Combination of cells

A torch needs more push than one cell gives, a car battery needs more push than one cell can supply at all — so real batteries are built by wiring several cells together. The two basic wiring plans, series and parallel, do very different jobs: series stacks up the EMF for a high-voltage push, while parallel keeps the EMF the same but shrinks the effective internal resistance, letting more current flow without the terminal voltage sagging. This section builds the effective-EMF and effective-internal-resistance formulas for both plans, for identical cells and for unequal cells, and shows how to fold the result into a single current calculation for an external resistor. It closes the ISC Class 12 Current Electricity sequence that starts with a single cell's EMF and internal resistance.

The core insight

Combining $n$ cells in series multiplies both EMF and internal resistance by $n$ , while combining them in parallel keeps the EMF the same (for identical cells) and divides internal resistance by $n$ , so series suits a large external resistance and parallel suits a small one.

The core idea

Series, identical cells. Chain $n$ identical cells (EMF $\varepsilon$ , internal resistance $r$ each) positive-to-negative, driving current through external resistance $R$ . The EMFs add and the internal resistances add, because current flows through every cell in turn:

\varepsilon_{\text{eff}} = n\varepsilon, \qquad r_{\text{eff}} = nr, \qquad I = \frac{n\varepsilon}{R + nr}

where: $\varepsilon_{\text{eff}}$ is the combination's effective EMF, $r_{\text{eff}}$ its effective internal resistance, and $I$ the current through $R$ . Series pays off when $R \gg r$ : the EMF grows $n$ -fold while the extra internal resistance barely dents the current, since $R$ dominates the denominator.

Parallel, identical cells. Join $n$ identical cells terminal-to-terminal (all positives together, all negatives together) across $R$ . Each cell now only has to supply a fraction of the total current, so the pack behaves like one cell with the same EMF but $n$ times less internal resistance:

\varepsilon_{\text{eff}} = \varepsilon, \qquad r_{\text{eff}} = \frac{r}{n}, \qquad I = \frac{\varepsilon}{R + r/n}

Parallel pays off when $R \ll r$ : shrinking the internal-resistance bottleneck lets more current flow than a single cell could deliver, and it also spreads the load so each cell only supplies $I/n$ , extending battery life.

Unequal cells in series. Two cells of EMF $\varepsilon_1, \varepsilon_2$ and internal resistance $r_1, r_2$ still add their resistances directly, but the EMFs add only if the cells "aid" each other (same polarity sense around the loop); if one is reversed relative to the other, the EMFs subtract:

\varepsilon_{\text{eff}} = \varepsilon_1 + \varepsilon_2 \ \text{(aiding)} \quad\text{or}\quad \varepsilon_1 - \varepsilon_2 \ \text{(opposing, } \varepsilon_1 > \varepsilon_2), \qquad r_{\text{eff}} = r_1 + r_2

Resistance has no direction, so $r_{\text{eff}}$ is always the plain sum regardless of polarity.

Unequal cells in parallel. Here the effective EMF is not a plain average — the branch with lower internal resistance pulls the combined EMF closer to its own value, an effect only Kirchhoff's laws can pin down exactly (derived below):

\varepsilon_{\text{eff}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}, \qquad \frac{1}{r_{\text{eff}}} = \frac{1}{r_1} + \frac{1}{r_2}

Figure

Two circuit sketches showing identical cells wired in series with EMFs adding along the loop and the same cells wired in parallel with EMF unchanged but internal resistance shrinking

Derivation AEffective EMF and internal resistance of two unequal cells in parallel

The goal is to find, purely from Kirchhoff's current and voltage laws, the single effective EMF $\varepsilon_{\text{eff}}$ and effective internal resistance $r_{\text{eff}}$ that make an external circuit behave exactly as if it were connected to two unequal parallel cells with EMFs $\varepsilon_1, \varepsilon_2$ and internal resistances $r_1, r_2$ , without simply guessing that the answer is an unweighted average of the two EMFs.

Step 1. Set up the junction. Let the two cells' positive terminals meet at a common node A and negative terminals at node B, with the external resistor connected between A and B carrying total current $I$ . Let $I_1$ flow out of cell 1 and $I_2$ flow out of cell 2, so by Kirchhoff's current law at node A the total current splits as $I = I_1 + I_2$ .

Step 2. Write the terminal voltage seen by each cell. If $V$ is the potential difference between A and B (the terminal voltage, the same for both cells since they share both nodes), then for cell 1, $V = \varepsilon_1 - I_1 r_1$ , and for cell 2, $V = \varepsilon_2 - I_2 r_2$ . Both statements just say EMF minus the internal voltage drop equals terminal voltage.

Step 3. Solve each equation for its branch current. Rearranging Step 2 gives $I_1 = \dfrac{\varepsilon_1 - V}{r_1}$ and $I_2 = \dfrac{\varepsilon_2 - V}{r_2}$ .

Step 4. Substitute into the junction equation from Step 1. $I = \dfrac{\varepsilon_1 - V}{r_1} + \dfrac{\varepsilon_2 - V}{r_2} = \left(\dfrac{\varepsilon_1}{r_1} + \dfrac{\varepsilon_2}{r_2}\right) - V\left(\dfrac{1}{r_1} + \dfrac{1}{r_2}\right)$ . This single equation already has the shape of a cell's terminal-voltage law, $I = \varepsilon_{\text{eff}}/r_{\text{eff}} - V/r_{\text{eff}}$ , once it is matched term by term.

Step 5. Match coefficients to read off the effective quantities. Comparing $V\left(\frac{1}{r_1}+\frac{1}{r_2}\right)$ with $V/r_{\text{eff}}$ gives $\dfrac{1}{r_{\text{eff}}} = \dfrac{1}{r_1} + \dfrac{1}{r_2}$ , exactly like resistors in parallel. Comparing the constant terms, $\dfrac{\varepsilon_{\text{eff}}}{r_{\text{eff}}} = \dfrac{\varepsilon_1}{r_1} + \dfrac{\varepsilon_2}{r_2}$ , and multiplying through by $r_{\text{eff}}$ gives $\varepsilon_{\text{eff}} = \dfrac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}$ .

Step 6. Read the physics. The effective EMF is a weighted average of $\varepsilon_1$ and $\varepsilon_2$ , weighted by the other cell's resistance, so the cell with the smaller internal resistance pulls the combined EMF closer to its own value — it dominates because it can push more current for the same voltage drop. Setting $r_1 = r_2 = r$ collapses this to the plain average $(\varepsilon_1+\varepsilon_2)/2$ , and setting $\varepsilon_1 = \varepsilon_2 = \varepsilon$ collapses it to $\varepsilon_{\text{eff}} = \varepsilon$ , matching the identical-cells result above.

Worked example

Five identical cells, each of EMF $1.5\ \text{V}$ and internal resistance $0.2\ \Omega$ , are connected in series to an external resistor $R = 5.5\ \Omega$ . Find the current, and the terminal voltage across the external resistor.

Step 1. Effective EMF and internal resistance: $\varepsilon_{\text{eff}} = n\varepsilon = 5 \times 1.5 = 7.5\ \text{V}$ and $r_{\text{eff}} = nr = 5 \times 0.2 = 1.0\ \Omega$ .

Step 2. Current: $I = \dfrac{\varepsilon_{\text{eff}}}{R + r_{\text{eff}}} = \dfrac{7.5}{5.5 + 1.0} = \dfrac{7.5}{6.5} = 1.15\ \text{A}$ .

Step 3. Terminal voltage across $R$ (also the voltage across the whole external circuit): $V = IR = 1.15 \times 5.5 = 6.34\ \text{V}$ , which is less than $\varepsilon_{\text{eff}} = 7.5\ \text{V}$ because $1.15\ \text{A}$ also drops $I r_{\text{eff}} = 1.15\ \text{V}$ across the combined internal resistance. The series stack delivers $I = 1.15\ \text{A}$ and a terminal voltage of $6.34\ \text{V}$ .

Competitive stretch

Two cells of $\varepsilon_1 = 6\ \text{V}$ , $r_1 = 1\ \Omega$ and $\varepsilon_2 = 4\ \text{V}$ , $r_2 = 2\ \Omega$ are connected in parallel across $R = 2\ \Omega$ . Using the derivation above, $\varepsilon_{\text{eff}} = \dfrac{(6)(2)+(4)(1)}{1+2} = \dfrac{16}{3} = 5.33\ \text{V}$ , and $\dfrac{1}{r_{\text{eff}}} = \dfrac{1}{1}+\dfrac{1}{2} = \dfrac{3}{2}$ , so $r_{\text{eff}} = 0.667\ \Omega$ . The current is $I = \dfrac{5.33}{2+0.667} = 2.0\ \text{A}$ . The parallel pair behaves like one cell of EMF $5.33\ \text{V}$ and internal resistance $0.667\ \Omega$ , driving $2.0\ \text{A}$ through $R$ — closer to the $6\ \text{V}$ cell's value than a plain average of $5\ \text{V}$ , exactly as the weighting by the other resistance predicts.

Common traps

Using $\varepsilon_{\text{eff}} = \varepsilon$ for unequal parallel cells. That shortcut only holds for identical cells; unequal cells need the weighted formula from the derivation, or a full Kirchhoff solve.
Adding internal resistances the wrong way. They always add directly in series and add reciprocally (like parallel resistors) in parallel — regardless of whether the EMFs aid or oppose.
Sign errors with opposing series cells. The effective EMF is the difference, not the sum, and the larger EMF sets the current direction.
Assuming series is always "better." For a low-resistance external load ( $R \ll r$ ), parallel reduces the internal-resistance bottleneck and delivers more current than series would.

Ask the tutor: Not sure whether your circuit calls for the series or parallel cell formulas, or whether two cells are aiding or opposing — walk through your specific diagram with the tutor and check your effective-EMF calculation.

Why does connecting cells in parallel not increase the voltage the way series does?

In parallel, all the positive terminals are tied to one node and all the negatives to another, so the potential difference between those two nodes is set by a single cell's EMF minus its own internal drop — adding more identical cells alongside it does not add another EMF in the loop, it only adds another path for current to flow. What changes is the internal resistance: with $n$ identical branches, current has $n$ parallel paths through $r$ , so the combined internal resistance falls to $r/n$ , letting more total current flow for the same EMF.

If two unequal cells are connected in parallel, why is the effective EMF not just the average of the two EMFs?

Because the two cells do not contribute equally — the cell with the smaller internal resistance can push more current for a given voltage drop, so it has more influence over the shared terminal voltage. Working through Kirchhoff's current law at the junction shows the effective EMF is $\varepsilon_1$ and $\varepsilon_2$ weighted by $r_2$ and $r_1$ respectively (the other cell's resistance), which only reduces to the plain average when $r_1 = r_2$ .

How do I know whether to add or subtract EMFs when two cells are in series?

Trace the loop in one direction and note which terminal of each cell you meet first. If both cells are oriented the same way around the loop (each cell's EMF pushing current the same direction), the EMFs add: $\varepsilon_{\text{eff}} = \varepsilon_1 + \varepsilon_2$ . If one cell is connected "backwards" relative to the other, it opposes the first, and the EMFs subtract: $\varepsilon_{\text{eff}} = \varepsilon_1 - \varepsilon_2$ (taking the larger EMF first), with the current flowing in the direction the larger cell dictates. Internal resistances always add regardless of orientation, since resistance has no direction.

Combination of cells

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads