Two cells, having the same emf, are connected in series through an external resistance `R`. Cells have internal resistance `r_(1)` and `r_(2) (r_(1) gt r_(2))` respectively. When the circuit is closed, the potentail difference across the first cell is zero the value of `R` is

Options

A. `R_(1)-R_(2)`
B. `(R_(1)+R_(2))/2`
C. `(R_(1)-R_(2))/2`
D. `R_(1)+R_(2)`

Answer

A

Explanation

According to question
`E-Ir_(1)=0&I=(E+E)/(r_(1)+r_(2)+R)` `therefore (E)/(r_(1))=(2E)/(r_(1)+r_(2)+R)`
`r_(1)+r_(2)=R=2r_(1) implies R=r_(1)-r_
(2)`