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. `10/9Omega`
B. `9/10Omega`
C. `11/9Omega`
D. `5/9Omega`

Answer

A

Explanation

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