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3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell is
- a)2.5Ω
- b)2.25Ω
- c)1.12Ω
- d)3.2Ω
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
3V poteniometer used for the determination of internal resistance of a...
Internal resistance of cell
= 3.05 V
l2 = 68.3cm or
= 1.12Ω
= 3.05 V
l2 = 68.3cm or
= 1.12Ω
Most Upvoted Answer
3V poteniometer used for the determination of internal resistance of a...
Ohm is connected in series with the cell, the balanced point is found to be 64.5cm. Determine the internal resistance of the cell.
Solution:
Let the internal resistance of the cell be r ohm.
When the cell is in open circuit, the potential difference across the potentiometer wire is equal to the emf of the cell, i.e., 2.4V. At the balanced point, the potential difference across the potentiometer wire is equal to the potential difference across the cell, which is 2.4V.
Therefore, using the potential divider rule,
2.4 = (75.8 / 100) × 3
Solving for 3, we get:
3 = (2.4 × 100) / 75.8
3 = 3.1744V
Now, when a resistor of 10.2 ohm is connected in series with the cell, the total resistance of the circuit becomes (r + 10.2) ohm. The potential difference across the potentiometer wire is still 3.1744V. At the balanced point, the potential difference across the resistor is equal to the potential difference across the potentiometer wire.
Therefore, using Ohm's law,
V = IR
where V is the potential difference across the resistor, I is the current flowing through the circuit, and R is the resistance of the resistor.
We can write:
3.1744 = I × 10.2
Solving for I, we get:
I = 0.3113A
Now, the potential difference across the cell is given by:
Vcell = IRcell
where Rcell is the internal resistance of the cell.
We can write:
2.4 = I × Rcell
Substituting the value of I, we get:
2.4 = 0.3113 × Rcell
Solving for Rcell, we get:
Rcell = 7.712 ohm
Therefore, the internal resistance of the cell is 7.712 ohm.
Solution:
Let the internal resistance of the cell be r ohm.
When the cell is in open circuit, the potential difference across the potentiometer wire is equal to the emf of the cell, i.e., 2.4V. At the balanced point, the potential difference across the potentiometer wire is equal to the potential difference across the cell, which is 2.4V.
Therefore, using the potential divider rule,
2.4 = (75.8 / 100) × 3
Solving for 3, we get:
3 = (2.4 × 100) / 75.8
3 = 3.1744V
Now, when a resistor of 10.2 ohm is connected in series with the cell, the total resistance of the circuit becomes (r + 10.2) ohm. The potential difference across the potentiometer wire is still 3.1744V. At the balanced point, the potential difference across the resistor is equal to the potential difference across the potentiometer wire.
Therefore, using Ohm's law,
V = IR
where V is the potential difference across the resistor, I is the current flowing through the circuit, and R is the resistance of the resistor.
We can write:
3.1744 = I × 10.2
Solving for I, we get:
I = 0.3113A
Now, the potential difference across the cell is given by:
Vcell = IRcell
where Rcell is the internal resistance of the cell.
We can write:
2.4 = I × Rcell
Substituting the value of I, we get:
2.4 = 0.3113 × Rcell
Solving for Rcell, we get:
Rcell = 7.712 ohm
Therefore, the internal resistance of the cell is 7.712 ohm.
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3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell isa)2.5Ωb)2.25Ωc)1.12Ωd)3.2ΩCorrect answer is option 'C'. Can you explain this answer?
Question Description
3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell isa)2.5Ωb)2.25Ωc)1.12Ωd)3.2ΩCorrect answer is option 'C'. 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 3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell isa)2.5Ωb)2.25Ωc)1.12Ωd)3.2ΩCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell isa)2.5Ωb)2.25Ωc)1.12Ωd)3.2ΩCorrect answer is option 'C'. Can you explain this answer?.
3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell isa)2.5Ωb)2.25Ωc)1.12Ωd)3.2ΩCorrect answer is option 'C'. 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 3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell isa)2.5Ωb)2.25Ωc)1.12Ωd)3.2ΩCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell isa)2.5Ωb)2.25Ωc)1.12Ωd)3.2ΩCorrect answer is option 'C'. Can you explain this answer?.
Solutions for 3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell isa)2.5Ωb)2.25Ωc)1.12Ωd)3.2ΩCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell isa)2.5Ωb)2.25Ωc)1.12Ωd)3.2ΩCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for 3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell isa)2.5Ωb)2.25Ωc)1.12Ωd)3.2ΩCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of 3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell isa)2.5Ωb)2.25Ωc)1.12Ωd)3.2ΩCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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