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A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf's is :
- a)3:2
- b)5:1
- c)5:4
- d)3:4
Correct answer is option 'A'. Can you explain this answer?
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A potentiometer wire is 100 cm long and a constant potential differenc...
Given information:
- Length of the potentiometer wire = 100 cm
- Balance point 1 (when cells are connected in series to support each other) = 50 cm from the positive end of the wire
- Balance point 2 (when cells are connected in opposite direction) = 10 cm from the positive end of the wire
Explanation:
The balance point on a potentiometer wire is the point where the potential difference across that point is zero. In this case, we have two balance points, one when the cells are connected in series to support each other, and the other when the cells are connected in opposite directions.
Balance point 1: (cells connected in series)
- Using the concept of potential gradient, we can say that the potential difference across the balance point is directly proportional to the distance of the balance point from the positive end of the wire.
- Let the emf of the first cell be E1 and its internal resistance be r1.
- Let the emf of the second cell be E2 and its internal resistance be r2.
- At the balance point 1 (50 cm from the positive end), the potential difference is zero. So, the potential difference contributed by the first cell is equal to the potential difference contributed by the second cell.
- Using the formula of potential difference, we can write:
- E1 - (r1/L1) * 50 = E2 - (r2/L2) * (100 - 50)
- E1 - (r1/100) * 50 = E2 - (r2/100) * 50
- E1 - (r1/2) = E2 - (r2/2)
- E1 - E2 = r1/2 - r2/2
- E1 - E2 = (r1 - r2)/2
- E1 - E2 = (r1 - r2)/2
Balance point 2: (cells connected in opposite direction)
- Using the same concept of potential gradient, at the balance point 2 (10 cm from the positive end), the potential difference contributed by the first cell is equal in magnitude but opposite in sign to the potential difference contributed by the second cell.
- Using the same formula of potential difference, we can write:
- E1 - (r1/L1) * 10 = -(E2 - (r2/L2) * (100 - 10))
- E1 - (r1/100) * 10 = -(E2 - (r2/100) * 90)
- E1 - (r1/10) = -(E2 - (9r2/10))
- E1 - E2 = -9r2/10 + r1/10
- E1 - E2 = r1/10 - 9r2/10
- E1 - E2 = (r1 - 9r2)/10
- E1 - E2 = (r1 - 9r2)/10
Comparing the two balance points:
- From the equations obtained from balance point 1 and balance point 2, we can equate the expressions for E1 - E2:
- (
- Length of the potentiometer wire = 100 cm
- Balance point 1 (when cells are connected in series to support each other) = 50 cm from the positive end of the wire
- Balance point 2 (when cells are connected in opposite direction) = 10 cm from the positive end of the wire
Explanation:
The balance point on a potentiometer wire is the point where the potential difference across that point is zero. In this case, we have two balance points, one when the cells are connected in series to support each other, and the other when the cells are connected in opposite directions.
Balance point 1: (cells connected in series)
- Using the concept of potential gradient, we can say that the potential difference across the balance point is directly proportional to the distance of the balance point from the positive end of the wire.
- Let the emf of the first cell be E1 and its internal resistance be r1.
- Let the emf of the second cell be E2 and its internal resistance be r2.
- At the balance point 1 (50 cm from the positive end), the potential difference is zero. So, the potential difference contributed by the first cell is equal to the potential difference contributed by the second cell.
- Using the formula of potential difference, we can write:
- E1 - (r1/L1) * 50 = E2 - (r2/L2) * (100 - 50)
- E1 - (r1/100) * 50 = E2 - (r2/100) * 50
- E1 - (r1/2) = E2 - (r2/2)
- E1 - E2 = r1/2 - r2/2
- E1 - E2 = (r1 - r2)/2
- E1 - E2 = (r1 - r2)/2
Balance point 2: (cells connected in opposite direction)
- Using the same concept of potential gradient, at the balance point 2 (10 cm from the positive end), the potential difference contributed by the first cell is equal in magnitude but opposite in sign to the potential difference contributed by the second cell.
- Using the same formula of potential difference, we can write:
- E1 - (r1/L1) * 10 = -(E2 - (r2/L2) * (100 - 10))
- E1 - (r1/100) * 10 = -(E2 - (r2/100) * 90)
- E1 - (r1/10) = -(E2 - (9r2/10))
- E1 - E2 = -9r2/10 + r1/10
- E1 - E2 = r1/10 - 9r2/10
- E1 - E2 = (r1 - 9r2)/10
- E1 - E2 = (r1 - 9r2)/10
Comparing the two balance points:
- From the equations obtained from balance point 1 and balance point 2, we can equate the expressions for E1 - E2:
- (
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A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Twocells are connected in series first to support one another and then in opposite direction. The balancepoints are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio ofemfs is :a)3:2b)5:1c)5:4d)3:4Correct answer is option 'A'. Can you explain this answer?
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A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Twocells are connected in series first to support one another and then in opposite direction. The balancepoints are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio ofemfs is :a)3:2b)5:1c)5:4d)3:4Correct answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Twocells are connected in series first to support one another and then in opposite direction. The balancepoints are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio ofemfs is :a)3:2b)5:1c)5:4d)3:4Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Twocells are connected in series first to support one another and then in opposite direction. The balancepoints are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio ofemfs is :a)3:2b)5:1c)5:4d)3:4Correct answer is option 'A'. Can you explain this answer?.
A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Twocells are connected in series first to support one another and then in opposite direction. The balancepoints are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio ofemfs is :a)3:2b)5:1c)5:4d)3:4Correct answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Twocells are connected in series first to support one another and then in opposite direction. The balancepoints are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio ofemfs is :a)3:2b)5:1c)5:4d)3:4Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Twocells are connected in series first to support one another and then in opposite direction. The balancepoints are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio ofemfs is :a)3:2b)5:1c)5:4d)3:4Correct answer is option 'A'. Can you explain this answer?.
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