A cell supplies a current of 0.9 A through a `1 Omega` resistor and a current of 0.3 A through a `2Omega` resistor. Calculate the internal resistance of the cell.

Explanation

(i) Measurement of internal resistance of a cell by potentiometer: To measure the internal resistance of a cell, the circuit connections are made as shown in figure. The end of the potentiometer wire is connected to the positive terminal of the battery Bt and the negative terminal of the battery is connected to the end D through a key `K_1` This forms the primary circuit.
The positive terminal of the cell `xi`whose internal resistance is to be determined is also connected to the end of the wire. The negative terminal of the cell `xi` is connected to a jockey through a galvanometer and a high resistance. A resistance box R and key `K_2` are connected across the cell `xi` . With `K_2` open, the balancing point J is obtained and the balancing length `CJ=l_1`, is measured. Since the cell is in open circuit, its emf is
`xi prop l_1` ...(1)
A suitable resistance (say, 10 `Omega`) is included in the resistance box and key `K_2` is closed. Let R be the internal resistance of the cell. The current passing through the cell and the resistance R is given by
`I=xi/(R+r)`
The potential difference across R is `V=(xiR)/(R+r)`
When this potential difference is balanced on the potentiometer wire, let `l_2` be the balancing length.
Then `(xiR)/(R+r) prop l_2` ...(2)
From equations (1) and (2)
`(R+r)/R=l_1/l_2` ..(3)
`1+r/R=l_1/l_2,r=R [l_1/l_2-1]`
`therefore r=R ((l_1-l_2)/l_2)` ...(4)
Substituting the values of the R, `l_1` and `l_2` the internal resistance of the cell is determined. The experiment can be repeated for different values of R. It is found that the internal resistance of the cell is not constant but increases with increase of external resistance connected across its terminals.
(ii) Current from the cell 1, `l_1`=0.9 A and Resistor , `R_1 = 1Omega`
Current from the cell 2, `l_2`=0.3 A and Resistor , `R_2=2Omega`
Current in the circuit ,`I_1=xi/(r+R_1)`
`xi=I_1r+I_1R_1` …(1)
Current in the circuit , `I_2=xi/(r+R_2)`
`xi=I_2r+I_2R_2` ....(2)
From equation (1) and (2),
`r=(I_2R_2-I_1R_1)/(I_1-I_2)=((0.3xx2)-(0.9xx1))/(0.9-0.3)=(0.6-0.9)/0.6`
Magnitude of internal resistance `r=0.5 Omega` .