3V poteniometer used for the determination of internal resistance of a...
Internal resistance of cell
= 3.05 V
l
2 = 68.3cm or
= 1.12Ω
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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.
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