For a cell, the terminal potential difference is `2.2 V`, when circuit is open and reduces to `1.8 V `. When cell is connected to a resistance `R=5Omega`, the internal resistance of cell `(R)` is

Options

A. `(10)/(9) Omega`
B. `(9)/(10) Omega`
C. `(11)/(9) Omega`
D. `(5)/(9) Omega`

Answer

A

Explanation

In an open circuit, emd of cell
`E = 2.2 V`
In a closed circuit terminal potential difference
`V =1.8 V`
External resistance, `R = 5 Omega`
Thus, internal resistance of cell is
`r = ((E)/(V) -1) R = ((2.2)/(1.8)-1)5`
`=((11)/(9)-1)5=(2)/(9)xx5 =(10)/(9)Omega`.