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

Answer

B

Explanation

As emfs `E_(1) and E_(2)` are in opposition and `E_(1) gt E_(2) ,` so equivalent emf is
`E_(eq) = E_(1) - E_(2)`
The equilvalent resistance of the circuit is `R_(eq) = r_(1) + r_(2) +R`
The current in the circuit is
`I = (E_(eq))/(R_(eq)) = (E_(1)-E_(2))/(r_(1)+r_(2)+R)`
The terminal potential difference is
` V = IR = (E_(1)-E_(2))/(r_(1)+r_(2)+R) xx R `