Cells in Series & Parallel

The equivalent-cell idea

Like resistors, cells can be combined in a circuit, and like resistors a combination of cells can be replaced — for the purpose of calculating currents and voltages — by a single equivalent cell. That equivalent cell carries an EMF $\varepsilon_{eq}$ and an internal resistance $r_{eq}$. The whole topic reduces to one question: given the way the cells are joined, what are $\varepsilon_{eq}$ and $r_{eq}$?

The starting point is the terminal-voltage relation derived in EMF and internal resistance: for a cell delivering current $I$ from its positive to negative terminal externally, the potential difference across its terminals is $V = \varepsilon - Ir$. NCERT applies this single relation, cell by cell, to assemble the equivalent cell for both arrangements. There is no new physics here — only careful bookkeeping of potentials.

The motivation is practical. A single dry cell supplies about $1.5\ \text{V}$; many devices need more, or need a larger current than one cell can deliver before its internal resistance throttles the flow. Combining cells lets a designer trade one quantity for another — raise the driving EMF by stacking cells in series, or lower the effective internal resistance by paralleling them. The equivalent-cell description makes that trade-off quantitative, and once $\varepsilon_{eq}$ and $r_{eq}$ are known the rest of the circuit is solved exactly as if a single cell were present.

Cells in series

Consider two cells in series (Figure 1), where one terminal of each cell is joined and the other two are left free at points A and C. Let $V(A)$, $V(B)$, $V(C)$ be the potentials at the three points. Applying $V = \varepsilon - Ir$ to each cell, NCERT writes the two terminal differences as

$$V_{AB} = V(A) - V(B) = \varepsilon_1 - I r_1, \qquad V_{BC} = V(B) - V(C) = \varepsilon_2 - I r_2.$$

Adding them gives the potential difference across the whole combination:

$$V_{AC} = V(A) - V(C) = (\varepsilon_1 + \varepsilon_2) - I(r_1 + r_2).$$

If this is to look like a single cell, $V_{AC} = \varepsilon_{eq} - I r_{eq}$, comparison gives the headline result of the section:

$$\boxed{\;\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2, \qquad r_{eq} = r_1 + r_2.\;}$$

The rule extends to any number of cells: the equivalent EMF of $n$ cells in series is the sum of the individual EMFs, and the equivalent internal resistance is the sum of the internal resistances. This holds when the current leaves each cell from its positive electrode — that is, when every cell aids the others.

When a cell is reversed

If instead the two negative terminals are joined, the second cell opposes the first. The relation for the second cell becomes $V_{BC} = -\varepsilon_2 - I r_2$, and the equivalent EMF reduces to

$$\varepsilon_{eq} = \varepsilon_1 - \varepsilon_2 \qquad (\varepsilon_1 > \varepsilon_2).$$

In words: if, in the combination, the current leaves a cell from its negative electrode, that cell's EMF enters the expression for $\varepsilon_{eq}$ with a minus sign. The internal resistances, however, still add — orientation never changes $r_{eq}$ for a series chain.

Cells in parallel

Now join the two positive terminals together and the two negative terminals together, so both cells sit between the same pair of nodes $B_1$ and $B_2$ (Figure 2). Let $I_1$ and $I_2$ be the currents leaving the positive electrodes; by the junction rule the total current drawn is

$$I = I_1 + I_2.$$

Because both cells are connected between the same two nodes, the terminal voltage $V = V(B_1) - V(B_2)$ is common. For each cell, $V = \varepsilon - I r$ rearranges to give its branch current, $I_1 = (\varepsilon_1 - V)/r_1$ and $I_2 = (\varepsilon_2 - V)/r_2$. Substituting into $I = I_1 + I_2$ and solving for $V$, NCERT obtains

$$V = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} - I\,\frac{r_1 r_2}{r_1 + r_2}.$$

Matching this to the equivalent-cell form $V = \varepsilon_{eq} - I r_{eq}$ gives the two parallel results:

$$\boxed{\;\varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}, \qquad r_{eq} = \frac{r_1 r_2}{r_1 + r_2}.\;}$$

NCERT also casts these in a cleaner, more memorable form. Writing them in terms of reciprocals,

$$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}, \qquad \frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}.$$

The first is just the parallel-resistor rule. The second says the quantity $\varepsilon/r$ — the short-circuit current each cell can supply — adds across branches. The equivalent EMF is therefore a resistance-weighted average of the individual EMFs, never their sum. If the negative terminal of the second cell is connected to the positive terminal of the first, the formulae still hold with $\varepsilon_2 \rightarrow -\varepsilon_2$.

Extending to n cells in parallel

The reciprocal form extends immediately. For $n$ cells in parallel,

$$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} + \dots + \frac{1}{r_n}, \qquad \frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} + \dots + \frac{\varepsilon_n}{r_n}.$$

Series vs parallel master table

Both arrangements are captured in one comparison. Note how the structure mirrors resistor combinations: series adds, parallel takes the reciprocal sum.

n identical cells driving R

A favourite numerical takes $n$ identical cells, each of EMF $\varepsilon$ and internal resistance $r$, driving an external resistance $R$. The two extreme groupings give very different currents.

All in series. The equivalent cell has EMF $n\varepsilon$ and internal resistance $nr$, so the current is

$$I_{\text{series}} = \frac{n\varepsilon}{R + nr}.$$

All in parallel. The equivalent EMF stays $\varepsilon$ (identical cells), while the internal resistance falls to $r/n$, giving

$$I_{\text{parallel}} = \frac{\varepsilon}{R + r/n} = \frac{n\varepsilon}{nR + r}.$$

Comparing the denominators $R + nr$ and $R + r/n$ shows the trade-off cleanly:

Mixed grouping

In practice cells are often arranged in a mixed array: $m$ identical cells in each series row, with $n$ such rows in parallel. Each row is an equivalent cell of EMF $m\varepsilon$ and internal resistance $mr$; the $n$ rows in parallel keep the EMF at $m\varepsilon$ but divide the internal resistance by $n$, giving $mr/n$. The current through an external $R$ is therefore

$$I = \frac{m\varepsilon}{R + \dfrac{mr}{n}} = \frac{mn\,\varepsilon}{nR + mr}.$$

For a fixed total number of cells $mn$, the current is maximum when the external resistance matches the equivalent internal resistance, $R = mr/n$. This is the same maximum-power-transfer condition that motivates choosing series versus parallel in the first place; the pure-series and pure-parallel cases are simply the limits $n = 1$ and $m = 1$.

The practical reading of this is worth holding on to. A torch that needs several volts uses cells in series, accepting the higher internal resistance because the load is comparatively large. A device that must draw a heavy current from low-voltage cells favours parallel rows, so that the combined internal resistance does not collapse the terminal voltage under load. Real battery packs sit between these extremes, and the mixed-array formula tells the designer exactly how many rows and how many cells per row deliver the most current into a given load. The same arithmetic — equivalent EMF on top, equivalent internal resistance plus the load on the bottom — runs through every one of these cases.

Worked examples