Explain the effective internal resistance of cells connected in parallel combination. Compare the results to the external resistance.

Explanation

Cells in parallel: In parallel connection all the positive terminals of the cells are connected to one point and all the negative terminals to a second point. These two points form the positive and negative terminals of the battery.
Let n cells be connected in parallel between the points A and B resistance R is connected between the points A and B. Let `xi` be the emf and r the internal resistance of each cell.
The equivalent internal resistance of the battery is `(1)/(r_(eq))=(1)/(r )+(1)/(r )+....(1)/(r )"(n temrs)"=(n)/(r )`
So, `r_(eq)=(r )/(n)` and the total resistance in the circuit `=R+(r )/(n)`. The total emf is the potential difference between the points A and B, which is equal to `xi`. The current in the circuit is given by

`I=(xi)/((r )/(n)+R)`
`I=(n xi)/(r +n R)" " ....(1)`
Case (a) If `r lt lt R`, then,
`I= (n xi)/(R )=n I_(1)" " ...(2)`
where II is the current due to a single cell and is equal to `(xi)/(r)` when R is negligible. Thus, the current through the external resistance due to the whole battery is n times the current due to a single cell.
Case (b) If `r lt lt R, I=(xi)/(R)" " ....(3)`
The above equation implies that current due to the whole battery is the same as that due to a single cell. Hence it is advantageous to connect cells in parallel when the external resistance is very small compared to the internal resistance of the cells.