When the modified galvanometer of previous question is connected across te terminal of a battery, it shows a current of 4A. The current drops to 1A When resistance of `1.5Omega` is connected in series withh the modified galvanometer. Find the emf and the internal resistance of the battery.

Explanation

`4=(epsilon)/(0.01+r)` and `I=(epsilon)/(0.01+r+1.5)`
`therefore(4)/(1)=(r+1.51)/(r+0.01)` or `4r+0.04=r+1.51`
or `3r=1.47` or `r=0.49Omega`
and `epsilon=4xx(0.01xx0.49)=4xx0.5=2` volts