CELLS IN SERIES AND PARALLEL :
Derive expressions for equivalent e.m.f. and internal resistance of cells connected in series, parallel and mixed combination.
Cells can be connected in :
(i) series,
(ii) parallel,
(iii) mixed combination.
Cells Connected in Series :
In series combination of the cells, the negative terminal of a cell is connected to the positive terminal of an other cell, and so on.
(a) Identical Cells Connected in Series:
Consider $n$ identical cells such as $C_1$, $C_2$, $\dots$, $C_n$ each of e.m.f. $\varepsilon$ and internal resistance $r$ connected in series to an external resistance $R$.
Since cells are connected in series, so the total e.m.f. of $n$ cells = $n\varepsilon$.
That is,
$\varepsilon_{\text{eff}}= \varepsilon + \varepsilon + \varepsilon + \dots n = n\varepsilon$
$r_{\text{eff}} = r + r + r + \dots \text{upto } n \text{ terms} = nr$
Total resistance = $r_{\text{eff}} + R$
Total resistance = $nr + R$
Hence, eqn. $(i)$ becomes
$ I = \frac{n\varepsilon}{nr + R} $
(i) If $R >>> nr$, then $R + nr \approx R$
$ I = \frac{n\varepsilon}{R} $
$ I = n\left(\frac{\varepsilon}{R}\right) $
Now, $\frac{\varepsilon}{R}$ =Current due to a single cell of negligible internal resistance. Therefore, total current flowing in the circuit $= n$ times the current due to a single cell.
Thus, in order to get a large amount of current from the cells connected in series, the external resistance should be very large as compared to the net internal resistance of the cells.
(ii) If $R << nr$, then $R + nr \approx nr$
Hence, eqn. $(i)$ becomes
$ I = \frac{n\varepsilon}{nr} = \frac{\varepsilon}{r} $
which is equal to the current due to a single cell.
Thus, if large number of cells in series are connected to a very small external resistance, then the current from these cells will be equal to the current due to a single cell. It means, there is no use of such a combination of cells.
(b) Different Cells Connected in Series :
STEP 1. Consider two cells having e.m.fs. $\varepsilon_1$, $\varepsilon_2$ and internal resistances $r_1$, $r_2$ respectively connected in series . If $V_X$, $V_Y$ and $V_Z$ are potentials at X, Y and Z respectively, then potential difference between X and Y is given by
$ V_{XY} = V_X - V_Y $
But $V_{XY} = \varepsilon_1 - I r_1$, where I is the current flowing in the series combination of cells from Y to X.
$ \therefore \qquad V_{XY} = V_X - V_Y = \varepsilon_1 - I r_1 $
Potential difference between Y and Z is given by :
$ V_{YZ} = V_Y - V_Z$
$ V_{YZ}= \varepsilon_2 - I r_2 $
Then
$V_{XZ} = V_{XY} + V_{YZ}
$V_{XZ} = (\varepsilon_1 - I r_1) + (\varepsilon_2 - I r_2)$
= $(\varepsilon_1 + \varepsilon_2) - I (r_1 + r_2) \qquad \dots (i)$
Let the series combination of the cells be replaced by single cell of equivalent e.m.f. $\varepsilon_{\text{eq}}$ and internal resistance $r_{\text{eq}}$, then
$ V_{XZ} = \varepsilon_{\text{eq}} - I r_{\text{eq}} \qquad \dots (ii) $
Comparing eqns. $(i)$ and $(ii)$, we have
$ \varepsilon_{\text{eq}} = \varepsilon_1 + \varepsilon_2 \quad \text{and} \quad r_{\text{eq}} = r_1 + r_2 $
STEP 2. If $n$ cells of e.m.fs. $\varepsilon_1$, $\varepsilon_2$, $\dots$, $\varepsilon_n$ and of internal resistance $r_1$, $r_2$, $\dots$, $r_n$ respectively be connected in series, then
Equivalent e.m.f., of all cells,
$ \varepsilon_{\text{eq}} = \varepsilon_1 + \varepsilon_2 + \dots + \varepsilon_n = \sum_{i=1}^{n} \varepsilon_i $
and internal resistance,
$ r_{\text{eq}} = r_1 + r_2 + \dots + r_n = \sum_{i=1}^{n} r_i $
Cells Connected in Parallel :
In parallel combination of cells, the positive terminal of a cell is connected to positive terminal of other cell and negative terminal of one cell is connected to negative terminal of the other cell.
(a) Identical Cells Connected in Parallel :
Consider $m$ identical cells such as $C_1$, $C_2$, $\dots$, $C_m$, each of e.m.f. $\varepsilon$ and internal resistance $r$ connected in parallel to an external resistance $R$.
Since cells are connected in parallel, so the total e.m.f. of all cells = e.m.f. of a single cell
$ \varepsilon_{\text{eff}} = \varepsilon $
That is,
Equivalent internal resistance of $m$ cells connected in parallel is given by
$ \frac{1}{r_{\text{eff}}} = \frac{1}{r} + \frac{1}{r} + \dots \text{upto } m $
$ \frac{1}{r_{\text{eff}}} = \frac{m}{r} $
or
$ r_{\text{eff}} = \frac{r}{m} $
Now $r_{\text{eff}}$ and $R$ are in series, therefore, the total resistance of the circuit
$ = r_{\text{eff}} + R = \frac{r}{m} + R $
The current flowing through the circuit is given by
$ I = \frac{\text{Total e.m.f.}}{\text{Total resistance}} $
$ I = \frac{\varepsilon}{\frac{r}{m} + R} = \frac{m\varepsilon}{r + mR} \qquad \dots (ii) $
Special Cases :
(i) If $R >> r$, then $\frac{r}{m} + R \approx R$
$ \therefore \text{ eqn. (ii) becomes, } I = \frac{m\varepsilon}{mR} = \frac{\varepsilon}{R} $
$ I $= current due to a single cell of negligible internal resistance
Therefore, such an arrangement of cells is of no significance.
(ii) If $R << r$, then $\frac{r}{m} + R \approx \frac{r}{m}$
$\therefore$ eqn. (ii) becomes
$I = \frac{\varepsilon}{r/m} =m\frac{\varepsilon}{r} $
Thus, $I = m$ times the current due to a single cell.
Therefore, in order to get a large current in the circuit, cells may be connected in parallel to a small external resistance.
(b) Different Cells in Parallel :
Let two cells having e.m.fs $\varepsilon_1$, $\varepsilon_2$ and internal resistances $r_1$, $r_2$ respectively be connected in parallel
If $I_1$ and $I_2$ are the currents supplied by the cells, then the main current is given by
$ I = I_1 + I_2 \qquad \dots (i) $
Terminal voltage of the cells connected in parallel is same. Let it be $V$.
For first cell,
$V = \varepsilon_1 - I_1 r_1 \quad \text{or} \quad I_1 = \frac{\varepsilon_1 - V}{r_1} $
and for second cell,
$ V = \varepsilon_2 - I_2 r_2 \quad \text{or} \quad I_2 = \frac{\varepsilon_2 - V}{r_2} $
Substituting the values of $I_1$ and $I_2$ in eqn. (i), we get
$I = \frac{\varepsilon_1 - V}{r_1} + \frac{\varepsilon_2 - V}{r_2} = \frac{\varepsilon_1}{r_1} - \frac{V}{r_1} + \frac{\varepsilon_2}{r_2} - \frac{V}{r_2} $
$I = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} - V \left( \frac{1}{r_1} + \frac{1}{r_2} \right)$
$I= \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 r_2} - V \frac{r_1 + r_2}{r_1 r_2} $
or
$V \frac{r_1 + r_2}{r_1 r_2} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 r_2} - I$
$V = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}-I ( \frac{r_1 r_2}{r_1 + r_2}) \qquad \dots (ii) $
Let the parallel grouping of cells be replaced with a single of e.m.f. $\varepsilon_{\text{eq}}$ and internal resistance $r_{\text{eq}}$, then
$V = \varepsilon_{\text{eq}} - I r_{\text{eq}} \qquad \dots (iii) $
Comparing eqns. (ii) and (iii), we get
$\varepsilon_{\text{eq}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} $
and
$ r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}$ or
$\frac{1}{r_{\text{eq}}} = \frac{r_1 + r_2}{r_1 r_2}$
$\frac{1}{r_{\text{eq}}}= \frac{1}{r_1} + \frac{1}{r_2}$
Also
$\frac{\varepsilon_{\text{eq}}}{r_{\text{eq}}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 . r_2}$
$\frac{\varepsilon_{\text{eq}}}{r_{\text{eq}}} = \frac{\varepsilon_{1}}{r_1} + \frac{\varepsilon_{2}}{r_2}$
or
$ \varepsilon_{\text{eq}} = \left[ \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} \right] r_{\text{eq}} $
Generalising, let $m$ cell of e.m.fs $\varepsilon_1$, $\varepsilon_2$, $\dots$, $\varepsilon_m$ and internal resistances $r_1$, $r_2$, $\dots$, $r_m$ be in parallel grouping, then
$\frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2} + \dots + \frac{1}{r_m}$
$\varepsilon_{\text{eq}} = \left[ \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} + \dots + \frac{\varepsilon_m}{r_m} \right] r_{\text{eq}}$