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

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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 and a resistance R is connected between the points A and B. Let ξ, be the emf and r the internal resistance of each cell.

The equivalent internal resistance of the battery is  \(\frac{1}{r_{eq}}\) = \(\frac{1}{r}\)\(\frac{1}{r}\) + ......... \(\frac{1}{r}\) (n terms) = \(\frac{n}{r}\).

So reg = \(\frac{r}{n}\) and the total resistance in the circuit = R +\(\frac{r}{n}\)

The total emf is the potential difference between the points A and B, which is equal to ξ. The current in the circuit is given by

Case (a) If r << R, then,

I = \(\frac{nξ}{R}\) = nI1 …….. (2)

where II is the current due to a single cell and is equal to \(\frac{ξ}{R}\) 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 << R. I = \(\frac{ξ}{R}\) …… (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.

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