Grouping of cells
Series:
Eeq = ∑E1
And req = ∑ri; \(i=\frac{\sum{{{E}_{i}}}}{R+\sum{{{r}_{i}}}}\).
If S number of identical cells each of marking (E, r) are connected in series across an external resistor R.
Eeq = SE
And req = Sr ⇒ \(I=\frac{SE}{R+Sr}\).
Out of these S cells if x number of cells are reversely connected then Eeq = (S – 2x)E, \({{r}_{eq}}=Sr\Rightarrow I=\frac{(S-2x)E}{Sr+R}\).
Parallel: If P number of identical cells each of marking (E, r) are connected in parallel across an external resistor R.
\({{E}_{eq}}=E\,{{r}_{eq}}=\frac{r}{P}\Rightarrow I=\frac{E}{\frac{r}{P}+R}\).
For un-identical cells in parallel
\({{E}_{eq}}=\left( \sum{\frac{{{E}_{i}}}{{{r}_{i}}}} \right){{\left( \sum{\frac{1}{{{r}_{i}}}} \right)}^{-1}}\).
And
\({{r}_{eq}}={{\left( \sum{\frac{1}{{{r}_{i}}}} \right)}^{-1}}\Rightarrow I=\frac{\left( \sum{\frac{{{E}_{i}}}{{{r}_{i}}}} \right){{\left( \sum{\frac{1}{{{r}_{i}}}} \right)}^{-1}}}{{{\left( \sum{\frac{1}{{{r}_{i}}}} \right)}^{-1}}+R}\).
Mixed Grouping: Let S number of identical cells having marking (E, r) are connected in series in a row and there areP such rows are connected in parallel across an external resistor R. Then
Eeq = SE
\({{r}_{eq}}=\frac{Sr}{P}\Rightarrow I=\frac{SE}{\frac{Sr}{P}+R}=\frac{E}{\frac{r}{P}+\frac{R}{S}}\).
This current I is maximum when \(\frac{S}{P}=\frac{R}{r}\).
So, \({{I}_{\max }}=\frac{PE}{2r}\) or \(\frac{SE}{2R}\).