When a battery 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_(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."`