drop across armature and field circuits is equal to the supply voltage, i.e., V = Vra + If x Ra
Therefore, R = (V - If x Ra)/Ist
Substituting the given values, we get:
R = (500 - 5 x 0.125)/132 = 3 ohm
Therefore, the resistance to be inserted in the armature circuit to limit starting current to 1.65 times the full load current is 3 ohm (Option D).
Voltage, V = 500 V
Power, P = 40 kW
Speed, N = 1000 rpm
Efficiency, η = 80%
Armature circuit resistance, Ra = 0.125 ohm
Field current, If = 5 A
To find: Resistance to be inserted in the armature circuit to limit starting current to 1.65 times the full load current.
Solution:
1. Calculation of full load current (Ifl):
Power, P = VIfl
Ifl = P/V = 40000/500 = 80 A
2. Calculation of starting current (Ist):
Ist = 1.65 x Ifl = 1.65 x 80 = 132 A
3. Calculation of armature circuit resistance required to limit starting current:
Let R be the resistance to be inserted in the armature circuit.
Voltage drop across armature circuit, Vra = Ist x R
At starting, the voltage drop across armature and field circuits is equal to the supply voltage, i.e., V = Vra + If x Ra
Therefore, R = (V - If x Ra)/Ist
Substituting the given values, we get:
R = (500 - 5 x 0.125)/132 = 3 ohm
Therefore, the resistance to be inserted in the armature circuit to limit starting current to 1.65 times the full load current is 3 ohm (Option D).
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Similar Electrical Engineering (EE) Doubts
Top Courses for Electrical Engineering (EE)View all
A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer?
Question Description
A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer?.
A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup