A galvanometer, together with an unknown resistance in series, is connected across two identical batteries of each `1.5 V`. When the batteries are connected in series,the galvanometer records a current of `1 A`, and when the batteries are connected in parallel, the current is `0.6 A`. In this case, the internal resistance of the battery is `1//'**' Omega`.
What is the value of '*'?

Explanation

Here, emf of each cell, `epsilon`= 1.5 V
Let r be the internal resistance of each cell and G be the resistance of galvanometer.
The circuits are shows in figure

In circuit (a), effective emf = 1.5 +1.5 = 3.0V
total resistance `= G + R + r +r= G +R +2r`
Current, `I = 1A`
`:.` Current, `I = ("total emf")/("total resistance")`
or `1 = (3.0)/( G + R + 2r)`
or `G + R + 2r = 3` or `G + R = 3 - 2r`
In circuit (b), effective emf = 1.5 V
`:'` When cells are in parallel, only the size of the electrodes increases, the total emf does not change.
Total internal resistance of the two cells will be `R_(p)= (r xx r)/(r + r)= r/2 = 0.5 r`
Total resistance of the circuit `= G +R + 0.5 r`
Here, current, I=0.6A
As, current `= ("effective emf")/("total resistance")`
`:. 0.6 = (1.5)/(G +R + 0.5r)`
or `0.6 (G + R + 0.5 r)=1.5`
or `G + R + 0.5 r= 1.5//0.6= 2.5`
or `G + R = 2.5 - 0.5 r`
From (i) and (ii), `3 - 2r= 2.5 - 0.5r`
or `0.5 = 1.5r or r = 0.5//1.5 = 0.333 Omega`