In an experiment to measure the internal resistance of a cell, by a potentiometer, it is found that the balance point is at a length of 2 m, when the cell is shunted by a 5 Ω resistance and is at a length of 3 m when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell is then(a) 1.5 Ω(b) 10 Ω(c) 15 Ω(d) 1 Ω
Answers
As, A potentiometer is a device used to measure the internal resistance of a cell .
The internal resistance r of a cell is given by -
r = R (E - V)/V
where, R = an external resistance
E = emf of cell
V = potential difference across cell less than E .
As, E = kl₁
V = kl₂
where, k is the potential gradient along the wire.
This can be modified as -
r = R * (l₁ - l₂)/l₂
Here, the balance point is at a length of 2 m, when the cell is shunted by a 5 Ω resistance .
=> r = 5 * (l₁/2 - 1) ......................(1)
Also, the balance point is at a length of 3 m when the cell is shunted by a 10 Ω resistance .
=> r = 10 * (l₁/3 - 1) ......................(2)
=> 5 * (l₁/2 - 1) = 10 * (l₁/3 - 1)
=> l₁ = 6 m
Thus, r = 5 * (6/2 - 1)
= 10 Ω
Thus, The internal resistance of the cell is 10 Ω .
Thus, option (b) is correct .
Answer:
b) 10Ω
Explanation:
Length of balance point = 2m (Given)
Resistance = 5Ω (Given)
Length of balance point after shunting = 3m (Given)
Resistance after shunting = 10Ω (Given)
Thus,
r = ( i1/i2 -1)R = ( i1-i2/i2)R
r = (i1-2/i2) × 5 -- (1)
r = (i1-3/i2) × 10 -- (2)
On solving 1 and 2 we will get = 10
Thus, the internal resistance of the cell is 10 Ω