Describe the formula for the equivalent EMF and internal resistance for the parallel combination of two cells with EMF E₁ and E₂ and internal resistances r₁ and r₂respectively.
What is the corresponding formula for the series combination? Two cells of EMF 1 V, 2 V and internal resistances 2 Ω and 1 Ω respectively are connected in (i) series (ii) parallel. What should be the external resistance in the circuit so that the current through the resistance be the same in the two cases? In which case more heat is generated in the cells?

Let,
In parallel combination, the combined emf is E_eq and,
Combined internal resistance be r_eq . 

$\therefore$                   ${\mathrm{r}}_{\mathrm{eq}} = \frac{{\mathrm{r}}_1{\mathrm{r}}_2}{{\mathrm{r}}_1+{\mathrm{r}}_2}$

and                 ${\mathrm{E}}_{\mathrm{eq}} = \frac{{\mathrm{E}}_1{\mathrm{r}}_2+{\mathrm{E}}_2{\mathrm{r}}_1}{{\mathrm{r}}_1+{\mathrm{r}}_2}$

In series combination,

Let E_eq and r_eq respectively be the equivalent emf and resistance in series combination,
then,
               E_eq = E_{1 +} E₂         and,    

               r_eq = r₁ + r₂

Numerical: 

Given, 

Emf across cell 1, E₁ = 1V
Emf across cell 2, E₂ = 2 V
Internal resistance of cell 1, r₁ = 2 Ω
Internal resistance of cell 2, r₂ = 1Ω 

Let the external resistance be R 

$\therefore$   In series combination, 

                   $$\begin{gathered} {\mathrm{E}}_{\mathrm{S}} = 1+ 2 = 3 \mathrm{V} \\ {\mathrm{R}}_{\mathrm{S}} = \mathrm{R} + 2 + 1 = \mathrm{R} + 3 \mathrm{\Omega } \end{gathered}$$   
            $\therefore {\mathrm{I}}_{\mathrm{s}} = \frac{3}{\mathrm{R}+3}$

Now, in parallel combination 

                  $$\begin{gathered} {\mathrm{E}}_{\mathrm{P}} = 2 - 1 = 1 \mathrm{V} \\ {\mathrm{R}}_{\mathrm{P}} = \mathrm{R} + \frac{{\mathrm{r}}_1{\mathrm{r}}_2}{{\mathrm{r}}_1+{\mathrm{r}}_2} \\ {\mathrm{R}}_{\mathrm{P}} = \mathrm{R}+\frac{2\times 1}{2+1} \\ {\mathrm{R}}_{\mathrm{P}} = \mathrm{R}+\frac{2}{3} = \frac{3\mathrm{R}+2}{3} \end{gathered}$$ 

$\therefore$            ${\mathrm{I}}_{\mathrm{P}} = \frac{1}{{\displaystyle \frac{3\mathrm{R}+2}{3}}} = \frac{3}{3\mathrm{R}+2}$ 
Now since, I_P = I_S we have,

             $\frac{3}{\mathrm{R}+3} = \frac{3}{3\mathrm{R}+2}$

$\therefore$             $3\mathrm{R}+2 = \mathrm{R}+3$

             $$\begin{gathered} \Rightarrow 3\mathrm{R}-\mathrm{R} = 3-2 \\ \Rightarrow 2\mathrm{R} = 1 \\ \Rightarrow \mathrm{R} = \frac{1}{2}\mathrm{\Omega } \end{gathered}$$  

$\therefore$ In series combination, 
          ${\mathrm{E}}_{\mathrm{S}} = 3\mathrm{V} \mathrm{and} {\mathrm{R}}_{\mathrm{S}} = \frac{1}{2}+3 = 3.5$ 

          $\mathrm{Heat} \mathrm{generated} = \frac{{\mathrm{E}}_{\mathrm{S}}^2}{{\mathrm{R}}_{\mathrm{S}}} = \frac{9}{3.5}$
and In parallel combination,

  ${\mathrm{E}}_{\mathrm{P}} = 1 \mathrm{V}, {\mathrm{R}}_{\mathrm{P}} = \frac{3\times 0.5+2}{3} = \frac{3.5}{3}$ 
 
   $$\begin{gathered} \mathrm{Heat} \mathrm{generated} = \frac{{\mathrm{E}}_{\mathrm{P}}^2}{{\mathrm{R}}_{\mathrm{P}}} \\ = \frac{1}{3.5/3} \\ =\frac{3}{3.5} \end{gathered}$$ 


Therefore, heat generated in the series combination will be more than the heat generated in parallel combination.