The emf of a storage battery is `90 V` before charging and `100 V` after charging. When charging began the current was `10 A`. What is the current at the end of charging if the internal resistance of the storge battery during the whole process of charging may be taken as constant and equal to `2Omega`?

Explanation

The voltage supplied by the charging plant is here constant which is equal to
`V=E_i+i_i.r=(90)+(10)(2)`
`=10V` ltbergt Let `i_f` be the current at the end of charging
Then, `V=E-f+i_fr`
`i_f=(V-E_f)/r`
`=(110-100)/2`
`=5A`