n identical cells are joined in series with its two cells A and B in the loop with reversed polarities. EMF of each shell is

Q: $n$ identical cells are joined in series with its two cells $A$ and $B$ in the loop with reversed polarities. $EMF$ of each shell is $E$ and internal resistance $r$. Potential difference across cell $A$ or $B$ is (here $n > 4$)

d The two opposite cells $A$ and $B$ will cancel two more cells, so net emf will be $n-4 .$ So current is $I=\frac{(n-4) \varepsilon}{n r}$ Now $pd$ across $A$ or $B$ is $I \varepsilon+I r$ (as they will be in charging state), brgt $=\varepsilon=\frac{(n-4) \varepsilon}{n}=2 \varepsilon\left(1-\frac{2}{n}\right)=\frac{2 \varepsilon(n-2)}{n}$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

$n$ identical cells are joined in series with its two cells $A$ and $B$ in the loop with reversed polarities. $EMF$ of each shell is $E$ and internal resistance $r$. Potential difference across cell $A$ or $B$ is (here $n > 4$)

A$\frac{{2\,E}}{n}$
B$2E\left( {1 - \frac{1}{n}} \right)$
C$\frac{{4\,E}}{n}$
D$2E\left( {1 - \frac{2}{n}} \right)$

Medium

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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