A battery of emf `epsilon` and internal resistance r sends currents `I_(1)` and `I_(2)`, when connected to external resistance `R_(1)` and `R_(2)` respectively. Find the emf and internal resistance of the battery.

Options

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

Answer

D

Explanation

`I_(1) = E/(r +R(1)), I_(2) =E/(r +R_(2))`
`rArr I_(1)(r + R_(1)) = I_(2) (r+R_(2))`
`rArr r(I_(1)-I_(2)) =I_(2)R_(2) - R_(1)I_(1)`
`r = (I_(1)R_(1) - I_(2)R_(2))/(I_(2)-I_(1))`