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

Answer

`(3)`

Explanation

Let `R` be the combined resistance of galvanometer and an unknown resistance and `r` the internal resistance of each battery.
When the batteries, each of emf `E` are connected in series, the net emf `= 2E` and net internal resistance `= 2r`.
Current `i_(1) = (2E)/(R + 2r)` or `1.0 = (2 xx 1.5)/(R + 2r)`
or `R + 2r = 3` (i)
When the batteries are connected in parallel, the emf remains `E` and net internal resistance become `r//2`. Therefore, current is
`i_(2) = (E)/(R + (r)/(2)) = (2E)/(2R + r)` or `2R + r = (2E)/(i_(2)) = (2 xx 1.5)/(0.6) = 5.0` (ii)
Solving Eqs.(i) and (ii), we get `r = (1)/(3) Omega`