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:-

Answer

`Oc = (2pi)/(3(4 - pi))`

Explanation

`x_(cm) = ((rho2r^(2)) (r )/(2) - (rho(pir^(2))/(2)) ((4r)/(3pi)))/((rho2r^(2)) - (rho(pir^(2))/(2)))= (2r)/(3(4 - pi))`