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A 6-pole lap wound de shunt generator supplies 290 amperes to a load and the field current of I the generator is 10 amperes. The current in each! parallel path of the armature winding is 1) 40 amp 2) 45 amp 3) 50 amp 4) 55 amp?
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A 6-pole lap wound de shunt generator supplies 290 amperes to a load a...
Solution:
Given data:
P = 6 poles
IL = 290 A
Ish = 10 A
The total current in the armature conductors, Ia = IL + Ish = 290 + 10 = 300 A.
To find the current in each parallel path, we need to find the number of parallel paths in the armature winding.
Number of parallel paths in a lap wound generator is given by the formula,
Z/2P
where Z is the total number of armature conductors and P is the number of poles.
Let's assume that the generator has 1000 armature conductors, then
Number of parallel paths, A = Z / 2P = 1000 / (2 x 6) = 83.33
Round off to the nearest integer, A = 83
Therefore, the current in each parallel path is given by,
Ia / A = 300 / 83 = 3.614 A
Rounded off to the nearest integer, the current in each parallel path is 4A.
Therefore, the correct answer is option 4) 55 A.
Explanation:
The armature conductors of a lap wound generator are grouped into a number of parallel paths, each consisting of a number of conductors in series. The current in each parallel path is determined by the total armature current divided by the number of parallel paths in the armature winding. In this case, the armature winding has 83 parallel paths, and the total armature current is 300 A. Therefore, the current in each parallel path is 3.614 A, which is rounded off to 4 A.
Given data:
P = 6 poles
IL = 290 A
Ish = 10 A
The total current in the armature conductors, Ia = IL + Ish = 290 + 10 = 300 A.
To find the current in each parallel path, we need to find the number of parallel paths in the armature winding.
Number of parallel paths in a lap wound generator is given by the formula,
Z/2P
where Z is the total number of armature conductors and P is the number of poles.
Let's assume that the generator has 1000 armature conductors, then
Number of parallel paths, A = Z / 2P = 1000 / (2 x 6) = 83.33
Round off to the nearest integer, A = 83
Therefore, the current in each parallel path is given by,
Ia / A = 300 / 83 = 3.614 A
Rounded off to the nearest integer, the current in each parallel path is 4A.
Therefore, the correct answer is option 4) 55 A.
Explanation:
The armature conductors of a lap wound generator are grouped into a number of parallel paths, each consisting of a number of conductors in series. The current in each parallel path is determined by the total armature current divided by the number of parallel paths in the armature winding. In this case, the armature winding has 83 parallel paths, and the total armature current is 300 A. Therefore, the current in each parallel path is 3.614 A, which is rounded off to 4 A.
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A 6-pole lap wound de shunt generator supplies 290 amperes to a load and the field current of I the generator is 10 amperes. The current in each! parallel path of the armature winding is 1) 40 amp 2) 45 amp 3) 50 amp 4) 55 amp?
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A 6-pole lap wound de shunt generator supplies 290 amperes to a load and the field current of I the generator is 10 amperes. The current in each! parallel path of the armature winding is 1) 40 amp 2) 45 amp 3) 50 amp 4) 55 amp? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 6-pole lap wound de shunt generator supplies 290 amperes to a load and the field current of I the generator is 10 amperes. The current in each! parallel path of the armature winding is 1) 40 amp 2) 45 amp 3) 50 amp 4) 55 amp? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 6-pole lap wound de shunt generator supplies 290 amperes to a load and the field current of I the generator is 10 amperes. The current in each! parallel path of the armature winding is 1) 40 amp 2) 45 amp 3) 50 amp 4) 55 amp?.
A 6-pole lap wound de shunt generator supplies 290 amperes to a load and the field current of I the generator is 10 amperes. The current in each! parallel path of the armature winding is 1) 40 amp 2) 45 amp 3) 50 amp 4) 55 amp? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 6-pole lap wound de shunt generator supplies 290 amperes to a load and the field current of I the generator is 10 amperes. The current in each! parallel path of the armature winding is 1) 40 amp 2) 45 amp 3) 50 amp 4) 55 amp? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 6-pole lap wound de shunt generator supplies 290 amperes to a load and the field current of I the generator is 10 amperes. The current in each! parallel path of the armature winding is 1) 40 amp 2) 45 amp 3) 50 amp 4) 55 amp?.
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