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

Let `epsilon` and r be the emf and internal resistance of a battery respectively.
In the first case,
Current flowing in the circuit,
`I_(1)=(epsilon)/(R_(1)+r)orepsilon=I_(1)(R_(1)+r)" "...(i)`
In the second case,
Current flowing in the circuit, `I_(2)=(epsilon)/(R_(2)+r)`
or `epsilon=I_(2)(R_(2)+r)" "...(ii)`
Equating equations (i) and (ii), we get
`I_(1)(R_(1)+r)=I_(2)(R_(2)+r)`
`I_(1)R_(1)+I_(1)r=I_(2)R_(2)+I_(2)r`
`(I_(2)-I_(1))r=I_(1)R_(1)-I_(2)R_(2)`
`r=(I_(1)R_(1)-I_(2)R_(2))/(I_(2)-I_(1))`