Infinite number of cells having emf and inter | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 =

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:

Similar Questions

One electric cell (having emf of $2 \ V$ and internal resistance of $0.1 \ \Omega$) and another electric cell (having emf of $4 \ V$ and internal resistance of $0.2 \ \Omega$) are connected in parallel to each other. The equivalent emf will be . . . . . . $V$.

If six identical cells each having an $e.m.f.$ of $6\,V$ are connected in parallel,the $e.m.f.$ of the combination is ................ $V$.

The variation of terminal potential difference $(V)$ with current $(I)$ flowing through a cell is as shown in the graph. The $EMF$ $(E)$ and internal resistance $(r)$ of the cell are:

Two identical cells,when connected either in parallel or in series,give the same current in an external resistance of $5\,\Omega$. The internal resistance of each cell will be $.............\,\Omega$.

The terminal potential difference of a cell when short-circuited is ($E$ = $E.M.F.$ of the cell).

Vedclass Test Series

Exam Paper Generator

Online Exam Module