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A 6-pole lap wound d.c shunt generator supplies 290 amperes to a load and the field current of the generator is 10 amperes. The current in each parallel path of the armature winding is?
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A 6-pole lap wound d.c shunt generator supplies 290 amperes to a load ...
Solution:
Given data:
Number of poles (p) = 6
Load current (IL) = 290 A
Field current (If) = 10 A
To find: Current in each parallel path of the armature winding
Formula used:
Armature current (Ia) = IL + If
Current per parallel path of armature winding (Ipp) = Ia / (p/2)
Where, p/2 is the number of parallel paths in a lap wound DC generator.
Calculation:
Armature current (Ia) = IL + If
= 290 A + 10 A
= 300 A
Current per parallel path of armature winding (Ipp) = Ia / (p/2)
= 300 A / (6/2)
= 300 A / 3
= 100 A
Therefore, the current in each parallel path of the armature winding is 100 A.
Explanation:
A lap wound DC generator has multiple parallel paths in the armature winding. The number of parallel paths is equal to half the number of poles in the generator. In this case, the generator has 6 poles, so there are 6/2 = 3 parallel paths in the armature winding.
The armature current is the sum of the load current and the field current. In this case, the load current is 290 A and the field current is 10 A, so the armature current is 300 A.
To find the current in each parallel path, we divide the armature current by the number of parallel paths. In this case, the armature current is 300 A and the number of parallel paths is 3, so the current in each parallel path is 100 A.
Therefore, the current in each parallel path of the armature winding is 100 A.
Given data:
Number of poles (p) = 6
Load current (IL) = 290 A
Field current (If) = 10 A
To find: Current in each parallel path of the armature winding
Formula used:
Armature current (Ia) = IL + If
Current per parallel path of armature winding (Ipp) = Ia / (p/2)
Where, p/2 is the number of parallel paths in a lap wound DC generator.
Calculation:
Armature current (Ia) = IL + If
= 290 A + 10 A
= 300 A
Current per parallel path of armature winding (Ipp) = Ia / (p/2)
= 300 A / (6/2)
= 300 A / 3
= 100 A
Therefore, the current in each parallel path of the armature winding is 100 A.
Explanation:
A lap wound DC generator has multiple parallel paths in the armature winding. The number of parallel paths is equal to half the number of poles in the generator. In this case, the generator has 6 poles, so there are 6/2 = 3 parallel paths in the armature winding.
The armature current is the sum of the load current and the field current. In this case, the load current is 290 A and the field current is 10 A, so the armature current is 300 A.
To find the current in each parallel path, we divide the armature current by the number of parallel paths. In this case, the armature current is 300 A and the number of parallel paths is 3, so the current in each parallel path is 100 A.
Therefore, the current in each parallel path of the armature winding is 100 A.
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A 6-pole lap wound d.c shunt generator supplies 290 amperes to a load and the field current of the generator is 10 amperes. The current in each parallel path of the armature winding is?
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A 6-pole lap wound d.c shunt generator supplies 290 amperes to a load and the field current of the generator is 10 amperes. The current in each parallel path of the armature winding is? 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 6-pole lap wound d.c shunt generator supplies 290 amperes to a load and the field current of the generator is 10 amperes. The current in each parallel path of the armature winding is? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 6-pole lap wound d.c shunt generator supplies 290 amperes to a load and the field current of the generator is 10 amperes. The current in each parallel path of the armature winding is?.
A 6-pole lap wound d.c shunt generator supplies 290 amperes to a load and the field current of the generator is 10 amperes. The current in each parallel path of the armature winding is? 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 6-pole lap wound d.c shunt generator supplies 290 amperes to a load and the field current of the generator is 10 amperes. The current in each parallel path of the armature winding is? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 6-pole lap wound d.c shunt generator supplies 290 amperes to a load and the field current of the generator is 10 amperes. The current in each parallel path of the armature winding is?.
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