Physics - Current Electricity Question with Solution | TestHub

Total number of cells =$\mathrm{n}$
Total resistance =$\mathrm{nr}$
$\mathrm{n}-2$cells provide a forward emf,$2$cells provide a reverse emf so:
Total emf$=\left(\text{n}-2\right)\text{E}-2\text{E}=\text{ne}-4\text{E}=\text{E}\left(\text{n}-4\right)$

Current,$\mathrm{I} =\frac{\mathrm{total} \mathrm{emf}}{\mathrm{total} \mathrm{resistance}}$

$\text{I}=\frac{\text{E}\left(\text{n}-4\right)}{\left(\text{nr}\right)}$

PD across A or B (voltage across internal resisitance)

$\text{V}=\text{Ir}+\text{E}$

$=\left[\text{E}\left(\text{n}-4\right)/\left(\text{nr}\right)\right]\text{r}+\text{E}$

$=\text{E}\left(\text{n}-4\right)\text{/n}+\text{E}$

$=\text{E}\left[\left(\text{n}-4)/\text{n}\right)+1\right]$

$=\text{E}\left[\left(\text{n}-4+\text{n}\right)/\text{n}\right]$

$=\text{E}\left(2\text{n}-4\right)/\text{n}$

$=2\text{E}\left(\text{n}-2\right)/\text{n}$

$=2\text{E}\left(1-2/\text{n}\right)$