When a bettery is connected to the resistance of `10Omega` the current in the circuit is 0.12 A the same battery gives `0.07A` current with `20Omega` calculate `e.m.f` and internal resistance of the battery.

Explanation

We know that `E=Ir+IR`
`I_(1)r+I_(1)R_(1)=I_(2)r+I_(2)r+I_(2)R_(2),r=(I_(2)R_(2)-I_(1)R_(1))/(I_(1)-I_(2))`
`r=(0.07xx20-0.12xx10)/(0.12-0.07)=(1.4-1.2)/(0.05)=(0.2)/(0.05)=4Omega`
Internal resistance `r=4Omega`
`e.m.f` `E=Ir+IR`
`0.12xx4+0.12xx10=0.48+1.2`, `E=1.68`volt