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If an AC main supply is given to be 220 V then the average emf during a positive half cycle will be
- a)198 V
- b)220 V
- c)240 V
- d)220 √2V
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
If an AC main supply is given to be 220 V then the average emf during...
Ems=220V
∴ 
Average e.m.f over half cycle
= 
= 198 V
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If an AC main supply is given to be 220 V then the average emf during...
Average EMF during a positive half cycle of an AC main supply can be calculated using the formula:
Average EMF = (2/π) * maximum EMF
Here, the maximum EMF is given to be 220 V.
To calculate the average EMF, we substitute this value into the formula:
Average EMF = (2/π) * 220 V
Simplifying this expression:
Average EMF = 2 * (220/π) V
Now, we need to calculate the value of (220/π) to find the average EMF.
The value of π is approximately 3.14.
So, (220/π) ≈ 70.06 V
Substituting this value back into the expression for average EMF:
Average EMF = 2 * 70.06 V
Simplifying further:
Average EMF ≈ 140.12 V
Therefore, the average EMF during a positive half cycle of the AC main supply is approximately 140.12 V.
Since none of the given options match this value, it seems that there might be an error in the options provided.
However, if we consider the value closest to the calculated average EMF, option 'A' which is 198 V, it is clear that this option is incorrect.
Therefore, none of the given options (A, B, C, D) is the correct answer to this question.
Average EMF = (2/π) * maximum EMF
Here, the maximum EMF is given to be 220 V.
To calculate the average EMF, we substitute this value into the formula:
Average EMF = (2/π) * 220 V
Simplifying this expression:
Average EMF = 2 * (220/π) V
Now, we need to calculate the value of (220/π) to find the average EMF.
The value of π is approximately 3.14.
So, (220/π) ≈ 70.06 V
Substituting this value back into the expression for average EMF:
Average EMF = 2 * 70.06 V
Simplifying further:
Average EMF ≈ 140.12 V
Therefore, the average EMF during a positive half cycle of the AC main supply is approximately 140.12 V.
Since none of the given options match this value, it seems that there might be an error in the options provided.
However, if we consider the value closest to the calculated average EMF, option 'A' which is 198 V, it is clear that this option is incorrect.
Therefore, none of the given options (A, B, C, D) is the correct answer to this question.
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If an AC main supply is given to be 220 V then the average emf during a positive half cycle will bea)198 Vb)220 Vc)240 Vd)220 √2VCorrect answer is option 'A'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If an AC main supply is given to be 220 V then the average emf during a positive half cycle will bea)198 Vb)220 Vc)240 Vd)220 √2VCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If an AC main supply is given to be 220 V then the average emf during a positive half cycle will bea)198 Vb)220 Vc)240 Vd)220 √2VCorrect answer is option 'A'. Can you explain this answer?.
If an AC main supply is given to be 220 V then the average emf during a positive half cycle will bea)198 Vb)220 Vc)240 Vd)220 √2VCorrect answer is option 'A'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If an AC main supply is given to be 220 V then the average emf during a positive half cycle will bea)198 Vb)220 Vc)240 Vd)220 √2VCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If an AC main supply is given to be 220 V then the average emf during a positive half cycle will bea)198 Vb)220 Vc)240 Vd)220 √2VCorrect answer is option 'A'. Can you explain this answer?.
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