Infinite number of cells having emf and inter | vedclass

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(D) The total $emf$ of the series combination is the sum of the individual $emf$s: $E_{eq} = E + \frac{E}{n} + \frac{E}{n^2} + \frac{E}{n^3} + \dots =

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$\frac{(n+1)E}{2nr}$

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(D) The total $emf$ of the series combination is the sum of the individual $emf$s: $E_{eq} = E + \frac{E}{n} + \frac{E}{n^2} + \frac{E}{n^3} + \dots = E(1 + \frac{1}{n} + \frac{1}{n^2} + \dots)$. This is a geometric series with sum $S = \frac{a}{1-r_{ratio}} = \frac{E}{1 - 1/n} = \frac{nE}{n-1}$.The total internal resistance is $r_{eq} = r + \frac{r}{n} + \frac{r}{n^2} + \dots = r(1 + \frac{1}{n} + \frac{1}{n^2} + \dots) = \frac{nr}{n-1}$.The current $I$ flowing through the external resistance $R = \frac{nr}{n+1}$ is given by $I = \frac{E_{eq}}{r_{eq} + R}$.Substituting the values: $I = \frac{\frac{nE}{n-1}}{\frac{nr}{n-1} + \frac{nr}{n+1}} = \frac{\frac{nE}{n-1}}{nr(\frac{1}{n-1} + \frac{1}{n+1})} = \frac{\frac{E}{n-1}}{\frac{(n+1) + (n-1)}{(n-1)(n+1)}} = \frac{E}{n-1} 	imes \frac{(n-1)(n+1)}{2n} = \frac{(n+1)E}{2nr}$.

Infinite number of cells having $emf$ and internal resistance $(E, r)$,$(\frac{E}{n}, \frac{r}{n})$,$(\frac{E}{n^2}, \frac{r}{n^2})$,$(\frac{E}{n^3}, \frac{r}{n^3})$... are connected in series across an external resistance of $\frac{nr}{n+1}$. The current flowing through the external resistor is:

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