n. cells, each of emf .e. and internal resistance .r. are joined in series to form a row. .m. such rows are connected in parallel to form a battery of N = mn cells. This battery is connected to an external resistance .R..
(i) What is the emf of this battery and how much is its internal resistance ?
Show that current .l. flowing through the external resistance .R. is given by-
`I=(Ne)/(mR+nr)`

Explanation

To solve the problem, we need to analyze the configuration of the cells and derive the expressions for the emf and internal resistance of the battery formed by connecting the cells in series and parallel.

### Step 1: Determine the emf of the battery
When \( n \) cells, each with an emf \( e \), are connected in series, the total emf \( E \) of the series combination is given by:
\[ E = n \cdot e \]
This is because the emf adds up in series.
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