$N$ cells, each of emf $E$ and internal resistance $r$ , are connected in a closed ring so that the positive terminal of each cell is joined to the negaive terminal of the next cell as shown in the adjoining diagram. Any two points of this ring are connected through an external resistance $R$ . Find the current in $R$ .

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Solution

If $n$ be the number of cells connected on one side of resistor $R$ , then the number of cells connected on the other side of $R$ is $(N - n)$ . Current distribution in the equivalent circuit is shown in the adjoining circuit diagram,
Here $n$ cells are shown as equivalent to a single cell of emf $n E$ and internal resistance $n R$ . Similarly, $(n - n)$ cells are shown equivalent to a single cell of emf $(N - n) E$ and internal resistance $(N - n) r$
Applying Kirchoff's second law to closed aboda $x [N - n] r + y R = (N - n) E$
$x = \frac{(N - n) E - y R}{(N - n) r} \Rightarrow x = \frac{E}{r} - (\frac{y}{N - n}) \frac{R}{r} . . . . (1)$
Applying Kirchoff's second law to closed mesh cdafebc
$x (N - n) r + (x - y) n r = (N - n) E + n E$
$x [(N - n) r + n r] = N E + y n r$
$x N r = N E + y n r$
$x = \frac{N E + y n r}{N r}$
$x = \frac{E}{r} + \frac{y n}{N} . . . . . (2)$
Combining (1) and (2) we get
$\frac{E}{r} - (\frac{y}{N - n}) \frac{R}{r} = \frac{E}{r} + \frac{y n}{N}$
$y [\frac{n}{N} + \frac{R}{(N - n) r}] = 0$
$∴$ $y = 0$
i.e., current through resistor $R$ is zero
Alternate
Current in the ring $I = E / r$
$∴ V_{a} + n E - i n r = V_{b}$
$V_{a} - V_{b} = - n E + n r (E / r) = 0$
No current through the resistor $R$ .

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