Question
Question: In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in a...
In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in an open circuit is 350cm. When a resistance of 9Ω is used in the external circuit of the cell, the balance point shifts to 300cm. Determine the internal resistance of the cell.
Solution
Use the relation of the potentiometer given below and substitute the formula of the potential difference in the circuit in it. Substitute the known values of the parameters in the obtained relation to find the value of the internal resistance of the circuit.
Useful formula:
(1) In the potentiometer, the following relation exists.
VintE=l2l1
Where E is the emf, Vint is the internal voltage, l1 is the length of the initial null point and l2 is the final null point in the circuit.
(2) The potential difference is given by
V=R+rER
Where V is the potential difference, R is the external resistance and r is the internal resistance of the circuit.
Complete step by step solution:
It is given that the
Length of the initial null point, l1=350cm
Length of the final null point, l2=300cm
The external resistance, R=9Ω
Using the relation (1),
VintE=l2l1
Substituting the formula (2) in the above step.
R+rERE=l2l1
By cancelling the similar terms,
RR+r=l2l1
By cross multiplying both the sides of the equation.
(R+r)l2=Rl1
By grouping the similar terms in one side of the equation.
r=l2(l1−l2)R
Substituting the known values in the above equation.
r=300(350−300)9
By performing the simple arithmetic operation.
r=1.5Ω
Hence the internal resistance of the cell is obtained as 1.5Ω.
Note: Potentiometer is the device with the three variable pins which may be of resistor and the wiper. It also works as the voltage divider or the rheostat. Remember the formula of the potentiometer relation which has the ratio of the initial length to the final length.