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If the speed of rotation of a separately excited dc machine is halved and the filed strength is also reduced to half of the original. Then the emf induced in the armature
- a)reduces to 1/4th of original
- b)reduces to half of original
- c)remains same
- d)doubles
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
If the speed of rotation of a separately excited dc machine is halved ...
Emf is proportional to speed of rotation and flux.
Most Upvoted Answer
If the speed of rotation of a separately excited dc machine is halved ...
Explanation:
To understand why the emf induced in the armature of a separately excited DC machine reduces to 1/4th of the original value when the speed of rotation is halved and the field strength is reduced to half, let's break down the process step by step.
1. Emf Induced in the Armature:
The emf induced in the armature of a DC machine is given by the equation:
E = kΦNZ/60A
Where:
- E is the induced emf
- k is a constant
- Φ is the flux per pole
- N is the speed of rotation in RPM
- Z is the total number of armature conductors
- A is the number of parallel paths
2. Effect of Halving the Speed of Rotation:
When the speed of rotation is halved, the value of N in the equation is reduced to half. Therefore, the induced emf is also reduced to half of its original value.
3. Effect of Reducing the Field Strength to Half:
The field strength is determined by the flux per pole (Φ). When the field strength is reduced to half, the value of Φ is also reduced to half.
4. Combined Effect:
Now, let's consider the combined effect of halving the speed of rotation and reducing the field strength to half.
- The induced emf is reduced to half due to the halving of the speed of rotation.
- Additionally, the induced emf is reduced to half again due to the reduction in field strength.
Therefore, the combined effect is such that the induced emf reduces to 1/4th (half * half) of its original value.
Conclusion:
Hence, the correct answer is option 'A' - the emf induced in the armature reduces to 1/4th of the original when the speed of rotation is halved and the field strength is reduced to half.
To understand why the emf induced in the armature of a separately excited DC machine reduces to 1/4th of the original value when the speed of rotation is halved and the field strength is reduced to half, let's break down the process step by step.
1. Emf Induced in the Armature:
The emf induced in the armature of a DC machine is given by the equation:
E = kΦNZ/60A
Where:
- E is the induced emf
- k is a constant
- Φ is the flux per pole
- N is the speed of rotation in RPM
- Z is the total number of armature conductors
- A is the number of parallel paths
2. Effect of Halving the Speed of Rotation:
When the speed of rotation is halved, the value of N in the equation is reduced to half. Therefore, the induced emf is also reduced to half of its original value.
3. Effect of Reducing the Field Strength to Half:
The field strength is determined by the flux per pole (Φ). When the field strength is reduced to half, the value of Φ is also reduced to half.
4. Combined Effect:
Now, let's consider the combined effect of halving the speed of rotation and reducing the field strength to half.
- The induced emf is reduced to half due to the halving of the speed of rotation.
- Additionally, the induced emf is reduced to half again due to the reduction in field strength.
Therefore, the combined effect is such that the induced emf reduces to 1/4th (half * half) of its original value.
Conclusion:
Hence, the correct answer is option 'A' - the emf induced in the armature reduces to 1/4th of the original when the speed of rotation is halved and the field strength is reduced to half.
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Question Description
If the speed of rotation of a separately excited dc machine is halved and the filed strength is also reduced to half of the original. Then the emf induced in the armaturea)reduces to 1/4th of originalb)reduces to half of originalc)remains samed)doublesCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2026 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about If the speed of rotation of a separately excited dc machine is halved and the filed strength is also reduced to half of the original. Then the emf induced in the armaturea)reduces to 1/4th of originalb)reduces to half of originalc)remains samed)doublesCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the speed of rotation of a separately excited dc machine is halved and the filed strength is also reduced to half of the original. Then the emf induced in the armaturea)reduces to 1/4th of originalb)reduces to half of originalc)remains samed)doublesCorrect answer is option 'A'. Can you explain this answer?.
If the speed of rotation of a separately excited dc machine is halved and the filed strength is also reduced to half of the original. Then the emf induced in the armaturea)reduces to 1/4th of originalb)reduces to half of originalc)remains samed)doublesCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2026 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about If the speed of rotation of a separately excited dc machine is halved and the filed strength is also reduced to half of the original. Then the emf induced in the armaturea)reduces to 1/4th of originalb)reduces to half of originalc)remains samed)doublesCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the speed of rotation of a separately excited dc machine is halved and the filed strength is also reduced to half of the original. Then the emf induced in the armaturea)reduces to 1/4th of originalb)reduces to half of originalc)remains samed)doublesCorrect answer is option 'A'. Can you explain this answer?.
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