For a cell terminal potential difference is 2.2 V when circuit is open and reduces to 1.8V when cell is connected to a resistance of R=`5Omega` then determine internal resistance of cell is:-

Options

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

Answer

1

Explanation

T.P.D(V)=E-Ir(Remember it)
`V=E-((E)/(R+r))r=(ER)/((R+r))`
from given condition `E=2.2 &` when R=5 then
TPD V=1.8V
therefore `1.8=(2.2xx5)/( 5+r)rArrr=(10)/(9)Omega`