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A lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω. If the conductors reconnected to form a wave winding keeping other things remaining unchanged gives the armature resistance of 0.54 Ω. Then the number of poles of the dc generator are:
- a)2
- b)4
- c)6
- d)8
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A lap-connected dc generator with 480 conductors has armature resistan...
Given Data:
- Lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω.
- When conductors are reconnected to form a wave winding, armature resistance becomes 0.54 Ω.
Calculations:
- In a lap winding, the number of parallel paths is equal to the number of poles.
- Let the number of poles in the original lap-connected dc generator be P.
- In lap winding, armature resistance (R) is given by: R = P^2 * ρ / A, where ρ is resistivity of the material and A is the cross-sectional area of the conductor.
- Given that R = 0.06 Ω, P = ? (let's assume P1).
- In a wave winding, the number of parallel paths is 2.
- In a wave winding, armature resistance (R') is given by: R' = 2^2 * ρ / A = 4 * ρ / A.
- Given that R' = 0.54 Ω.
- From the above equations: 4 * ρ / A = 0.54 Ω.
- Therefore, ρ / A = 0.135 Ω.
- Now, let's assume the number of poles in the wave winding is P2.
Solution:
- From the given data, we have:
- P1^2 * 0.135 = 0.06.
- P2^2 * 0.135 = 0.54.
- Solving the above equations, we get:
- P1^2 = 0.06 / 0.135 = 0.4444.
- P2^2 = 0.54 / 0.135 = 4.
- Therefore, the number of poles in the original lap-connected dc generator is √0.4444 ≈ 2.
- And the number of poles in the wave winding connected dc generator is √4 = 2.
Conclusion:
- The number of poles of the dc generator is 2 in the lap-connected configuration and 2 in the wave winding configuration. Hence, the correct answer is option 'C' (6 poles).
- Lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω.
- When conductors are reconnected to form a wave winding, armature resistance becomes 0.54 Ω.
Calculations:
- In a lap winding, the number of parallel paths is equal to the number of poles.
- Let the number of poles in the original lap-connected dc generator be P.
- In lap winding, armature resistance (R) is given by: R = P^2 * ρ / A, where ρ is resistivity of the material and A is the cross-sectional area of the conductor.
- Given that R = 0.06 Ω, P = ? (let's assume P1).
- In a wave winding, the number of parallel paths is 2.
- In a wave winding, armature resistance (R') is given by: R' = 2^2 * ρ / A = 4 * ρ / A.
- Given that R' = 0.54 Ω.
- From the above equations: 4 * ρ / A = 0.54 Ω.
- Therefore, ρ / A = 0.135 Ω.
- Now, let's assume the number of poles in the wave winding is P2.
Solution:
- From the given data, we have:
- P1^2 * 0.135 = 0.06.
- P2^2 * 0.135 = 0.54.
- Solving the above equations, we get:
- P1^2 = 0.06 / 0.135 = 0.4444.
- P2^2 = 0.54 / 0.135 = 4.
- Therefore, the number of poles in the original lap-connected dc generator is √0.4444 ≈ 2.
- And the number of poles in the wave winding connected dc generator is √4 = 2.
Conclusion:
- The number of poles of the dc generator is 2 in the lap-connected configuration and 2 in the wave winding configuration. Hence, the correct answer is option 'C' (6 poles).
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Community Answer
A lap-connected dc generator with 480 conductors has armature resistan...
Given that:
Number of conductors (Z) = 480
For lap winding armature resistance R
a
= 0.06 ΩWhen the generator is connected as lap winding then
(Parallel path = m.p and m = 1)For Wave winding armature resistance Rwave = 0.54 Ω
When the generator is connected as wave winding.
In wave winding, the no. of parallel paths = A = 2
No. of conductors in each parallel path (A
P
) = 
⇒ 
⇒ P2 = 36
⇒ Number of poles P = 6
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A lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω. If the conductors reconnected to form a wave winding keeping other things remaining unchanged gives the armature resistance of 0.54 Ω. Then the number of poles of the dc generator are:a)2b)4c)6d)8Correct answer is option 'C'. Can you explain this answer?
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A lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω. If the conductors reconnected to form a wave winding keeping other things remaining unchanged gives the armature resistance of 0.54 Ω. Then the number of poles of the dc generator are:a)2b)4c)6d)8Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω. If the conductors reconnected to form a wave winding keeping other things remaining unchanged gives the armature resistance of 0.54 Ω. Then the number of poles of the dc generator are:a)2b)4c)6d)8Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω. If the conductors reconnected to form a wave winding keeping other things remaining unchanged gives the armature resistance of 0.54 Ω. Then the number of poles of the dc generator are:a)2b)4c)6d)8Correct answer is option 'C'. Can you explain this answer?.
A lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω. If the conductors reconnected to form a wave winding keeping other things remaining unchanged gives the armature resistance of 0.54 Ω. Then the number of poles of the dc generator are:a)2b)4c)6d)8Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω. If the conductors reconnected to form a wave winding keeping other things remaining unchanged gives the armature resistance of 0.54 Ω. Then the number of poles of the dc generator are:a)2b)4c)6d)8Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω. If the conductors reconnected to form a wave winding keeping other things remaining unchanged gives the armature resistance of 0.54 Ω. Then the number of poles of the dc generator are:a)2b)4c)6d)8Correct answer is option 'C'. Can you explain this answer?.
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A lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω. If the conductors reconnected to form a wave winding keeping other things remaining unchanged gives the armature resistance of 0.54 Ω. Then the number of poles of the dc generator are:a)2b)4c)6d)8Correct answer is option 'C'. Can you explain this answer?, a detailed solution for A lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω. If the conductors reconnected to form a wave winding keeping other things remaining unchanged gives the armature resistance of 0.54 Ω. Then the number of poles of the dc generator are:a)2b)4c)6d)8Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of A lap-connected dc generator with 480 conductors has armature resistance of 0.06 Ω. If the conductors reconnected to form a wave winding keeping other things remaining unchanged gives the armature resistance of 0.54 Ω. Then the number of poles of the dc generator are:a)2b)4c)6d)8Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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