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**Hint:**Recall the parallel connection and series connection of cells and how each affects the net emf of cells. To calculate current flowing through the external resistance, we will use Ohm’s law. It states that current flowing through a conductor is directly proportional to the potential difference across it. Here the net emf will be the potential difference.

**Complete step by step solution:**

Batteries or cells have a measure of potential or electric power of themselves. This physical quantity is known as electromotive-force or emf of a cell. Let’s first understand what emf is. Electromotive force of a cell is the total work done in moving a unit positive charge once around a closed circuit including the cell.

Mathematically, $E = \dfrac{W}{{{q_0}}}$

Where $E$ is the emf of the cell, $W$ is the total work done.

It can also be defined as the total potential difference between two terminals of a cell. It is also measured in the unit of $Volts$. A battery is a group of cells. We can group cells in two ways. Series and parallel.

A number of cells are said to be connected in series if the same amount of current passes through each cell. For $n$ number of cells with emf $E$ connected in series to an external resistance $R$ and net internal resistance of each cell $nr$, and the net emf for the combination will be \[nE\].

Amount of current is given by $I = \dfrac{{nE}}{{R + nr}}$

In parallel grouping of cells, potential difference across each cell is the same. So we can conclude that no matter how many cells we connect in parallel grouping, the net emf for the combination will be the same as one individual cell.

For the same number of cells $(n)$ having equal emf $E$, when connected in parallel, the net emf would be $E$. This combination of cells is connected to an external resistance $R$. Let us assign $r$ as internal resistance of each individual cell.

In parallel connection of $n$ number of identical cells, the net internal resistance would be $\dfrac{r}{n}$. From this we can establish the formula for current through external resistance in a circuit of parallel grouped cells.

So, $I = \dfrac{E}{{R + \dfrac{r}{n}}}$

We are given the problem that the total number of cells connected in parallel is 2 with emf $E$ each. As the internal resistance of the cells is negligible, then the net internal resistance will also be negligible, that is $\dfrac{r}{n} \approx 0$.

In other terms, we can also write that, $R \gg \dfrac{r}{n}$

In this case the current flowing in the circuit or through the register is given by,

$I = \dfrac{E}{R}$, This is the correct answer.

**Note:**Keep in mind that in series grouping, the current flowing through each cell is the same, in parallel grouping the potential difference across each cell is the same. Internal resistance of a cell simply means that it is the resistance the cell offers to current passing through it. It depends upon some factors like nature, temperature and concentration of the electrolytes used in a cell.

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