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In a potentiometer experiment, the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2Ω, the balancing length becomes 120 cm. The internal resistance of the cell, in ohm, is
Correct answer is '2'. Can you explain this answer?
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In a potentiometer experiment, the balancing with a cell is at length...
Problem Statement: In a potentiometer experiment, the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2Ω, the balancing length becomes 120 cm. Find the internal resistance of the cell, in ohms.
Solution:
The potentiometer experiment is based on the principle of a potential divider. In this experiment, a uniform wire of length L is connected in series with a resistance box. A galvanometer is connected in parallel with the resistance box. The cell whose internal resistance needs to be determined is connected across the ends of the wire.
When the circuit is balanced, the potential difference across the length of the wire is equal to the potential difference across the length of the wire plus the internal resistance of the cell.
Step 1: Identifying the known values
- Balancing length without shunting: L1 = 240 cm
- Balancing length with shunting: L2 = 120 cm
- Shunt resistance: R = 2Ω
Step 2: Understanding the concept
In a potentiometer experiment, the balancing length is inversely proportional to the potential difference across the wire. Hence, we can write:
L1/L2 = V2/V1
where V1 is the potential difference without shunting, and V2 is the potential difference with shunting.
Step 3: Applying the formula
Using the formula for potential difference in a potential divider:
V1 = (R + r) * I
where r is the internal resistance of the cell, and I is the current flowing through the circuit.
Similarly,
V2 = [R / (R + r)] * I
Substituting the values of V1 and V2, we get:
L1/L2 = [(R + r) * I] / [R * I]
Simplifying the equation, we find:
L1/L2 = (R + r) / R
Step 4: Solving for r
Given L1 = 240 cm, L2 = 120 cm, and R = 2Ω, we can substitute these values into the equation:
240/120 = (2 + r) / 2
Simplifying further, we get:
2 = 2 + r
r = 2Ω
Therefore, the internal resistance of the cell is 2Ω.
Solution:
The potentiometer experiment is based on the principle of a potential divider. In this experiment, a uniform wire of length L is connected in series with a resistance box. A galvanometer is connected in parallel with the resistance box. The cell whose internal resistance needs to be determined is connected across the ends of the wire.
When the circuit is balanced, the potential difference across the length of the wire is equal to the potential difference across the length of the wire plus the internal resistance of the cell.
Step 1: Identifying the known values
- Balancing length without shunting: L1 = 240 cm
- Balancing length with shunting: L2 = 120 cm
- Shunt resistance: R = 2Ω
Step 2: Understanding the concept
In a potentiometer experiment, the balancing length is inversely proportional to the potential difference across the wire. Hence, we can write:
L1/L2 = V2/V1
where V1 is the potential difference without shunting, and V2 is the potential difference with shunting.
Step 3: Applying the formula
Using the formula for potential difference in a potential divider:
V1 = (R + r) * I
where r is the internal resistance of the cell, and I is the current flowing through the circuit.
Similarly,
V2 = [R / (R + r)] * I
Substituting the values of V1 and V2, we get:
L1/L2 = [(R + r) * I] / [R * I]
Simplifying the equation, we find:
L1/L2 = (R + r) / R
Step 4: Solving for r
Given L1 = 240 cm, L2 = 120 cm, and R = 2Ω, we can substitute these values into the equation:
240/120 = (2 + r) / 2
Simplifying further, we get:
2 = 2 + r
r = 2Ω
Therefore, the internal resistance of the cell is 2Ω.
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In a potentiometer experiment, the balancing with a cell is at length...
The internal resistance of a cell is given by
Hence:
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In a potentiometer experiment, the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2Ω, the balancing length becomes 120 cm. The internal resistance of the cell, in ohm, isCorrect answer is '2'. Can you explain this answer?
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