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

Options

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

Answer

A

Explanation

Current :I `= (E_(1) - E_(2))/(r_(1) + r_(2) + R)`
`therefore ` Terminal P.DV= `IR= ((E_(1)-E_(2))/(r_(1)+ r-(2) + R))R`