Question

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.

Answer 1

In first case, current is
`I_(1)=epsilon/R_(1)+r or epsilon = I_(1)(R_(1)+r)`....(i)
In second case, current, `I_(2)=epsilon/(R_(2)+r)`
or `epsilon = I_(2)(R_(2)+r)`....(ii)
From (i) and (ii), we have
`I_(1)(R_(1) +r)=I_(2)(R_(2) +r)`
or `r (I_(1)- I_(2))=I_(2)R_(2) - I_(1)R_(1) or r =(I_(1)R_(2)-I_(1)R_(1))/(I_(1)-I_(2))`
Putting this value of r in equation (i), we get,
`epsilon= I_(1)[R_(1) +(I_(2)R_(2) - I_(1)R_(1))/(I_(1)-I_(2))]`
`=I_(1)[(I_(1)R_(1)-I_(2)R_(1)+I_(2)R_(2)- I_(1)R_(1))/(I_(1)-I_(2))]`
`=(I_(1)I_(2)(R_(2)-R_(1)))/(I_(1)-I_(2))`