When the resistance connected in series with a cell is halved, the current is equal to or slightly less ro slighty greater than double. Why?

Explanation

When the resistance R is connected to cell of emf, `epsilon` in series, the current is given by
`I = (epsilon)/(R + r)`
Where r is internal resistance of cell.
when the resistance is halved `((R )/(2))` the current flows
through the circuit is `I^(1) = (epsilon)/((R )/(2) + r)`
If r is negligible comparision with `(R )/(2), I^(1) = (2 epsilon)/(R )`
`:. I^(1) = 2I [:' (epsilon)/(R ) "also equal to" 1]`
(ii) If `r lt lt (R )/(2)`, the current `I^(1)` slightly greater than 2.
(iii) If r ust slightly greater than R, the current `(I^(1))` is slightly less than 2.