How cells are combined in parallel? Derive the expression for the current flowing in the external circuit. When is this combination useful?

Hint: When cells are connected in parallel, the current is divided among various cells. In a parallel combination, all the positive ends are connected together and all the negative ends are connected together.

Formula used: In this solution, we’ll be using the following formula:
Ohm’s law: $ V = IR $ where $ V $ is the EMF of the cell, $ I $ is the current in the circuit which has resistance $ R $ .

Complete step by step solution:
When cells are connected in parallel, the EMF of the combination will be equal to the EMF of any one of the cells that are connected in parallel. Let us assume that $ n $ cells of emf $ E $ are connected in parallel that has internal resistance $ r $ .
Since we know that the net EMF of all the cells in parallel is $ E $ , let us now calculate the net internal resistance of all the cells. Since all the cells are combined in parallel, the internal resistances will be in parallel and we can calculate the net resistance as
 $\Rightarrow \dfrac{1}{{{r_{net}}}} = \dfrac{1}{r} + \dfrac{1}{r} + ....{\text{n}}\,{\text{times}} $
 $\Rightarrow {r_{net}} = \dfrac{r}{n} $
Hence the net internal resistance will be $ {r_{net}} $ . Using Ohm’s law $ V = IR $ , we can determine the current flowing in the circuit as
 $\Rightarrow I = \dfrac{V}{R} $
Since the potential/EMFs of the combination of all the cells is $ E $ and we calculated the internal resistance as $ {r_{net}} = \dfrac{r}{n} $ we can calculate the current in the circuit as
 $\Rightarrow I = \dfrac{E}{{r/n}} = \dfrac{{nE}}{r} $
The current in the circuit is hence $ n $ times the current in the circuit if only one cell were present. This is because the internal resistance decreases when the cells are attached in parallel. This is advantageous to us in situations when the internal resistance of the cell is very high and if we want to reduce the internal resistance, we can attach these cells in parallel and hence reduce the internal resistance of the circuit

Note:
If we also have an external resistance in the circuit of resistance $ R $ , it will be connected in series with the internal resistances and the net resistance will be
 $\Rightarrow {r_{net}} = \dfrac{r}{n} + R $ and then the current can be calculated from ohm’s law again as:
 $\Rightarrow I = \dfrac{E}{{r/n + R}} $