Combining Cells: Series & Parallel

Every battery pack you own is an $m \times n$ answer to this module's question. A laptop pack might run three parallel groups of four series cells (4s3p in industry shorthand): four in series to reach a working voltage near 15 V, three in parallel to triple the capacity and cut the internal resistance to a third. Electric-vehicle packs scale the same grammar to thousands of cells — around a hundred series groups to reach 400–800 V, with many cells parallel inside each group for capacity and stiffness. The pack designer's spreadsheet is equations (5) and (6) with thermal limits attached. The identical-cells warning is live engineering: a pack's battery-management system continuously balances its groups, because a group whose emf drifts low starts receiving charge from its neighbours — exactly this article's circulating current. "Never mix old and new cells", printed on every blister pack of AAs, is the consumer edition of the same physics.

Every combination question is answered by following one coulomb and keeping the budget from EMF, Terminal Voltage & Internal Resistance: each cell it passes through pays it $\varepsilon$ and charges it the internal toll.

Series means the coulomb passes through every cell, one after another, like climbing a staircase of pumps. Each pump lifts it by its own emf — the lifts add. But each pump also makes it squeeze through its own interior — the tolls add too. Four 1.5 V cells with $r = 0.2\ \Omega$ each behave as one source of 6.0 V with $r_{\text{eq}} = 0.8\ \Omega$. Voltage is bought at the price of internal resistance.

Parallel means the cells stand side by side between the same two rails, and the coulomb passes through only one of them. It gets one lift — the emf is unchanged. But the traffic now splits: with $m$ identical cells, each carries only $I/m$, so each pays a smaller toll, and the bank as a whole behaves as one cell with $r_{\text{eq}} = r/m$. Parallel buys no voltage; it buys stiffness — less sag — and endurance, each cell draining $m$ times slower.

One warning shapes everything about parallel banks: the sharing argument assumed identical cells. Cells of different emf disagree about the rail voltage and settle it by driving current through each other — even with no load. We set that problem up at the end of this article and hand it, by name, to Kirchhoff's Rules.

Follow one coulomb through a chain of cells connected + to −, all pointing the same way. Each cell lifts it by its own emf, so the lifts accumulate:

where: the sum runs over every cell in the chain, identical or not. Six 2.1 V lead–acid cells make the 12.6 V car battery; for $n$ identical cells, $\varepsilon_{\text{eq}} = n\varepsilon$ — which is why battery voltages come in multiples of the single cell. The tolls have no direction — resistance is resistance whichever way the current crosses it — so the internal resistances add exactly like resistors in series (the rule from Resistors in Series & Parallel, applied inside the battery case):

where: for $n$ identical cells, $r_{\text{eq}} = nr$. The trade-off in one sentence: chaining multiplies push and toll together — a long chain is high-voltage but comparatively soft. Now reverse one cell — the everyday torch accident. Its chemistry pushes against the traffic: its emf enters equation (1) with a minus sign. The total drops by $\varepsilon$ for the missing lift, and by $\varepsilon$ again for the fight it now puts up — a $2\varepsilon$ swing. Its internal resistance, direction-blind, adds as before.

Stand $m$ identical cells between two common rails and connect a load drawing total current $I$. By symmetry — same emf, same $r$, same rails — no cell carries more than its share: each carries $I/m$. What is the bank's terminal voltage? Pick any one cell (they all agree, sharing the same rails) and apply the budget equation from EMF, Terminal Voltage & Internal Resistance to it alone:

where: the second form is the point — the bank behaves exactly like a single cell of emf $\varepsilon$ and internal resistance $r/m$. Nothing about the voltage improved; what improved is the sag per ampere. Read equation (3) against the fingerprint line of the previous module: same intercept, slope $m$ times shallower. So the parallel rules for identical cells are

where: $m$ — the number of cells in the bank. The $r/m$ is just the parallel-resistor rule from Resistors in Series & Parallel applied to $m$ equal internal resistances; the constant $\varepsilon_{\text{eq}}$ is new physics — rails cannot sit at two different voltages, and identical cells all want the same one. When does parallel actually help? Compare the drives for a load $R$: three in series give $I = 3\varepsilon/(R + 3r)$; three in parallel give $I = \varepsilon/(R + r/3)$. With $\varepsilon = 1.5\ \mathrm{V}$, $r = 1.0\ \Omega$:

The pattern generalises and is worth carrying into every exam: series wins when $R$ is large (the load dominates the loop, so stack up emf), parallel wins when $R$ is small (the internal resistance dominates, so attack $r$). The crossover logic — make the internal resistance comparable to the external — is about to become a formula.

Real packs do both at once: $m$ parallel rows, each a chain of $n$ series cells ($N = mn$ identical cells). Each row is a source $(n\varepsilon, nr)$ by (1)–(2); $m$ identical rows in parallel keep the emf and divide the internal resistance by (4):

where: $I$ — the current delivered to an external load $R$. Every symbol is now a design choice: same $N$ cells, different $(m, n)$ splits, different performance. JEE's favourite question — which split maximises the current? — has a beautiful answer you can find numerically (no calculus needed): the current in equation (5) is largest when the arrangement's internal resistance matches the load,

where: the matching echoes the maximum-power teaser from EMF, Terminal Voltage & Internal Resistance — a source delivers hardest into a load comparable to its own internal resistance. Too series-heavy ($nr/m \gg R$) and the pack throttles itself; too parallel-heavy ($nr/m \ll R$) and there isn't enough emf to push with. The competitive example below shows the sweep.

Board level — the torch with a backwards cell (CBSE/ICSE pattern, 3 marks)

A torch uses four cells in series, each of emf $1.5\ \mathrm{V}$ and internal resistance $0.20\ \Omega$, driving a bulb of resistance $5.2\ \Omega$. Find the current through the bulb (a) with all cells correctly placed, (b) with one cell accidentally reversed. (c) State the terminal voltage of the battery in case (a).

Step 1. Part (a). Equivalent source: $\varepsilon_{\text{eq}} = 4 \times 1.5 = 6.0\ \mathrm{V}$, $r_{\text{eq}} = 4 \times 0.20 = 0.80\ \Omega$. The loop equation gives $I = \dfrac{6.0}{5.2 + 0.80} = \dfrac{6.0}{6.0} = \boxed{1.0\ \mathrm{A}}$.

Step 2. Part (b). One reversed cell: $\varepsilon_{\text{eq}} = (4 - 2\times 1)\times 1.5 = 3.0\ \mathrm{V}$ — the double penalty — while $r_{\text{eq}}$ stays $0.80\ \Omega$. So $I = \dfrac{3.0}{6.0} = \boxed{0.50\ \mathrm{A}}$: the bulb current halves.

Step 3. Part (c). Terminal voltage of the whole battery in case (a): $V = \varepsilon_{\text{eq}} - I r_{\text{eq}} = 6.0 - 1.0 \times 0.80 = \boxed{5.2\ \mathrm{V}}$ — equal to $IR = 1.0 \times 5.2$ ✓, as the budget demands.

Step 4. The sentence the scheme wants in (b): "A reversed cell contributes $-\varepsilon$ to the net emf (it opposes the others), so the emf falls by $2\varepsilon$, while the internal resistances still add."

Competitive level — arranging 24 cells for maximum current (JEE-Main pattern, ~3 minutes)

You have 24 identical cells, each of emf $2.0\ \mathrm{V}$ and internal resistance $0.30\ \Omega$, to be arranged as $m$ parallel rows of $n$ series cells ($mn = 24$) driving a load $R = 0.80\ \Omega$. Which arrangement gives the maximum current, and what is that current?

Step 1. Write the delivered current from equation (5): $I = \dfrac{n\varepsilon}{R + nr/m} = \dfrac{2.0\,n}{0.80 + 0.30\,n/m}$.

Step 2. Apply the matching condition (6): maximum current wants $\dfrac{nr}{m} = R$, i.e. $\dfrac{0.30\,n}{m} = 0.80$, so $\dfrac{n}{m} = \dfrac{8}{3}$. With $mn = 24$: $n = 8$, $m = 3$ — three parallel rows of eight cells. Check: $nr/m = 8 \times 0.30 / 3 = 0.80\ \Omega = R$ ✓.

Step 3. The maximum current: $I = \dfrac{8 \times 2.0}{0.80 + 0.80} = \dfrac{16}{1.6} = \boxed{10\ \mathrm{A}}$.

Step 4. Sanity sweep of the rival arrangements (equation (5) each time): all-series $m=1, n=24$: $I = 48/(0.8+7.2) = 6.0\ \mathrm{A}$. $m=2, n=12$: $24/2.6 = 9.2\ \mathrm{A}$. $m=4, n=6$: $12/1.25 = 9.6\ \mathrm{A}$. $m=6, n=4$: $8/1.0 = 8.0\ \mathrm{A}$. All-parallel $m=24, n=1$: $2/0.8125 = 2.5\ \mathrm{A}$. The peak sits exactly at the matched split — and note how flat the top is (9.2, 9.6, 10 for neighbouring splits): matching is forgiving. Timing note: under 3 minutes if you go straight to $nr/m = R$; a full sweep of divisor pairs is the slow route — save it for checking. In an MCQ, combine $nr/m = R$ with $mn = N$ to get $n = \sqrt{NR/r}$ mentally: $\sqrt{24 \times 0.8/0.3} = \sqrt{64} = 8$. That square root is the whole question.

Everything above leaned on symmetry. Break it — put a fresh $\varepsilon_1 = 1.6\ \mathrm{V}$ cell in parallel with a tired $\varepsilon_2 = 1.4\ \mathrm{V}$ one, each with $r = 0.10\ \Omega$ — and the sharing argument dies: the two cells want different rail voltages. What happens instead: with nothing connected, the emf difference drives a current around the internal loop the two cells form,

where: the fresh cell force-charges the tired one, each dissipating $I^2 r = 0.10\ \mathrm{W}$ internally — the pair quietly drains itself. This is why a mixed old-and-new-cell torch mysteriously eats batteries. Now connect a load as well, and ask for the three branch currents. Series-parallel rules have nothing to say: there are two loops, three unknown currents, and the answer needs systematic junction-and-loop bookkeeping. Set it up, label the currents, and you are standing at the exact doorstep of Kirchhoff's Rules — the next module, which turns precisely this circuit into its flagship worked example.

Forgetting that internal resistances add in series. Half the lost marks in this topic. The chain's $r_{\text{eq}} = nr$ belongs in the loop equation: $I = n\varepsilon/(R + nr)$, not $n\varepsilon/(R + r)$.

The reversed cell: $-\varepsilon$ once in the sum, but a $2\varepsilon$ swing. Net emf $(n-2m)\varepsilon$; internal resistance still $nr$. If your answer to a reversed-cell question changed $r$, re-read the chain.

"Parallel doubles the voltage." Never. Parallel identical cells keep $\varepsilon$ exactly; what halves (or divides by $m$) is the internal resistance. If a question reports a doubled voltage from two cells, they are in series.

Averaging unequal parallel emfs. Two cells of 1.6 V and 1.4 V in parallel do not make a 1.5 V source by arithmetic mean — the correct combination is current-weighted and, for exams, out of scope until Kirchhoff's Rules. What you can be asked: the circulating current $\Delta\varepsilon/(r_1 + r_2)$.

Applying $r/m$ to non-identical cells. The division-by-$m$ rule assumed identical cells sharing symmetrically. Different $r$ values combine by the parallel-resistor formula; different $\varepsilon$ values break the framework entirely.

Blind series-maximising in arrangement problems. "More cells in series = more current" fails the moment $nr/m$ outgrows $R$. The matched split $nr/m = R$ — equivalently $n = \sqrt{NR/r}$ — is the intended answer, and options are built to punish the all-series reflex.

Constants and conventions. SI throughout; no $g$ anywhere in this module. CBSE/ISC quote emf and $r$ per cell — read carefully whether "battery" in the stem means one cell or the whole combination. JEE arrangement problems state $N$, $r$, $R$: go straight for the matching condition.

Texts. NCERT Physics Class 12, Ch. 3 (cells in series and parallel) · HC Verma, Concepts of Physics Vol. 2, Ch. 32 · ISC Physics Class 12 syllabus unit on current electricity (combination of cells is an explicit numericals mandate).

Provenance. All results are standard. A verifier should recompute: torch example $I = 6.0/6.0 = 1.0\ \mathrm{A}$, reversed $I = 3.0/6.0 = 0.50\ \mathrm{A}$, $V = 5.2\ \mathrm{V}$; arrangement sweep for $N = 24$, $\varepsilon = 2.0$, $r = 0.3$, $R = 0.8$: currents 6.0, 9.2, 10.0, 9.6, 8.0, 2.5 A for $m = 1, 2, 3, 4, 6, 24$, max at $m = 3, n = 8$ where $nr/m = 0.80 = R$ and $n = \sqrt{NR/r} = \sqrt{64} = 8$; comparison table at $R = 0.5$: 1.00/1.29/1.80 A, at $R = 10$: 0.14/0.35/0.15 A; circulating current $0.20/0.20 = 1.0\ \mathrm{A}$, $0.10\ \mathrm{W}$ per cell.

Prerequisites. EMF, Terminal Voltage & Internal Resistance · Resistors in Series & Parallel. Leads to. Kirchhoff's Rules (which inherits the unequal-parallel setup) · the battery-problems archetype (maximum power and efficiency — coming later in E3).