When a celll is connected to an external resistance `R_(1)`, current through it is `i_(1)`. When the same cell is connected to an external resistance `R_(2)`, current through it is `i_(2)`. Calculate the emf and internal resisance of the cell.

Explanation

Using KVL in closed loop ABCDA
`epsilon= iR + ir`
when `R = R_(1), epsilon= i_(1) R_(1) + i_(1) r ` ….(i)
when `R= R_(2) , epsilon_(2) R_(2) + i_(2) r ` …..(ii)
From (i) and (ii) ,
`i_(1) R_(1) + i_(1) r = i_(2) R_(2) + i_(2) r`
`implies r= ( i_(1)R_(1)- i_(2)R_(2))/(i_(2)-i_(1))`
Also, `epsilon-i_(2) R_(2) - i_(2) r ` [from (i)]
`epsilon= i_(2) R_(2) = = i_(2)r ` [from (2)]
`= ( epsilon-i_(1)R_(1))/(epsilon-i_(2)R_(2))= (i_(1))/(i_(2))implies epsilon = (i_(1)i_(2) (R_(1)-R_(2)))/((i_(2)-i_(1)))`