Two cells of emf `E_(1)` and `E_(2)` are joined in opposition (such that `E_(1) gt E_(2)`) . If `r_(1)` be the internal resistances and R be the external resistance, then the terminal potential difference is

Options

A. `(E_(1)-E_(2))/(r_(1)+r_(2))xxR`
B. `(E_(1)+E_(2))/(r_(1)+r_(2))xxR`
C. `(E_(1)-E_(2))/(r_(1)+r_(2)+R)xxR`
D. `(E_(1)+E_(2))/(r_(1)+r_(2)+R)xxR`

Answer

C

Explanation

Given, two cells of emf `E_(1)` and `E_(2)` are joined in opposition such that, `E_(1)gt E_(2)`.

Current, `l=(E_(1)-E_(2))/(r_(1)+r_(2)+R)`
Now, terminal potential difference across a circuit,
`V=lR=((E_(1)-E_(2))/(r_(1)+r_(2)+R))R`