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In a potentiometer experiment the balancing length of the cell was found to be 80 cm. Now, a resistance of 8 Ω is connected across the terminal of this cell the balancing length becomes 60 cm. The internal resistance of this cell is
- a)8/3 Ω
- b)4/3 Ω
- c)2 Ω
- d)3 Ω
Correct answer is option 'A'. Can you explain this answer?
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Balancing Length in Potentiometer Experiment:
In a potentiometer experiment, the balancing length is the position along the wire where the potential difference is zero. It is used to determine the unknown emf or internal resistance of a cell.
Given Information:
- Balancing length without any external resistance connected: 80 cm
- Balancing length with a resistance of 8 Ω connected: 60 cm
Determination of Internal Resistance:
We can use the concept of potential drop in a potentiometer wire to determine the internal resistance of the cell.
1. Potential drop across the wire without any external resistance:
The potential drop across the wire is directly proportional to the length of the wire. Therefore, the potential drop across the balancing length (80 cm) is equal to the emf of the cell.
2. Potential drop across the wire with an external resistance of 8 Ω:
When the external resistance (8 Ω) is connected, the potential drop across the wire decreases. The new potential drop is equal to the emf minus the potential drop across the internal resistance.
Using the formula: V = IR
- Potential drop across the wire without any external resistance: V1 = emf
- Potential drop across the wire with an external resistance: V2 = emf - (8 Ω) * I
where I is the current flowing through the circuit.
3. Determination of the internal resistance:
Since the balancing length decreases from 80 cm to 60 cm, the potential drop across the wire decreases. Therefore, we can equate the potential drops in the two cases:
V1 = V2
emf = emf - (8 Ω) * I
Simplifying the equation:
(8 Ω) * I = 0
I = 0
This implies that when the balancing length is 60 cm, no current is flowing through the circuit.
4. Calculation of internal resistance:
Using Ohm's Law: emf = (internal resistance) * I
Since I = 0 in this case, the internal resistance of the cell is equal to the external resistance connected across the terminal, which is 8 Ω.
Therefore, the correct answer is option 'A': 8/3 Ω.
In a potentiometer experiment, the balancing length is the position along the wire where the potential difference is zero. It is used to determine the unknown emf or internal resistance of a cell.
Given Information:
- Balancing length without any external resistance connected: 80 cm
- Balancing length with a resistance of 8 Ω connected: 60 cm
Determination of Internal Resistance:
We can use the concept of potential drop in a potentiometer wire to determine the internal resistance of the cell.
1. Potential drop across the wire without any external resistance:
The potential drop across the wire is directly proportional to the length of the wire. Therefore, the potential drop across the balancing length (80 cm) is equal to the emf of the cell.
2. Potential drop across the wire with an external resistance of 8 Ω:
When the external resistance (8 Ω) is connected, the potential drop across the wire decreases. The new potential drop is equal to the emf minus the potential drop across the internal resistance.
Using the formula: V = IR
- Potential drop across the wire without any external resistance: V1 = emf
- Potential drop across the wire with an external resistance: V2 = emf - (8 Ω) * I
where I is the current flowing through the circuit.
3. Determination of the internal resistance:
Since the balancing length decreases from 80 cm to 60 cm, the potential drop across the wire decreases. Therefore, we can equate the potential drops in the two cases:
V1 = V2
emf = emf - (8 Ω) * I
Simplifying the equation:
(8 Ω) * I = 0
I = 0
This implies that when the balancing length is 60 cm, no current is flowing through the circuit.
4. Calculation of internal resistance:
Using Ohm's Law: emf = (internal resistance) * I
Since I = 0 in this case, the internal resistance of the cell is equal to the external resistance connected across the terminal, which is 8 Ω.
Therefore, the correct answer is option 'A': 8/3 Ω.
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In a potentiometer experiment the balancing length of the cell was found to be 80 cm. Now, a resistance of 8 Ω is connected across the terminal of this cell the balancing length becomes 60 cm. The internal resistance of this cell isa)8/3 Ωb)4/3 Ωc)2 Ωd)3 ΩCorrect answer is option 'A'. Can you explain this answer?
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